5.1 Areas and Distances
1. (a) Since is decreasing , we can obtain a lower estimate by using right endpoints. We are instructed to use five rectangles, so
= 5.
5 =
5
=1
()∆ ∆ = − = 105− 0 = 2
=
...
5.1 Areas and Distances
1. (a) Since is decreasing , we can obtain a lower estimate by using right endpoints. We are instructed to use five rectangles, so
= 5.
5 =
5
=1
()∆ ∆ = − = 105− 0 = 2
= (1) · 2 + (2) · 2 + (3) · 2 + (4) · 2 + (5) · 2
= 2[(2) + (4) + (6) + (8) + (10)]
≈ 2(32 + 18 + 08 + 02 + 0)
= 2(6) = 12
Since is decreasing , we can obtain an upper estimate by using left endpoints.
5 =
5
=1
(−1)∆
= (0) · 2 + (1) · 2 + (2) · 2 + (3) · 2 + (4) · 2
= 2[(0) + (2) + (4) + (6) + (8)]
≈ 2(5 + 32 + 18 + 08 + 02)
= 2(11) = 22
(b) 10 =
10
=1
()∆ ∆ = 1010 − 0 = 1
= 1[(1) + (2) + · · · + (10)]
= (1) + (2) + · · · + (10)
≈ 4 + 32 + 25 + 18 + 13 + 08 + 05 + 02 + 01 + 0
= 144
10 =
10
=1
(−1)∆
= (0) + (1) + · · · + (9)
= 10 + 1 · (0) − 1 · (10) subtract rightmost lower rectangle add leftmost upper rectangle,
= 144 + 5 − 0
= 194
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2. (a) (i) 6 =
6
=1
(−1)∆ [∆ = 126− 0 = 2]
= 2[(0) + (1) + (2) + (3) + (4) + (5)]
= 2[(0) + (2) + (4) + (6) + (8) + (10)]
≈ 2(9 + 88 + 82 + 73 + 59 + 41)
= 2(433) = 866
(ii) 6 = 6 + 2 · (12) − 2 · (0)
≈ 866 + 2(1) − 2(9) = 706
[Add area of rightmost lower rectangle
and subtract area of leftmost upper rectangle.]
(iii) 6 =
6
=1
()∆
= 2[(1) + (3) + (5) + (7) + (9) + (11)]
≈ 2(89 + 85 + 78 + 66 + 51 + 28)
= 2(397) = 794
(b) Since is decreasing, we obtain an overestimate by using left endpoints; that is, 6.
(c) Since is decreasing, we obtain an underestimate by using right endpoints; that is, 6.
(d) 6 gives the best estimate, since the area of each rectangle appears to be closer to the true area than the overestimates and
underestimates in 6 and 6.
3. (a) 4 =
4
=1
()∆ ∆ = 2 −4 1 = 1 4 = =1 4 ()∆
= [(1) + (2) + (3) + (4)] ∆
= 514 + 614 + 714 + 81 4 14 = 4 5 + 2 3 + 4 7 + 1 2 1 4 ≈ 06345
Since is decreasing on [12], an underestimate is obtained by using the
right endpoint approximation, 4.
(b) 4 =
4
=1
(−1)∆ = =1 4 (−1) ∆
= [(0) + (1) + (2) + (3)] ∆
= 11 + 514 + 614 + 714 1 4 = 1 + 4 5 + 2 3 + 4 7 1 4 ≈ 07595
4 is an overestimate. Alternatively, we could just add the area of the
leftmost upper rectangle and subtract the area of the rightmost lower
rectangle; that is, 4 = 4 + (1) · 1 4 − (2) · 1 4.
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4. (a) 4 =
4
=1
() ∆ ∆ = 24− 0 = 8 = =1 4 () ∆
= [(1) + (2) + (3) + (4)] ∆
= sin 8 + sin 28 + sin 38 + sin 48 8
≈ 11835
Since is increasing on 0 2 , 4 is an overestimate.
(b) 4 =
4
=1
(−1) ∆ = =1 4 (−1) ∆
= [(0) + (1) + (2) + (3)] ∆
= sin 0 + sin 8 + sin 28 + sin 38 8
≈ 07908
Since is increasing on 0 2 , 4 is an underestimate.
5. (a) () = 1 + 2 and ∆ = 2 − (−1)
3
= 1 ⇒
3 = 1 · (0) + 1 · (1) + 1 · (2) = 1 · 1 + 1 · 2 + 1 · 5 = 8.
∆ = 2 − (−1)
6
= 05 ⇒
6 = 05[(−05) + (0) + (05) + (1) + (15) + (2)]
= 05(125 + 1 + 125 + 2 + 325 + 5)
= 05(1375) = 6875
(b) 3 = 1 · (−1) + 1 · (0) + 1 · (1) = 1 · 2 + 1 · 1 + 1 · 2 = 5
6 = 05[(−1) + (−05) + (0) + (05) + (1) + (15)]
= 05(2 + 125 + 1 + 125 + 2 + 325)
= 05(1075) = 5375
(c) 3 = 1 · (−05) + 1 · (05) + 1 · (15)
= 1 · 125 + 1 · 125 + 1 · 325 = 575
6 = 05[(−075) + (−025) + (025)
+ (075) + (125) + (175)]
= 05(15625 + 10625 + 10625 + 15625 + 25625 + 40625)
= 05(11875) = 59375
(d) 6 appears to be the best estimate.
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6. (a)
(b) () = − 2ln and ∆ = 5 − 1
4 = 1 ⇒
(i) 4 = 1 · (2) + 1 · (3) + 1 · (4) + 1 · (5)
= (2 − 2ln2) + (3 − 2ln3) + (4 − 2ln4) + (5 − 2ln5)
≈ 4425
(ii) 4 = 1 · (15) + 1 · (25) + 1 · (35) + 1 · (45)
= (15 − 2ln 15) + (25 − 2ln25)
+ (35 − 2ln35) + (45 − 2ln45)
≈ 3843
(c) (i) 8 = 1 2[(15) + (2) + · · · + (5)]
=
1 2
[(15 − 2ln15) + (2 − 2ln2) + · · · + (5 − 2ln5)]
≈ 4134
(ii) 8 = 1 2[(125) + (175) + · · · + (475)]
=
1 2
[(125 − 2ln125) + (175 − 2ln175) + · · ·
+ (475 − 2ln475)]
≈ 3889
7. () = 2 + sin, 0 ≤ ≤ , ∆ = .
= 2: The maximum values of on both subintervals occur at = 2 , so
upper sum = 2 · 2 + 2 · 2 = 3 · 2 + 3 · 2
= 3 ≈ 942.2
The minimum values of on the subintervals occur at = 0 and
= , so
lower sum = (0) · 2 + () · 2 = 2 · 2 + 2 · 2 = 2 ≈ 628.
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= 4: upper sum = 4 + 2 + 2 + 34 4
= 2 + 1 2√2 + (2 + 1) + (2 + 1) + 2 + 1 2√2 4
= 10 + √2 4 ≈ 896
lower sum = (0) + 4 + 34 + () 4
= (2 + 0) + 2 + 1 2√2 + 2 + 1 2√2 + (2 + 0) 4
= 8 + √2 4 ≈ 739
= 8: upper sum = 8 + 4 + 38 + 2 + 2
+ 58 + 34 + 78 8
≈ 865
lower sum = (0) + 8 + 4 + 38 + 58
+ 34 + 78 + () 8
≈ 786
8. () = 1 + 2, −1 ≤ ≤ 1, ∆ = 2.
= 3: upper sum = (−1) + 1 3 + (1) 2 3
= 2 + 10 9 + 2 2 3
=
92
27 ≈ 341
lower sum = − 1 3 + (0) + 1 3 2 3
= 10 9 + 1 + 10 9 2 3
=
58
27 ≈ 215
= 4: upper sum = (−1) + − 1 2 + 1 2 + (1) 2 4
= 2 + 5 4 + 5 4 + 2 1 2
=
13
4 = 325
lower sum = − 1 2 + (0) + (0) + 1 2 2 4
= 5 4 + 1 + 1 + 5 4 1 2
=
9 4
= 225
9. Here is one possible algorithm (ordered sequence of operations) for calculating the sums:
1 Let SUM = 0, X_MIN = 0, X_MAX = 1, N = 10 (depending on which sum we are calculating),
DELTA_X = (X_MAX - X_MIN)/N, and RIGHT_ENDPOINT = X_MIN + DELTA_X.
2 Repeat steps 2a, 2b in sequence until RIGHT_ENDPOINT X_MAX.
2a Add (RIGHT_ENDPOINT)^4 to SUM.
Add DELTA_X to RIGHT_ENDPOINT.
At the end of this procedure, (DELTA_X)·(SUM) is equal to the answer we are looking for. We find that
10 = 1
10
10
=1 10 4 ≈ 02533, 30 = 30 1 30 =1 30 4 ≈ 02170, 50 = 50 1 50 =1 50 4 ≈ 02101 and
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100 = 1
100
100
=1 100 4 ≈ 02050. It appears that the exact area is 02. The following display shows the program
SUMRIGHT and its output from a TI-83/4 Plus calculator. To generalize the program, we have input (rather than
assign) values for Xmin, Xmax, and N. Also, the function, 4, is assigned to Y1, enabling us to evaluate any right sum
merely by changing Y1 and running the program.
10. We can use the algorithm from Exercise 9 with X_MIN = 0, X_MAX = 2, and cos(RIGHT_ENDPOINT) instead of
(RIGHT_ENDPOINT)^4 in step 2a. We find that 10 = 2
10
10
=1
cos20 ≈ 09194, 30 = 302 30 =1cos60 ≈ 09736,
and 50 = 2
50
50
=1
cos100 ≈ 09842, and 100 = 1002 100 =1cos200 ≈ 09921. It appears that the exact area is 1.
11. In Maple, we have to perform a number of steps before getting a numerical answer. After loading the student package
[command: with(student);] we use the command
left_sum:=leftsum(1/(xˆ2+1),x=0..1,10 [or 30, or 50]); which gives us the expression in summation
notation. To get a numerical approximation to the sum, we use evalf(left_sum);. Mathematica does not have a special
command for these sums, so we must type them in manually. For example, the first left sum is given by
(1/10)*Sum[1/(((i-1)/10)ˆ2+1)],{i,1,10}], and we use the N command on the resulting output to get a
numerical approximation.
In Derive, we use the LEFT_RIEMANN command to get the left sums, but must define the right sums ourselves.
(We can define a new function using LEFT_RIEMANN with ranging from 1 to instead of from 0 to − 1.)
(a) With () = 1
2 + 1, 0 ≤ ≤ 1, the left sums are of the form = 1 =1 −112 + 1. Specifically, 10 ≈ 08100,
30 ≈ 07937, and 50 ≈ 07904. The right sums are of the form = 1
=1
1
2 + 1. Specifically, 10 ≈ 07600,
30 ≈ 07770, and 50 ≈ 07804.
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(b) In Maple, we use the leftbox (with the same arguments as left_sum) and rightbox commands to generate the
graphs.
left endpoints, = 10 left endpoints, = 30 left endpoints, = 50
right endpoints, = 10 right endpoints, = 30 right endpoints, = 50
(c) We know that since = 1(2 + 1) is a decreasing function on (01), all of the left sums are larger than the actual area,
and all of the right sums are smaller than the actual area. Since the left sum with = 50 is about 07904 0791 and the
right sum with = 50 is about 07804 0780, we conclude that 0780 50 exact area 50 0791, so the
exact area is between 0780 and 0791.
12. See the solution to Exercise 11 for the CAS commands for evaluating the sums.
(a) With () = ln, 1 ≤ ≤ 4, the left sums are of the form = 3
=1
ln1 + 3(− 1). In particular, 10 ≈ 23316,
30 ≈ 24752, and 50 ≈ 25034. The right sums are of the form = 3
=1
ln1 + 3. In particular,
10 ≈ 27475, 30 ≈ 26139, and 50 ≈ 25865.
(b) In Maple, we use the leftbox (with the same arguments as left_sum) and rightbox commands to generate the
graphs.
left endpoints, = 10 left endpoints, = 30 left endpoints, = 50
right endpoints, = 10 right endpoints, = 30 right endpoints, = 50
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(c) We know that since = ln is an increasing function on (1 4), all of the left sums are smaller than the actual area, and all
of the right sums are larger than the actual area. Since the left sum with = 50 is about 2503 250 and the right sum
with = 50 is about 2587 259, we conclude that 250 50 exact area 50 259, so the exact area is
between 250 and 259.
13. Since is an increasing function, 6 will give us a lower estimate and 6 will give us an upper estimate.
6 = (0 fts)(05 s) + (62)(05) + (108)(05) + (149)(05) + (181)(05) + (194)(05) = 05(694) = 347 ft
6 = 05(62 + 108 + 149 + 181 + 194 + 202) = 05(896) = 448 ft
14. (a) The velocities are given with units mih, so we must convert the 10-second intervals to hours:
10 seconds = 10 seconds
3600 secondsh =
1
360
h
distance ≈ 6 = (1829 mih) 360 1 h + (1680) 360 1 + (1066) 360 1 + (998) 360 1
+ (1245) 360 1 + (1761) 360 1
=
8579
360
≈ 2383 miles
(b) Distance ≈ 6 = 360 1 (1680 + 1066 + 998 + 1245 + 1761 + 1756) = 850 3606 ≈ 2363 miles
(c) The velocity is neither increasing nor decreasing on the given interval, so the estimates in parts (a) and (b) are neither
upper nor lower estimates.
15. Lower estimate for oil leakage: 5 = (76 + 68 + 62 + 57 + 53)(2) = (316)(2) = 632 L.
Upper estimate for oil leakage: 5 = (87 + 76 + 68 + 62 + 57)(2) = (35)(2) = 70 L.
16. We can find an upper estimate by using the final velocity for each time interval. Thus, the distance traveled after 62 seconds
can be approximated by
=
6
=1
() ∆ = (185 fts)(10 s) + 319 · 5 + 447 · 5 + 742 · 12 + 1325 · 27 + 1445 · 3 = 54,694 ft
17. For a decreasing function, using left endpoints gives us an overestimate and using right endpoints results in an underestimate.
We will use 6 to get an estimate. ∆ = 1, so
6 = 1[(05) + (15) + (25) + (35) + (45) + (55)] ≈ 55 + 40 + 28 + 18 + 10 + 4 = 155 ft
For a very rough check on the above calculation, we can draw a line from (070) to (60) and calculate the area of the
triangle: 1 2(70)(6) = 210. This is clearly an overestimate, so our midpoint estimate of 155 is reasonable.
18. For an increasing function, using left endpoints gives us an underestimate and using right endpoints results in an overestimate.
We will use 6 to get an estimate. ∆ = 306− 0 = 5 s = 3600 5 h = 720 1 h.
6 = 720 1 [(25) + (75) + (125) + (175) + (225) + (275)]
=
1
720(3125 + 66 + 88 + 1035 + 11375 + 11925) = 720 1 (52175) ≈ 0725 km
For a very rough check on the above calculation, we can draw a line from (00) to (30 120) and calculate the area of the
triangle: 1 2(30)(120) = 1800. Divide by 3600 to get 05, which is clearly an underestimate, making our midpoint estimate of
0725 seem reasonable. Of course, answers will vary due to different readings of the graph.
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19. () = −( − 21)( + 1) and ∆ = 126− 0 = 2
6 = 2 · (1) + 2 · (3) + 2 · (5) + 2 · (7) + 2 · (9) + 2 · (11)
= 2 · 40 + 2 · 216 + 2 · 480 + 2 · 784 + 2 · 1080 + 2 · 1320
= 7840 (infected cellsmL) · days
Thus, the total amount of infection needed to develop symptoms of measles is about 7840 infected cells per mL of blood
plasma.
20. (a) Use ∆ = 14 days. The number of people who died of SARS in Singapore between March 1 and May 24, 2003, using left
endpoints is
6 = 14(00079 + 00638 + 01944 + 04435 + 05620 + 04630) = 14(17346) = 242844 ≈ 24 people
Using right endpoints,
6 = 14(00638 + 01944 + 04435 + 05620 + 04630 + 02897) = 14(20164) = 282296 ≈ 28 people
(b) Let be the number of days since March 1, 2003, () be the number of deaths per day on day , and the graph of = ()
be a reasonable continuous function on the interval [0 84]. Then the number of SARS deaths from = to = is
approximately equal to the area under the curve = () from = to = .
21. () = 2
2 + 1, 1 ≤ ≤ 3. ∆ = (3 − 1) = 2 and = 1 + ∆ = 1 + 2.
= lim
→∞
= lim
→∞
=1
()∆ = lim
→∞
=1
2(1 + 2)
(1 + 2)2 + 1 ·
2
.
22. () = 2 + √1 + 2, 4 ≤ ≤ 7. ∆ = (7 − 4) = 3 and = 4 + ∆ = 4 + 3.
= lim
→∞
= lim
→∞
=1
()∆ = lim
→∞
=1 (4 + 3)2 + 1 + 2(4 + 3) · 3 .
23. () = √sin, 0 ≤ ≤ . ∆ = ( − 0) = and = 0 + ∆ = .
= lim
→∞
= lim
→∞
=1
()∆ = lim
→∞
=1
sin() ·
.
24. lim
→∞
=1
3
1 + 3 can be interpreted as the area of the region lying under the graph of = √1 + on the interval [03],
since for = √1 + on [03] with ∆ = 3 − 0
=
3
, = 0 + ∆ = 3
, and ∗ = , the expression for the area is
= lim
→∞
=1
(∗ )∆ = lim
→∞
=1 1 + 3 3 . Note that this answer is not unique. We could use = √ on [14] or,
in general, = √ − on [ + 1 + 4], where is any real number.
25. lim
→∞
=1
4
tan
4 can be interpreted as the area of the region lying under the graph of = tan on the interval 0 4 ,
since for = tan on 0 4 with ∆ = 4 − 0
=
4
, = 0 + ∆ =
4 , and ∗ = , the expression for the area is
= lim
→∞
=1
(∗ )∆ = lim
→∞
=1
tan4 4 . Note that this answer is not unique, since the expression for the area is
the same for the function = tan( − ) on the interval + 4 , where is any integer.
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26. (a) ∆ = 1 − 0
=
1
and = 0 + ∆ =
. = lim
→∞
= lim
→∞
=1
()∆ = lim
→∞
=1 3 · 1 .
(b) lim
→∞
=1
3
3 ·
1
= lim
→∞
1 4
=1
3 = lim
→∞
1 4
(2+ 1)2 = lim →∞ (4+ 1) 2 2 = 14 lim →∞1 + 1 2 = 14
27. (a) Since is an increasing function, is an underestimate of [lower sum] and is an overestimate of [upper sum].
Thus, , , and are related by the inequality .
(b) = (1)∆ + (2)∆ + · · · + ()∆
= (0)∆ + (1)∆ + · · · + (−1)∆
− = ()∆ − (0)∆
= ∆[() − (0)]
=
−
[() − ()]
In the diagram, − is the sum of the areas of the shaded rectangles. By sliding the shaded rectangles to the left so
that they stack on top of the leftmost shaded rectangle, we form a rectangle of height () − () and width −
.
(c) , so − − ; that is, − −
[() − ()].
28. − −
[() − ()] = 3 − 1
[(3) − (1)] = 2
(3 − )
Solving 2
(3 − ) 00001 for gives us 2(3 − ) 00001 ⇒ 2(3 − )
00001
⇒ 3473451. Thus,
a value of that assures us that − 00001 is = 347346. [This is not the least value of .]
29. (a) = () = 5. ∆ = 2 − 0
=
2
and = 0 + ∆ = 2
.
= lim
→∞
= lim
→∞
=1
()∆ = lim
→∞
=125 · 2 = lim →∞ =1 32 55 · 2 = lim →∞ 646 =1 5.
(b)
=1
5 CAS = 2( + 1)222 + 2 − 1
12
(c) lim
→∞
64
6 ·
2( + 1)222 + 2 − 1
12
=
64
12
lim
→∞
2 + 2 + 122 + 2 − 1
2 · 2
=
16
3
lim
→∞1 + 2 + 12 2 + 2 − 12 = 16 3 · 1 · 2 = 32 3
30. From Example 3(a), we have = lim
→∞
2
=1
−2. Using a CAS,
=1
−2 = −22 − 1
2 − 1 and
lim
→∞
2
·
−22 − 1
2 − 1 = −22 − 1 ≈ 08647, whereas the estimate from Example 3(b) using 10 was 08632.
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31. = () = cos. ∆ = − 0
=
and = 0 + ∆ =
.
= lim
→∞
= lim
→∞
=1
()∆ = lim
→∞
=1
cos ·
CAS
= lim
→∞
sin21 + 1
2sin2 −
2
CAS
= sin
If =
2 , then = sin 2 = 1.
32. (a) The diagram shows one of the congruent triangles, 4, with central angle
2. is the center of the circle and is one of the sides of the polygon.
Radius is drawn so as to bisect ∠. It follows that intersects at
right angles and bisects . Thus, 4 is divided into two right triangles with
legs of length 1 2() = sin() and cos(). 4 has area
2 · 1
2[ sin()][ cos()] = 2 sin() cos() = 1 2 2 sin(2),
so = · area(4) = 1 2 2 sin(2).
(b) To use Equation 3.3.2, lim
→0
sin
= 1, we need to have the same expression in the denominator as we have in the argument
of the sine function—in this case, 2.
lim
→∞
= lim
→∞
1 2
2 sin(2) = lim
→∞
1 2
2 sin(2)
2 ·
2
= lim
→∞
sin(2)
2 2. Let = 2 .
Then as → ∞, → 0, so lim
→∞
sin(2)
2 2 = lim →0 sin 2 = (1)2 = 2.
5.2 The Definite Integral
1. () = − 1, −6 ≤ ≤ 4. ∆ = −
=
4 − (−6)
5
= 2.
Since we are using right endpoints, ∗ = .
5 =
5
=1
() ∆
= (∆)[(1) + (2) + (3) + (4) + (5) + (6)]
= 2[(−4) + (−2) + (0) + (2) + (4)]
= 2[−5 + (−3) + (−1) + 1 + 3]
= 2(−5) = −10
The Riemann sum represents the sum of the areas of the two rectangles above the -axis minus the sum of the areas of the
three rectangles below the -axis; that is, the net area of the rectangles with respect to the -axis.
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2. () = cos, 0 ≤ ≤ 34 . ∆ = −
=
34 − 0
6
=
8
.
Since we are using left endpoints, ∗ = −1.
6 =
6
=1
(−1)∆
= (∆)[(0) + (1) + (2) + (3) + (4) + (5)]
=
8
[(0) + 8 + 28 + 38 + 48 + 58]
≈ 1033186
The Riemann sum represents the sum of the areas of the four rectangles above the -axis minus the area of the rectangle below
the -axis; that is, the net area of the rectangles with respect to the -axis. A sixth rectangle is degenerate, with height 0, and
has no area.
3. () = 2 − 4, 0 ≤ ≤ 3. ∆ = −
=
3 − 0
6
=
1 2
.
Since we are using midpoints, ∗ = = 1 2(−1 + ).
6 =
6
=1
() ∆
= (∆)[(1) + (2) + (3) + (4) + (5) + (6)]
=
1 2
1 4 + 3 4 + 5 4 + 7 4 + 9 4 + 11 4
=
1 2
− 63 16 − 55 16 − 39 16 − 15 16 + 17 16 + 57 16 = 1 2− 98 16 = − 49 16
The Riemann sum represents the sum of the areas of the two rectangles above the -axis minus the sum of the areas of the four
rectangles below the -axis; that is, the net area of the rectangles with respect to the -axis.
4. (a) () = 1
, 1 ≤ ≤ 2. ∆ = −
=
2 − 1
4 =
1 4
.
Since we are using right endpoints, ∗ = .
4 =
4
=1
() ∆
= (∆)[(1) + (2) + (3) + (4)]
=
1 4
[ 5 4 + 6 4 + 7 4 + 8 4]
=
1 4
4 5 + 2 3 + 4 7 + 1 2
≈ 0634524
The Riemann sum represents the sum of the areas of the four rectangles.
(b) Since we are using midpoints, ∗ = = 1 2(−1 + ).
4 =
4
=1
() ∆
= (∆)[(1) + (2) + (3) + (4)]
=
1 4
9 8 + 11 8 + 13 8 + 15 8
=
1 4
8 9 + 11 8 + 13 8 + 15 8 ≈ 0691220
The Riemann sum represents the sum of the areas of the four rectangles.
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5. (a) 010 () ≈ 5 = [(2) + (4) + (6) + (8) + (10)]∆
= [−1 + 0 + (−2) + 2 + 4](2) = 3(2) = 6
(b) 010 () ≈ 5 = [(0) + (2) + (4) + (6) + (8)] ∆
= [3 + (−1) + 0 + (−2) + 2](2) = 2(2) = 4
(c) 010 () ≈ 5 = [(1) + (3) + (5) + (7) + (9)]∆
= [0 + (−1) + (−1) + 0 + 3](2) = 1(2) = 2
6. (a) −42 () ≈ 6 = [(−1) + (0) + (1) + (2) + (3) + (4)]∆
= − 3 2 + 0 + 3 2 + 1 2 + (−1) + 1 2(1) = 0
(b) −42 () ≈ 6 = [(−2) + (−1) + (0) + (1) + (2) + (3)]∆
= 0 + − 3 2 + 0 + 3 2 + 1 2 + (−1)(1) = − 1 2
(c) −42 () ≈ 6 = − 3 2 + − 1 2 + 1 2 + 3 2 + 5 2 + 7 2 ∆
= −1 + (−1) + 1 + 1 + 0 + − 1 2(1) = − 1 2
7. Since is increasing, 5 ≤ 10 30 () ≤ 5.
Lower estimate = 5 =
5
=1
(−1)∆ = 4[(10) + (14) + (18) + (22) + (26)]
= 4[−12 + (−6) + (−2) + 1 + 3] = 4(−16) = −64
Upper estimate = 5 =
5
=1
()∆ = 4[(14) + (18) + (22) + (26) + (30)]
= 4[−6 + (−2) + 1 + 3 + 8] = 4(4) = 16
8. (a) Using the right endpoints to approximate 39 (), we have
3
=1
()∆ = 2[(5) + (7) + (9)] = 2 (−06 + 09 + 18) = 42.
Since is increasing, using right endpoints gives an overestimate.
(b) Using the left endpoints to approximate 39 (), we have
3
=1
(−1)∆ = 2[(3) + (5) + (7)] = 2(−34 − 06 + 09) = −62.
Since is increasing, using left endpoints gives an underestimate.
(c) Using the midpoint of each interval to approximate 39 (), we have
3
=1
()∆ = 2[(4) + (6) + (8)] = 2(−21 + 03 + 14) = −08.
We cannot say anything about the midpoint estimate compared to the exact value of the integral.
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9. ∆ = (8 − 0)4 = 2, so the endpoints are 0, 2, 4, 6, and 8, and the midpoints are 1, 3, 5, and 7. The Midpoint Rule gives
08 sin√ ≈ 4
=1
(¯ )∆ = 2sin√1 + sin√3 + sin√5 + sin√7 ≈ 2(30910) = 61820.
10. ∆ = (1 − 0)5 = 1 5 , so the endpoints are 0, 1 5 , 2 5 , 3 5 , 4 5 , and 1, and the midpoints are 10 1 , 10 3 , 10 5 , 10 7 , and 10 9 . The Midpoint
Rule gives
01 3 + 1 ≈ =1 5 (¯ )∆ = 1 5 10 1 3 + 1 + 10 3 3 + 1 + 10 5 3 + 1 + 10 7 3 + 1 + 10 9 3 + 1
≈ 11097
11. ∆ = (2 − 0)5 = 2 5 , so the endpoints are 0, 2 5 , 4 5 , 6 5 , 8 5 , and 2, and the midpoints are 1 5 , 3 5 , 5 5 , 7 5 and 9 5 . The Midpoint Rule
gives
02 + 1 ≈ =1 5 (¯ )∆ = 25 1 5 + 1 1 5 + 3 5 + 1 3 5 + 5 5 + 1 5 5 + 7 5 + 1 7 5 + 9 5 + 1 9 5 = 25127 56 = 127 140 ≈ 09071.
12. ∆ = ( − 0)4 = 4 , so the endpoints are 4 , 24 , 34 , and 44 , and the midpoints are 8 , 38 , 58 , and 78 . The Midpoint Rule
gives
0 sin2 ≈ =1 5 (¯ )∆ = 4 8 sin2 8 + 38 sin2 38 + 58 sin2 58 + 78 sin2 78 ≈ 24674
13. In Maple 14, use the commands with(Student[Calculus1]) and
ReimannSum(x/(x+1),0..2,partition=5,method=midpoint,output=plot). In some older versions of
Maple, use with(student) to load the sum and box commands, then m:=middlesum(x/(x+1),x=0..2), which
gives us the sum in summation notation, then M:=evalf(m) to get the numerical approximation, and finally
middlebox(x/(x+1),x=0..2) to generate the graph. The values obtained for = 5, 10, and 20 are 09071, 09029, and
09018, respectively.
14. For () = ( + 1) on [02], we calculate 100 ≈ 089469 and 100 ≈ 090802. Since is increasing on [02], 100 is
an underestimate of 02 + 1 and 100 is an overestimate. Thus, 08946 02 + 1 09081
15. We’ll create the table of values to approximate 0 sin by using the
program in the solution to Exercise 5.1.9 with Y1 = sin, Xmin = 0,
Xmax = , and = 5, 10, 50, and 100.
The values of appear to be approaching 2.
5 1933766
10 1983524
50 1999342
100 1999836
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16. 02 −2 with = 5, 10, 50, and 100.
5 1077467 0684794
10 0980007 0783670
50 0901705 0862438
100 0891896 0872262
The value of the integral lies between 0872 and 0892. Note that
() = −2 is decreasing on (0 2). We cannot make a similar statement
for −21 −2 since is increasing on (−10).
17. On [0 1], lim
→∞
=1
1 +
∆ = 01 1 + .
18. On [2 5], lim
→∞
=1
1 + 3 ∆ = 25 √1 + 3 .
19. On [2 7], lim
→∞
=1
[5(∗ )3 − 4∗ ]∆ = 27(53 − 4).
20. On [1 3], lim
→∞
=1
∗
(∗ )2 + 4 ∆ = 13 2+ 4 .
21. Note that ∆ = 5 − 2
=
3
and = 2 + ∆ = 2 + 3
.
25(4 − 2) = lim →∞ =1 ()∆ = lim →∞ =1 2 + 3 3 = lim →∞ 3 =1 4 − 22 + 3
= lim
→∞
3
=1 −6 = lim →∞ 3 −6 =1 = lim →∞ −182 (2+ 1)
= lim
→∞ −18 2 + 1 = −9 lim →∞ 1 + 1 = −9(1) = −9
22. Note that ∆ = 4 − 1
=
3
and = 1 + ∆ = 1 + 3
.
14(2 − 4 + 2) = lim →∞ =1 ()∆ = lim →∞ =1 1 + 3 3 = lim →∞ 3 =1 1 + 3 2 − 41 + 3 + 2
= lim
→∞
3
=1 1 + 6 + 922 − 4 − 12 + 2 = lim →∞ 3 =1 922 − 6 − 1
= lim
→∞
3
92 =1 2 − 6 =1 − =1 1
= lim
→∞ 273 ( + 1)(2 6 + 1) − 182 (2+ 1) − 3 · (1)
= lim
→∞ 9 2 ( + 1)(2 2 + 1) − 9 + 1 − 3 = lim →∞ 9 2 + 1 2+ 1 − 91 + 1 − 3
= lim
→∞ 9 21 + 1 2 + 1 − 91 + 1 − 3 = 9 2(1)(2) − 9(1) − 3 = −3
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23. Note that ∆ = 0 − (−2)
=
2
and = −2 + ∆ = −2 + 2
.
−02(2 + ) = lim →∞ =1 ()∆ = lim →∞ =1 −2 + 2 2 = lim →∞ 2 =1 −2 + 22 + −2 + 2
= lim
→∞
2
=1 4 − 8 + 422 − 2 + 2 = lim →∞ 2 =1 422 − 6 + 2
= lim
→∞
2
42 =1 2 − 6 =1 + =1 2 = lim →∞ 83 ( + 1)(2 6 + 1) − 122 (2+ 1) + 2 · (2)
= lim
→∞ 4 3 ( + 1)(2 2 + 1) − 6 + 1 + 4 = lim →∞ 4 3 + 1 2+ 1 − 61 + 1 + 4
= lim
→∞ 4 31 + 1 2 + 1 − 61 + 1 + 4 = 43(1)(2) − 6(1) + 4 = 23
24. Note that ∆ = 2 − 0
=
2
and = 0 + ∆ = 2
.
02(2 − 3) = lim →∞ =1 ()∆ = lim →∞ =1 2 2 = lim →∞ 2 =1 22 − 23
= lim
→∞
2
=1 4 − 833 = lim →∞ 2 4 =1 − 83 =1 3
= lim
→∞ 82 (2+ 1) − 164 (2+ 1)2 = lim →∞ 4 + 1 − 4 ( + 1) 2 2
= lim
→∞ 41 + 1 − 4 + 1 + 1 = lim →∞ 41 + 1 − 41 + 1 1 + 1
= 4(1) − 4(1)(1) = 0
25. Note that ∆ = 1 − 0
=
1
and = 0 + ∆ =
.
01(3 − 32) = lim →∞ =1 ()∆ = lim →∞ =1 ∆ = lim →∞ =1 3 − 3 2 1
= lim
→∞
1
=1 33 − 322 = lim →∞ 1 13 =1 3 − 32 =1 2
= lim
→∞ 14 (2+ 1)2 − 33 ( + 1)(2 6 + 1) = lim →∞ 1 4 + 1 + 1 − 1 2 + 1 2+ 1
= lim
→∞ 1 41 + 1 1 + 1 − 1 21 + 1 2 + 1 = 14(1)(1) − 1 2(1)(2) = −34
26. (a) ∆ = (4 − 0)8 = 05 and ∗ = = 05.
04(2 − 3) ≈ 8
=1
(∗ )∆
= 05052 − 3(05) + 102 − 3(10) + · · ·
+ 352 − 3(35) + 402 − 3(40)
=
1 2
− 5 4 − 2 − 9 4 − 2 − 5 4 + 0 + 7 4 + 4 = −15
(b)
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(c) 04(2 − 3) = lim →∞ =1 42 − 344
= lim
→∞
4
162 =1 2 − 12 =1
= lim
→∞643 · ( + 1)(2 6 + 1) − 482 · (2+ 1)
= lim
→∞32 3 1 + 1 2 + 1 − 241 + 1
=
32
3 · 2 − 24 = − 8 3
(d) 04(2 − 3) = 1 − 2, where 1
is the area marked + and 2 is the area
marked −.
27. = lim →∞ − =1 + − = lim →∞(− ) =1 1 + ( −2)2 =1
= lim
→∞(− ) + ( −2)2 · (2+ 1) = ( − ) + lim →∞ ( −2)2 1 + 1
= ( − ) + 1 2( − )2 = ( − ) + 1 2 − 1 2 = ( − ) 1 2( + ) = 1 22 − 2
28. 2 = lim →∞ − =1 + − 2 = lim →∞ − =1 2 + 2 − + ( −2)2 2
= lim
→∞( −3)3 =1 2 + 2(−2 )2 =1 + 2 (− ) =1 1
= lim
→∞( −3)3 ( + 1) (2 6 + 1) + 2(−2 )2 (2+ 1) + 2 (− )
= lim
→∞( −6)3 · 1 · 1 + 1 2 + 1 + ( − )2 · 1 · 1 + 1 + 2 ( − )
=
( − )3
3
+ ( − )2 + 2 ( − ) = 3 − 32 + 32 − 3
3
+ 2 − 22 + 3 + 2 − 3
=
3
3
−
3
3
− 2 + 2 + 2 − 2 = 3 − 3
3
29. () = √4 + 2, = 1, = 3, and ∆ = 3 − 1
=
2
. Using Theorem 4, we get ∗ = = 1 + ∆ = 1 + 2
, so
13 4 + 2 = lim →∞ = lim →∞ =1 4 + 1 + 22 · 2 .
30. () = 2 + 1
, = 2, = 5, and ∆ = 5 − 2
=
3
. Using Theorem 4, we get ∗ = = 2 + ∆ = 2 + 3
, so
25 2 + 1 = lim →∞ = lim →∞ =1 2 + 32 + 2 +1 3 · 3 .
31. ∆ = ( − 0) = and ∗ = = .
0 sin 5 = lim →∞ =1 (sin 5) = lim →∞ =1 sin 5 CAS = lim →∞ 1 cot5 2 CAS = 52 = 25
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32. ∆ = (10 − 2) = 8 and ∗ = = 2 + 8.
210 6 = lim →∞ =1 2 + 868 = 8 lim →∞ 1 =1 2 + 86
CAS
= 8 lim
→∞
1
·
6458,5936 + 164,0525 + 131,2084 − 27,7762 + 2048
215
CAS
= 81,2497,984 = 9,9997,872 ≈ 1,428,5531
33. (a) Think of 02 () as the area of a trapezoid with bases 1 and 3 and height 2. The area of a trapezoid is = 1 2( + ),
so 02 () = 1 2(1 + 3)2 = 4.
(b) 05 () = 02 ()
trapezoid
+ 23 ()
rectangle
+ 35 ()
triangle
=
1 2
(1 + 3)2 + 3 · 1 + 1 2 · 2 · 3 = 4 + 3 + 3 = 10
(c) 57 () is the negative of the area of the triangle with base 2 and height 3. 57 () = − 1 2 · 2 · 3 = −3.
(d) 79 () is the negative of the area of a trapezoid with bases 3 and 2 and height 2, so it equals
−
1 2
( + ) = − 1 2(3 + 2)2 = −5. Thus,
09 () = 05 () + 57 () + 79 () = 10 + (−3) + (−5) = 2.
34. (a) 02 () = 1 2 · 4 · 2 = 4 [area of a triangle]
(b) 26 () = − 1 2(2)2 = −2 [negative of the area of a semicircle]
(c) 67 () = 1 2 · 1 · 1 = 1 2 [area of a triangle]
07 () = 02 () + 26 () + 67 () = 4 − 2 + 1 2 = 45 − 2
35. −21(1 − ) can be interpreted as the difference of the areas of the two
shaded triangles; that is, 1 2(2)(2) − 1 2(1)(1) = 2 − 1 2 = 3 2 .
36. 09 1 3 − 2 can be interpreted as the difference of the areas of the two
shaded triangles; that is, − 1 2(6)(2) + 1 2(3)(1) = −6 + 3 2 = − 9 2 .
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37. −031 + √9 − 2 can be interpreted as the area under the graph of
() = 1 + √9 − 2 between = −3 and = 0. This is equal to one-quarter
the area of the circle with radius 3, plus the area of the rectangle, so
−031 + √9 − 2 = 1 4 · 32 + 1 · 3 = 3 + 9 4.
38. −55 − √25 − 2 = −55 − −55 √25 − 2 . By
symmetry, the value of the first integral is 0 since the shaded
area above the -axis equals the shaded area below the -axis.
The second integral can be interpreted as one half the area of a
circle with radius 5; that is, 1 2(5)2 = 25 2 . Thus, the value of
the original integral is 0 − 25 2 = − 25 2 .
39. −34 1 2 can be interpreted as the sum of the areas of the two shaded
triangles; that is, 1 2(4)(2) + 1 2(3) 3 2 = 4 + 9 4 = 25 4 .
40. 01 |2 − 1| can be interpreted as the sum of the areas of the two shaded
triangles; that is, 2 1 2 1 2 (1) = 1 2 .
41. 11 √1 + 4 = 0 since the limits of integration are equal.
42. 0 sin4 = − 0 sin4 [because we reversed the limits of integration]
= − 0 sin4 [we can use any letter without changing the value of the integral]
= −
3 8
[given value]
43. 01(5 − 62) = 01 5 − 601 2 = 5(1 − 0) − 6 1 3 = 5 − 2 = 3
44. 13 (2 − 1) = 213 − 13 1 = 2(3 − ) − 1(3 − 1) = 23 − 2 − 2
45. 13 + 2 = 13 · 2 = 2 13 = 2(3 − ) = 5 − 3
46. 02(2 cos − 5) = 02 2cos − 02 5 = 202 cos − 502
= 2(1) − 5(2)2 − 02
2
= 2 − 52
8
47. −22 () + 25 () − −−21 () = −52 () + −−12 () [by Property 5 and reversing limits]
= −51 () [Property 5]
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48. 24 () + 48 () = 28 (), so 48 () = 28 () − 24 () = 73 − 59 = 14.
49. 09[2() + 3()] = 209 () + 309 () = 2(37) + 3(16) = 122
50. If () = 3 for for ≥ 3 3 , then 05 () can be interpreted as the area of the shaded
region, which consists of a 5-by-3 rectangle surmounted by an isosceles right triangle
whose legs have length 2. Thus, 05 () = 5(3) + 1 2(2)(2) = 17.
51. 03 () is clearly less than −1 and has the smallest value. The slope of the tangent line of at = 1, 0(1), has a value
between −1 and 0, so it has the next smallest value. The largest value is 38 (), followed by 48 (), which has a
value about 1 unit less than 38 (). Still positive, but with a smaller value than 48 (), is 08 (). Ordering these
quantities from smallest to largest gives us
03 () 0(1) 08 () 48 () 38 () or B E A D C
52. (0) = 20 () = − 02 (), so (0) is negative, and similarly, so is (1). (3) and (4) are negative since they
represent negatives of areas below the -axis. Since (2) = 22 () = 0 is the only non-negative value, choice C is the
largest.
53. = −24[() + 2 + 5] = −24 () + 2−24 + −24 5 = 1 + 22 + 3
1 = −3 [area below -axis] + 3 − 3 = −3
2 = − 1
2(4)(4) [area of triangle, see figure] + 1 2(2)(2)
= −8 + 2 = −6
3 = 5[2 − (−4)] = 5(6) = 30
Thus, = −3 + 2(−6) + 30 = 15.
54. Using Integral Comparison Property 8, ≤ () ≤ ⇒ (2 − 0) ≤ 02 () ≤ (2 − 0) ⇒
2 ≤ 02 () ≤ 2.
55. 2 − 4 + 4 = ( − 2)2 ≥ 0 on [04], so 04(2 − 4 + 4) ≥ 0 [Property 6].
56. 2 ≤ on [01] so √1 + 2 ≤ √1 + on [01]. Hence, 01 √1 + 2 ≤ 01 √1 + [Property 7].
57. If −1 ≤ ≤ 1, then 0 ≤ 2 ≤ 1 and 1 ≤ 1 + 2 ≤ 2, so 1 ≤ √1 + 2 ≤ √2 and
1[1 − (−1)] ≤ −11 √1 + 2 ≤ √2[1 − (−1)] [Property 8]; that is, 2 ≤ −11 √1 + 2 ≤ 2√2.
58. If
6
≤ ≤
3
, then 1
2
≤ sin ≤
√3
2 sin is increasing on 6 3 , so
1 2
3 − 6 ≤ 63 sin ≤ √23 3 − 6 [Property 8]; that is, 12 ≤ 63 sin ≤ √12 3 .
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59. If 0 ≤ ≤ 1, then 0 ≤ 3 ≤ 1, so 0(1 − 0) ≤ 01 3 ≤ 1(1 − 0) [Property 8]; that is, 0 ≤ 01 3 ≤ 1.
60. If 0 ≤ ≤ 3, then 4 ≤ + 4 ≤ 7 and 1
7 ≤ + 4 1 ≤ 14, so 17(3 − 0) ≤ 03 + 4 1 ≤ 1 4(3 − 0) [Property 8]; that is,
3 7
≤ 03 + 4 1 ≤ 34.
61. If
4 ≤ ≤ 3 , then 1 ≤ tan ≤ √3, so 1 3 − 4 ≤ 43 tan ≤ √3 3 − 4 or 12 ≤ 43 tan ≤ 12 √3.
62. Let () = 3 − 3 + 3 for 0 ≤ ≤ 2. Then 0() = 32 − 3 = 3( + 1)( − 1), so is decreasing on (01) and
increasing on (12). has the absolute minimum value (1) = 1. Since (0) = 3 and (2) = 5, the absolute maximum
value of is (2) = 5. Thus, 1 ≤ 3 − 3 + 3 ≤ 5 for in [0 2]. It follows from Property 8 that
1 · (2 − 0) ≤ 02 3 − 3 + 3 ≤ 5 · (2 − 0); that is, 2 ≤ 02 3 − 3 + 3 ≤ 10.
63. The only critical number of () = − on [02] is = 1. Since (0) = 0, (1) = −1 ≈ 0368, and
(2) = 2−2 ≈ 0271, we know that the absolute minimum value of on [02] is 0, and the absolute maximum is −1. By
Property 8 0 ≤ − ≤ −1 for 0 ≤ ≤ 2 ⇒ 0(2 − 0) ≤ 02 − ≤ −1(2 − 0) ⇒ 0 ≤ 02 − ≤ 2.
64. Let () = − 2sin for ≤ ≤ 2. Then 0() = 1 − 2cos and 0() = 0 ⇒ cos = 1 2 ⇒ = 53 .
has the absolute maximum value 53 = 53 − 2sin 53 = 53 + √3 ≈ 697 since () = and (2) = 2 are both
smaller than 697. Thus, ≤ () ≤ 53 + √3 ⇒ (2 − ) ≤ 2 () ≤ 53 + √3(2 − ); that is,
2 ≤ 2( − 2sin) ≤ 5 32 + √3.
65. √4 + 1 ≥ √4 = 2, so 13 √4 + 1 ≥ 13 2 = 1 3 33 − 13 = 26 3 .
66. 0 ≤ sin ≤ 1 for 0 ≤ ≤ 2 , so sin ≤ ⇒ 02 sin ≤ 02 = 1 2 2 2 − 02 = 82 .
67. sin √ for 1 ≤ ≤ 2 and arctan is an increasing function, so arctan(sin) arctan√ arctan, and hence,
12 arctan(sin) 12 arctan√ 12 arctan . Thus, 12 arctan has the largest value.
68. 2 √ for 0 ≤ 05 and cosine is a decreasing function on [005], so cos(2) cos√, and hence,
005 cos(2) 005 cos√ . Thus, 005 cos(2) is larger.
69. Using right endpoints as in the proof of Property 2, we calculate
() = lim
→∞
=1
()∆ = lim
→∞
=1
()∆ = lim
→∞
=1
()∆ = ().
70. (a) Since − |()| ≤ () ≤ |()|, it follows from Property 7 that
− |()| ≤ () ≤ |()| ⇒ () ≤ |()|
Note that the definite integral is a real number, and so the following property applies: − ≤ ≤ ⇒ || ≤ for all
real numbers and nonnegative numbers .
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(b)
02 ()sin2 ≤ 02 |()sin2| [by part (a)] = 02 |()| |sin 2| ≤ 02 |()| by Property 7
since |sin 2| ≤ 1 ⇒ |()| |sin 2| ≤ |()|.
71. Suppose that is integrable on [01] that is, lim
→∞
=1
(∗ )∆ exists for any choice of ∗ in [−1 ]. Let n denote a
positive integer and divide the interval [01] into n equal subintervals 0 1 , 1 2 , , − 1 1. If we choose ∗ to be
a rational number in the ith subinterval, then we obtain the Riemann sum
=1
(∗ ) · 1
= 0, so
lim
→∞
=1
(∗ ) · 1
= lim
→∞
0 = 0. Now suppose we choose ∗ to be an irrational number. Then we get
=1
(∗ ) · 1
=
=1
1 · 1
= ·
1
= 1 for each , so lim
→∞
=1
(∗ ) · 1
= lim
→∞
1 = 1. Since the value of
lim
→∞
=1
(∗ )∆ depends on the choice of the sample points ∗ , the limit does not exist, and is not integrable on [0 1].
72. Partition the interval [01] into n equal subintervals and choose ∗ 1 = 1
2 . Then with () = 1,
=1
(∗ )∆ ≥ (∗ 1)∆ = 1
12 ·
1
= Thus,
=1
(∗ )∆ can be made arbitrarily large and hence, is not integrable
on [0 1].
73. lim
→∞
=1
4
5 = lim →∞
=1
4
4 ·
1
= lim
→∞
=1 4 1 . At this point, we need to recognize the limit as being of the form
lim
→∞
=1
()∆, where ∆ = (1 − 0) = 1, = 0 + ∆ = , and () = 4. Thus, the definite integral
is 01 4 .
74. lim
→∞
1
=1
1
1 + ()2 = lim →∞
=1
1
1 + ()2 ·
1
= lim
→∞
=1
()∆, where ∆ = (1 − 0) = 1,
= 0 + ∆ = , and () = 1
1 + 2 . Thus, the definite integral is 01 1 +2 .
75. Choose = 1 +
and ∗ = √−1 = 1 + − 11 + . Then
12 −2 = lim
→∞
1
=1
1
1 + − 11 + = lim →∞
=1
1
( + − 1)( + )
= lim
→∞
=1 +1 − 1 − +1 [by the hint] = lim →∞=0 −1 +1 − =1 +1
= lim
→∞
1 + + 1 1 + · · · + 21− 1 − + 1 1 + · · · + 21− 1 + 21
= lim
→∞
1 − 21 = lim →∞1 − 1 2 = 1 2
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© Cengage Learning. All Rights Reserved.DISCOVERY PROJECT AREA FUNCTIONS ¤ 23
DISCOVERY PROJECT Area Functions
1. (a)
Area of trapezoid = 1 2(1 + 2) = 1 2(3 + 7)2
= 10 square units
Or:
Area of rectangle + area of triangle
= + 1 2 = (2)(3) + 1 2(2)(4) = 10 square units
(b)
As in part (a),
() = 1 2[3 + (2 + 1)]( − 1) = 1 2(2 + 4)( − 1)
= ( + 2)( − 1) = 2 + − 2 square units
(c) 0() = 2 + 1. This is the -coordinate of the point (2 + 1) on the given line.
2. (a) (b) () = −1 1 + 2 = −1 1 + −1 2 [Property 2]
= 1[ − (−1)] + 3 − (−1)3
3 Exercise 5.2.28 Property 1 and
= + 1 + 1
3 3 + 1 3
=
1 3
3 + + 4
3
(c) 0() = 2 + 1. This is the -coordinate of the point 1 + 2
on the given curve.
(d)
( + ) − () is the area
under the curve = 1 + 2
from = to = + .
(e)
An approximating rectangle is shown in the figure.
It has height 1 + 2, width , and area (1 + 2), so
( + ) − () ≈ (1 + 2) ⇒ ( + ) − ()
≈ 1 + 2.
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© Cengage Learning. All Rights Reserved.24 ¤ CHAPTER 5 INTEGRALS
(f ) Part (e) says that the average rate of change of is approximately 1 + 2. As approaches 0, the quotient approaches the
instantaneous rate of change—namely, 0(). So the result of part (c), 0() = 2 + 1, is geometrically plausible.
3. (a) () = cos(2) (b) () starts to decrease at that value of where cos2 changes from
positive to negative; that is, at about = 125.
(c) () = 0 cos(2). Using an integration command, we find that
(0) = 0, (02) ≈ 0200, (04) ≈ 0399, (06) ≈ 0592,
(08) ≈ 0768, (10) ≈ 0905, (12) ≈ 0974, (14) ≈ 0950,
(16) ≈ 0826, (18) ≈ 0635, and (20) ≈ 0461.
(d) We sketch the graph of 0 using the method of Example 1 in Section 2.8.
The graphs of 0() and () look alike, so we guess that 0() = ().
4. In Problems 1 and 2, we showed that if () = (), then 0() = (), for the functions () = 2 + 1 and
() = 1 + 2. In Problem 3 we guessed that the same is true for () = cos(2), based on visual evidence. So we conjecture
that 0() = () for any continuous function . This turns out to be true and is proved in Section 5.3 (the Fundamental
Theorem of Calculus).
5.3 The Fundamental Theorem of Calculus
1. One process undoes what the other one does. The precise version of this statement is given by the Fundamental Theorem of
Calculus. See the statement of this theorem and the paragraph that follows it on page 398.
2. (a) () = 0 (), so (0) = 00 () = 0.
(1) = 01 () = 1 2 · 1 · 1 [area of triangle] = 1 2.
(2) = 02 () = 01 () + 12 () [below the t-axis]
=
1 2
−
1 2
· 1 · 1 = 0.
(3) = (2) + 23 () = 0 − 1 2 · 1 · 1 = − 1 2.
(4) = (3) + 34 () = − 1 2 + 1 2 · 1 · 1 = 0.
(5) = (4) + 45 () = 0 + 15 = 15.
(6) = (5) + 56 () = 15 + 25 = 4.
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© Cengage Learning. All Rights Reserved.SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 25
(b) (7) = (6) + 67 () ≈ 4 + 22 [estimate from the graph] = 62. (d)
(c) The answers from part (a) and part (b) indicate that has a minimum at
= 3 and a maximum at = 7. This makes sense from the graph of
since we are subtracting area on 1 3 and adding area on 3 7.
3. (a) () = 0 ().
(0) = 00 () = 0
(1) = 01 () = 1 · 2 = 2 [rectangle],
(2) = 02 () = 01 () + 12 () = (1) + 12 ()
= 2 + 1 · 2 + 1
2 · 1 · 2 = 5 [rectangle plus triangle],
(3) = 03 () = (2) + 23 () = 5 + 1 2 · 1 · 4 = 7,
(6) = (3) + 36 () [the integral is negative since lies under the -axis]
= 7 + − 1 2 · 2 · 2 + 1 · 2 = 7 − 4 = 3
(b) is increasing on (03) because as increases from 0 to 3, we keep
adding more area.
(d)
(c) has a maximum value when we start subtracting area; that is,
at = 3.
4. (a) () = 0 (), so (0) = 0 since the limits of integration are equal and (6) = 0 since the areas above and below the
-axis are equal.
(b) (1) is the area under the curve from 0 to 1, which includes two unit squares and about 80% to 90% of a third unit square,
so (1) ≈ 28. Similarly, (2) ≈ 49 and (3) ≈ 57. Now (3) − (2) ≈ 08, so (4) ≈ (3) − 08 ≈ 49 by the
symmetry of about = 3. Likewise, (5) ≈ 28.
(c) As we go from = 0 to = 3, we are adding area, so increases on the interval (03).
(d) increases on (03) and decreases on (36) [where we are subtracting area], so has a maximum value at = 3.
(e) A graph of must have a maximum at = 3, be symmetric about = 3,
and have zeros at = 0 and = 6.
(f) If we sketch the graph of 0 by estimating slopes on the graph of (as in Section 2.8), we get a graph that looks like (as
indicated by FTC1).
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© Cengage Learning. All Rights Reserved.26 ¤ CHAPTER 5 INTEGRALS
5. (a) By FTC1 with () = 2 and = 1, () = 1 2 ⇒
0() = () = 2.
(b) Using FTC2, () = 1 2 = 1 3 3 1 = 1 3 3 − 1 3 ⇒ 0() = 2.
6. (a) By FTC1 with () = 2 + sin and = 0, () = 0(2 + sin) ⇒
0() = () = 2 + sin.
(b) Using FTC2,
() = 0(2 + sin) = [2 − cos] 0 = (2 − cos) − (0 − 1)
= 2 − cos + 1 ⇒
0() = 2 − (−sin) + 0 = 2 + sin
7. () = √ + 3 and () = 0 √ + 3 , so by FTC1, 0() = () = √ + 3.
8. () = ln(1 + 2) and () = 1 ln(1 + 2) , so by FTC1, 0() = () = ln(1 + 2).
9. () = ( − 2)8 and () = 5( − 2)8 , so by FTC1, 0() = () = ( − 2)8.
10. () =
√
+ 1
and () = 0 √+ 1 , so by FTC1, 0() = () = √+ 1 .
11. () = 0 √1 + sec = − 0 √1 + sec ⇒ 0() = − 0 √1 + sec = −√1 + sec
12. () = 2 3 sin = − 2 3 sin ⇒ 0() = − 2 3 sin = −3 sin
13. Let = . Then
= . Also, = , so
0() =
1 ln = 1 ln · = ln = (ln) · = .
14. Let = √. Then
=
1
2√. Also, = , so
0() =
1√ 4+ 1 2 = 1 4+ 1 2 · = 4+ 1 2 = 2 + 1 2√1 = 2(√2 + 1) .
15. Let = 3 + 2. Then
= 3. Also, = , so
0 =
13+2 1 + 3 = 1 1 + 3 · = 1 +3 = 1 + (3 3+ 2 + 2)3 · 3 = 1 + (3 3(3+ 2) + 2)3
16. Let = 4. Then
= 43. Also, = , so
0 =
04 cos2 = 0 cos2 · = cos2 = cos2(4) · 43.
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© Cengage Learning. All Rights Reserved.SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 27
17. Let = √. Then
=
1
2√. Also, = , so
0 =
√ 4 tan = − √4 tan · = −tan = −√tan√ · 2√1 = −12 tan√
18. Let = sin. Then
= cos. Also, = , so
0 =
sin 1 1 + 2 = 1 1 + 2 · = − 1 1 + 2 ·
= −√1 + 2 cos = −1 + sin2 cos
19. 13(2 + 2 − 4) = 1 33 + 2 − 43 1 = (9 + 9 − 12) − 1 3 + 1 − 4 = 6 + 8 3 = 26 3
20. −11 100 = 101 1 1011 −1 = 101 1 − − 101 1 = 101 2
21. 02( 4 53 − 3 42 + 2 5) = 1 54 − 1 43 + 1 522 0 = 16 5 − 2 + 4 5 − 0 = 2
22. 01(1 − 83 + 167) = − 24 + 281 0 = (1 − 2 + 2) − 0 = 1
23. 19 √ = 19 12 = 3322 9 1 = 2 3329 1 = 2 3(932 − 132) = 2 3(27 − 1) = 52 3
24. 18 −23 = 1133 8 1 = 3138 1 = 3(813 − 113) = 3(2 − 1) = 3
25. 6 sin = −cos 6 = −cos − −cos 6 = −(−1) − −√32 = 1 + √32
26. −55 = 5 −5 = 5 − (−5) = 10
27. 01( + 2)( − 3) = 01(2 − − 6) = 1 33 − 1 22 − 61 0 = 1 3 − 1 2 − 6 − 0 = − 37 6
28. 04(4 − )√ = 04(4 − )12 = 04(412 − 32) = 8 332 − 2 5524 0 = 8 3(8) − 2 5(32) = 32015 −192 = 128 15
29. 14 2 + √2 = 14 √2 + √2 = 14(2−12 + 32)
= 412 + 2 5524 1 = 4(2) + 2 5(32) − 4 + 2 5 = 8 + 64 5 − 4 − 2 5 = 82 5
30. −21(3 − 2)( + 1) = −21(32 + − 2) = 3 + 1 22 − 22 −1 = (8 + 2 − 4) − −1 + 1 2 + 2 = 6 − 3 2 = 9 2
31. 62 csc cot = −csc 2 6 = −csc 2 − −csc 6 = −1 − (−2) = 1
32. 43 csc2 = −cot 3 4 = −cot 3 − −cot 4 = −√13 − (−1) = 1 − √13
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© Cengage Learning. All Rights Reserved.28 ¤ CHAPTER 5 INTEGRALS
33. 01(1 + )3 = 01(1 + 3 + 32 + 3) = + 3 2 2 + 3 + 1 4 41 0 = 1 + 3 2 + 1 + 1 4 − 0 = 15 4
34. 03(2 sin − ) = −2cos − 3 0 = (−2cos 3 − 3) − (−2 − 1) = 3 − 2cos 3 − 3
35. 12 3 + 3 4 6 = 12 1 + 32 = ln|| + 32 1 = (ln 2 + 8) − (ln 1 + 1) = ln 2 + 7
36. 118 3 = 118 √3−12 = √321218 1 = 2√3(1812 − 112) = 2√3(3√2 − 1)
37. 01( + ) = + 1 +1 + 1 0 = + 1 1 + − (0 + 1) = + 1 1 + − 1
38. 01 cosh = sinh1 0 = sinh 1 − sinh 0 = sinh 1 or 1 2( − −1)
39. 1√√33 1 +82 = 8arctan√ 13√3 = 8 3 − 6 = 8 6 = 43
40. 13 3 − 22 2 − = 13 − 2 − 1 = 1 2 2 − 2 − ln||3 1 = 9 2 − 6 − ln 3 − 1 2 − 2 − 0 = −ln 3
41. 04 2 = ln 2 1 24 0 = ln 2 16 − ln 2 1 = ln 2 15
42. 112√2 √1 4− 2 = 4arcsin1 1 √ 2 2 = 4 4 − 6 = 412 = 3
43. If () = sin cos if if 0≤2 ≤ ≤ 2 then
0 () = 02 sin + 2 cos = −cos 0 2 + sin 2 = −cos 2 + cos 0 + sin − sin 2
= −0 + 1 + 0 − 1 = 0
Note that is integrable by Theorem 3 in Section 5.2.
44. If () = 24 − 2 if if 0− 2 ≤≤2≤ 0 then
−22 () = −02 2 + 02(4 − 2) = 20 −2 + 4 − 1 3 32 0 = [0 − (−4)] + 16 3 − 0 = 28 3
Note that is integrable by Theorem 3 in Section 5.2.
45. Area = 04 √ = 04 12 = 2 3 324 0 = 2 3(8) − 0 = 16 3
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© Cengage Learning. All Rights Reserved.SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 29
46. Area = 01 3 = 1 441 0 = 1 4 − 0 = 1 4
47. Area = −22(4 − 2) = 4 − 1 332 −2 = 8 − 8 3 − −8 + 8 3 = 32 3
48. Area = 02(2 − 2) = 2 − 1 332 0 = 4 − 8 3 − 0 = 4 3
49. From the graph, it appears that the area is about 60. The actual area is
027 13 = 3 44327 0 = 3 4 · 81 − 0 = 243 4 = 6075. This is 3 4 of the
area of the viewing rectangle.
50. From the graph, it appears that the area is about 1 3. The actual area is
16 −4 = −−33 6 1 = 3−13 6 1 = −3 ·1216 + 13 = 215 648 ≈ 03318.
51. It appears that the area under the graph is about 2 3 of the area of the viewing
rectangle, or about 2 3 ≈ 21. The actual area is
0 sin = [−cos ] 0 = (−cos) − (−cos 0) = −(−1) + 1 = 2.
52. Splitting up the region as shown, we estimate that the area under the graph
is
3 + 1 43 · 3 ≈ 18. The actual area is
03 sec2 = [tan] 0 3 = √3 − 0 = √3 ≈ 173.
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© Cengage Learning. All Rights Reserved.30 ¤ CHAPTER 5 INTEGRALS
53. −21 3 = 1 442 −1 = 4 − 1 4 = 15 4 = 375
54. 26 cos = sin2 6 = 0 − 1 2 = − 1 2
55. () = −4 is not continuous on the interval [−21], so FTC2 cannot be applied. In fact, has an infinite discontinuity at
= 0, so −12 −4 does not exist.
56. () = 4
3 is not continuous on the interval [−1 2], so FTC2 cannot be applied. In fact, has an infinite discontinuity at
= 0, so −21 43 does not exist.
57. () = sec tan is not continuous on the interval [3 ], so FTC2 cannot be applied. In fact, has an infinite
discontinuity at = 2, so 3 sec tan does not exist.
58. () = sec2 is not continuous on the interval [0 ], so FTC2 cannot be applied. In fact, has an infinite discontinuity at
= 2, so 0 sec2 does not exist.
59. () = 23 2 2 − + 1 1 = 20 2 2 − + 1 1 + 03 2 2 − + 1 1 = − 02 2 2 − + 1 1 + 03 2 2 − + 1 1 ⇒
0() = −(2)2 − 1
(2)2 + 1 ·
(2) + (3 (3))2 2 − + 1 1 · (3) = −2 · 442 2 − + 1 1 + 3 · 992 2 − + 1 1
60. () = 1−1+2 2 sin = 1−0 2 sin + 01+2 sin = − 01−2 sin + 01+2 sin ⇒
0() = −(1 − 2)sin(1 − 2) ·
(1 − 2) + (1 + 2)sin(1 + 2) · (1 + 2)
= 2(1 − 2)sin(1 − 2) + 2(1 + 2)sin(1 + 2)
61. () = 2 2 = 0 2 + 02 2 = − 0 2 + 02 2 ⇒
0() = −2 + (2)2 ·
(2) = −2 + 24
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62. () = √2 arctan = √0 arctan + 02 arctan = − 0√ arctan + 02 arctan ⇒
0() = −arctan√ ·
(√) + arctan 2 · (2) = −2√1 arctan√ + 2 arctan 2
63. = cos sin ln(1 + 2) = cos 0 ln(1 + 2) + 0sin ln(1 + 2)
= − 0cos ln(1 + 2) + 0sin ln(1 + 2) ⇒
0 = −ln(1 + 2 cos ) ·
cos + ln(1 + 2 sin) · sin = sin ln(1 + 2 cos) + cos ln(1 + 2 sin)
64. () = 0(1 − 2)2 is increasing when 0() = (1 − 2)2 is positive.
Since 2 0, 0() 0 ⇔ 1 − 2 0 ⇔ || 1, so is increasing on (−11).
65. = 0 2 +2 + 2 ⇒ 0 = 2 +2 + 2 ⇒
00 = (2 + + 2)(2) − 2(2 + 1)
(2 + + 2)2 =
23 + 22 + 4 − 23 − 2
(2 + + 2)2 =
2 + 4
(2 + + 2)2 =
( + 4)
(2 + + 2)2 .
The curve is concave downward when 00 0; that is, on the interval (−4 0).
66. If () = 1 () , then by FTC1, 0() = (), and also, 00() = 0(). is concave downward where 00 is
negative; that is, where 0 is negative. The given graph shows that is decreasing ( 0 0) on the interval (−11).
67. () = 2 2 ⇒ 0() = 2, so the slope at = 2 is 22 = 4. The -coordinate of the point on at = 2 is
(2) = 22 2 = 0 since the limits are equal. An equation of the tangent line is − 0 = 4( − 2), or = 4 − 24.
68. () = 3 () ⇒ 0() = (). Since () = 0sin √1 + 2 , 00() = 0() = 1 + sin2 · cos,
so 00( 6 ) = 1 + sin2( 6 ) · cos 6 = 1 + ( 1 2)2 · √23 = √25 · √23 = √415.
69. By FTC2, 14 0() = (4) − (1), so 17 = (4) − 12 ⇒ (4) = 17 + 12 = 29.
70. (a) erf() = √2 0 −2 ⇒ 0 −2 = √2 erf() By Property 5 of definite integrals in Section 5.2,
0 −2 = 0 −2 + −2 , so
−2 = 0 −2 − 0 −2 = √2 erf() − √2 erf() = 1 2 √ [erf() − erf()].
(b) = 2 erf() ⇒ 0 = 22 erf() + 2 erf0() = 2 + 2 · √2 −2 [by FTC1] = 2 + √2 .
71. (a) The Fresnel function () = 0 sin 2 2 has local maximum values where 0 = 0() = sin 2 2 and
0 changes from positive to negative. For 0, this happens when 2 2 = (2 − 1) [odd multiples of ] ⇔
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2 = 2(2 − 1) ⇔ = √4 − 2, any positive integer. For 0, 0 changes from positive to negative where
2
2 = 2 [even multiples of ] ⇔ 2 = 4 ⇔ = −2√. 0 does not change sign at = 0.
(b) is concave upward on those intervals where 00() 0. Differentiating our expression for 0(), we get
00() = cos 2 22 2 = cos 2 2. For 0, 00() 0 where cos( 2 2) 0 ⇔ 0 2 2 2 or
2 − 1 2 2 2 2 + 1 2 , any integer ⇔ 0 1 or √4 − 1 √4 + 1, any positive integer.
For 0, 00() 0 where cos( 2 2) 0 ⇔ 2 − 3 2 2 2 2 − 1 2 , any integer ⇔
4 − 3 2 4 − 1 ⇔ √4 − 3 || √4 − 1 ⇒ √4 − 3 − √4 − 1 ⇒
−√4 − 3 −√4 − 1, so the intervals of upward concavity for 0 are −√4 − 1 −√4 − 3, any
positive integer. To summarize: is concave upward on the intervals (01), −√3 −1, √3 √5, −√7 −√5,
√7 3, .
(c) In Maple, we use plot({int(sin(Pi*tˆ2/2),t=0..x),0.2},x=0..2);. Note that
Maple recognizes the Fresnel function, calling it FresnelS(x). In Mathematica, we use
Plot[{Integrate[Sin[Pi*tˆ2/2],{t,0,x}],0.2},{x,0,2}]. In Derive, we load the utility file
FRESNEL and plot FRESNEL_SIN(x). From the graphs, we see that 0 sin 2 2 = 02 at ≈ 074.
72. (a) In Maple, we should start by setting si:=int(sin(t)/t,t=0..x);. In
Mathematica, the command is si=Integrate[Sin[t]/t,{t,0,x}].
Note that both systems recognize this function; Maple calls it Si(x) and
Mathematica calls it SinIntegral[x]. In Maple, the command to generate
the graph is plot(si,x=-4*Pi..4*Pi);. In Mathematica, it is
Plot[si,{x,-4*Pi,4*Pi}]. In Derive, we load the utility file EXP_INT and plot SI(x).
(b) Si() has local maximum values where Si0() changes from positive to negative, passing through 0. From the
Fundamental Theorem we know that Si0() =
0 sin = sin, so we must have sin = 0 for a maximum, and
for 0 we must have = (2 − 1), any positive integer, for Si0 to be changing from positive to negative at .
For 0, we must have = 2, any positive integer, for a maximum, since the denominator of Si0() is negative
for 0. Thus, the local maxima occur at = , −2, 3, −4, 5, −6, .
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© Cengage Learning. All Rights Reserved.SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 33
(c) To find the first inflection point, we solve Si00() = cos
−
sin
2 = 0. We can see from the graph that the first inflection
point lies somewhere between = 3 and = 5. Using a rootfinder gives the value ≈ 44934. To find the -coordinate
of the inflection point, we evaluate Si(44934) ≈ 16556. So the coordinates of the first inflection point to the right of the
origin are about (4493416556). Alternatively, we could graph 00() and estimate the first positive -value at which it
changes sign.
(d) It seems from the graph that the function has horizontal asymptotes at ≈ 15, with lim
→±∞
Si() ≈ ±15 respectively.
Using the limit command, we get lim
→∞
Si() = 2 . Since Si() is an odd function, lim
→−∞
Si() = − 2 . So Si() has the
horizontal asymptotes = ± 2 .
(e) We use the fsolve command in Maple (or FindRoot in Mathematica) to find that the solution is ≈ 11. Or, as in
Exercise 65(c), we graph = Si() and = 1 on the same screen to see where they intersect.
73. (a) By FTC1, 0() = (). So 0() = () = 0 at = 1 357, and 9. has local maxima at = 1 and 5 (since = 0
changes from positive to negative there) and local minima at = 3 and 7. There is no local maximum or minimum at
= 9, since is not defined for 9.
(b) We can see from the graph that
01 13 35 57 79 . So (1) = 01 ,
(5) = 05 = (1) − 13 + 35 , and (9) = 09 = (5) − 57 + 79 . Thus,
(1) (5) (9), and so the absolute maximum of () occurs at = 9.
(c) is concave downward on those intervals where 00 0. But 0() = (),
so 00() = 0(), which is negative on (approximately) 1 22, (4 6) and
(89). So is concave downward on these intervals.
(d)
74. (a) By FTC1, 0() = (). So 0() = () = 0 at = 2, 4, 6, 8, and 10. has local maxima at = 2 and 6 (since = 0
changes from positive to negative there) and local minima at = 4 and 8. There is no local maximum or minimum at
= 10, since is not defined for 10.
(b) We can see from the graph that
02 24 46 68 810 . So (2) = 02 ,
(6) = 06 = (2) − 24 + 46 , and (10) = 010 = (6) − 68 + 810 . Thus,
(2) (6) (10), and so the absolute maximum of () occurs at = 2.
(c) is concave downward on those intervals where 00 0. But 0() = (),
so 00() = 0(), which is negative on (13), (57) and (910). So is
concave downward on these intervals.
(d)
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© Cengage Learning. All Rights Reserved.34 ¤ CHAPTER 5 INTEGRALS
75. lim
→∞
=1 45 + 2 = lim →∞ =1 44 + 1 = lim →∞ 1 − 0 =1 4 + = 01(4 + )
= 1 55 + 1 221 0 = 1 5 + 1 2 − 0 = 10 7
76. lim
→∞
1
1 + 2 + · · · + = lim →∞ 1 − 0 =1 = 01 √ = 2332 1 0 = 2 3 − 0 = 23
77. Suppose 0. Since is continuous on [ + ], the Extreme Value Theorem says that there are numbers and in
[ + ] such that () = and () = , where and are the absolute minimum and maximum values of on
[ + ]. By Property 8 of integrals, (−) ≤ + () ≤ (−); that is, ()(−) ≤ − + () ≤ ()(−).
Since − 0, we can divide this inequality by −: () ≤ 1
+ () ≤ (). By Equation 2,
( + ) − ()
=
1
+ () for 6= 0, and hence () ≤ ( + ) − () ≤ (), which is Equation 3 in the
case where 0.
78.
(()) () = () () + () () [where is in the domain of ]
=
−() () + () () = −(()) 0() + (()) 0()
= (())0 () − (())0()
79. (a) Let () = √ ⇒ 0() = 1(2√) 0 for 0 ⇒ is increasing on (0 ∞). If ≥ 0, then 3 ≥ 0, so
1 + 3 ≥ 1 and since is increasing, this means that 1 + 3 ≥ (1) ⇒ √1 + 3 ≥ 1 for ≥ 0. Next let
() = 2 − ⇒ 0() = 2 − 1 ⇒ 0() 0 when ≥ 1. Thus, is increasing on (1 ∞). And since (1) = 0,
() ≥ 0 when ≥ 1. Now let = √1 + 3, where ≥ 0. √1 + 3 ≥ 1 (from above) ⇒ ≥ 1 ⇒ () ≥ 0 ⇒
1 + 3 − √1 + 3 ≥ 0 for ≥ 0. Therefore, 1 ≤ √1 + 3 ≤ 1 + 3 for ≥ 0.
(b) From part (a) and Property 7: 01 1 ≤ 01 √1 + 3 ≤ 01(1 + 3) ⇔
1 0 ≤ 01 √1 + 3 ≤ + 1 441 0 ⇔ 1 ≤ 01 √1 + 3 ≤ 1 + 1 4 = 125.
80. (a) For 0 ≤ ≤ 1 we have 2 ≤ Since () = cos is a decreasing function on [01],cos(2) ≥ cos
(b) 6 1 so by part (a), cos(2) ≥ cos on [0 6]. Thus,
06 cos(2) ≥ 06 cos = sin 0 6 = sin(6) − sin 0 = 1 2 − 0 = 1 2.
81. 0 2
4 + 2 + 1
2
4 =
1 2
on [5 10], so
0 ≤ 510 4 +22 + 1 510 12 = −110 5 = −10 1 − −15 = 10 1 = 01
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82. (a) If 0, then () = 0 () = 0 0 = 0.
If 0 ≤ ≤ 1, then () = 0 () = 0 = 1 22 0 = 1 22.
If 1 ≤ 2, then
() = 0 () = 01 () + 1 () = (1) + 1(2 − )
=
1 2
(1)2 + 2 − 1 22 1 = 1 2 + 2 − 1 22 − 2 − 1 2 = 2 − 1 22 − 1.
If 2, then () = 0 () = (2) + 2 0 = 1 + 0 = 1. So
() =
0 if 0
1 2
2 if 0 ≤ ≤ 1
2 − 1
22 − 1 if 1 ≤ 2
1 if 2
(b)
(c) is not differentiable at its corners at = 0, 1, and 2. is differentiable on (−∞0), (01), (12) and (2 ∞).
is differentiable on (−∞ ∞).
83. Using FTC1, we differentiate both sides of 6 + (2) = 2√ to get (2) = 2 2√1 ⇒ () = 32.
To find , we substitute = in the original equation to obtain 6 + (2) = 2√ ⇒ 6 + 0 = 2√ ⇒
3 = √ ⇒ = 9.
84. = 3 ⇒ 0 = 30 ⇒ [] 0 = 3 [] 0 ⇒ − 1 = 3( − 1) ⇒ = 3 − 2 ⇒
= ln(3 − 2)
85. (a) Let () = 0 (). Then, by FTC1, 0() = () = rate of depreciation, so () represents the loss in value over the
interval [0 ].
(b) () = 1
+ 0 () = +() represents the average expenditure per unit of during the interval [0 ],
assuming that there has been only one overhaul during that time period. The company wants to minimize average
expenditure.
(c) () = 1
+ 0 (). Using FTC1, we have 0() = −12 + 0 () + 1 ().
0() = 0 ⇒ () = + 0 () ⇒ () = 1 + 0 () = ().
86. (a) () = 1
0 [() + ()]. Using FTC1 and the Product Rule, we have
0() = 1
[() + ()] − 1
2 0 [() + ()]. Set 0() = 0: 1 [() + ()] − 12 0 [() + ()] = 0 ⇒
[() + ()] − 1
0 [() + ()] = 0 ⇒ [() + ()] − () = 0 ⇒ () = () + ().
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(b) For 0 ≤ ≤ 30, we have () = 015 − 450 = 15 − 900 2 0 = 15 − 900 2.
So () = ⇒
15
−
900
2 = ⇒ 60 − 2 = 900 ⇒ 2 − 60 + 900 = 0 ⇒
( − 30)2 = 0 ⇒ = 30. So the length of time is 30 months.
(c) () = 1
015 − 450 + 12,9002 = 1 15 − 900 2 + 38,7003 0
=
1
15 − 900 2 + 38,7003 = 15 − 900 + 38,7002 ⇒
0() = −
900
+
19,350
= 0 when 1
19,350
=
1
900
⇒ = 215.
(215) =
15
−
900
(215) +
38,700(215)2 ≈ 005472 , (0) = 15 ≈ 006667 , and
(30) =
15
−
900
(30) +
38,700(30)2 ≈ 005659 , so the absolute minimum is (215) ≈ 005472 .
(d) As in part (c), we have () =
15
−
900
+
38,7002, so () = () + () ⇔
15
−
900
+
38,7002 = 15 − 450 + 12,9002 ⇔
212,1900 − 38,1700 = 450 1 − 900 1 ⇔ = 2138 900 ,700 = 43 2 = 215.
This is the value of that we obtained as the critical number of in part (c), so we
have verified the result of (a) in this case.
5.4 Indefinite Integrals and the Net Change Theorem
1.
−√1 + 2 + = −(1 + 2)12 + = − · 1 2(1 + 2)−12((2 )2) − (1 + 2)12 · 1 + 0
= −
(1 + 2)−12 2 − (1 + 2)
2 = −
−1
(1 + 2)122 =
1
2√1 + 2
2.
21 + 41 sin 2 + = 1 2 + 14 cos 2 · 2 + 0 = 12 + 12 cos 2
=
1 2
+ 1
22cos2 − 1 = 1 2 + cos2 − 1 2 = cos2
3.
(tan − + ) = sec2 − 1 + 0 = tan2
4.
1522 (3 − 2)( + )32 + = 1522 (3 − 2) 3 2( + )12() + ( + )32(3) + 0
=
2
152 (3)( + )12 (3 − 2) 1 2 + ( + )
=
2 5
( + )125 2 = √ +
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5. (13 + 725) = 21323 + 37535 + = 21323 + 235 +
6. √4 5 = 54 = 4 994 +
7. 5 + 2 32 + 3 43 = 5 + 2 3 · 1 33 + 3 4 · 1 44 + = 5 + 2 93 + 16 3 4 +
8. (6 − 25 − 3 + 2 7) = 1 77 − 2 · 1 66 − 1 44 + 2 7 + = 1 77 − 1 36 − 1 44 + 2 7 +
9. ( + 4)(2 + 1) = (22 + 9 + 4) = 2 33 + 9 22 + 4 + = 2 33 + 922 + 4 +
10. √(2 + 3 + 2) = 12(2 + 3 + 2) = (52 + 332 + 212)
=
2 7
72 + 3 · 2
552 + 2 · 2 332 + = 2 772 + 6 552 + 4 332 +
11. 1 + √ + = 1 + √ + = 1 + −12 + 1
= ln|| + 212 + + = ln|| + 2√ + +
12. 2 + 1 + 21+ 1 = 33 + + tan−1 +
13. (sin + sinh) = −cos + cosh +
14. 1 + 2 = 1 + 22+ 2 = (−2 + 2−1 + 1) = −−1 + 2 ln|| + + = −1 + 2 ln|| + +
15. (2 + tan2 ) = [2 + (sec2 − 1)] = (1 + sec2 ) = + tan +
16. sec(sec + tan) = (sec2 + sec tan) = tan + sec +
17. 2(1 + 5) = (2 + 2 · 5) = (2 + 10) = ln 2 2 + ln 10 10 +
18. sin 2 sin = 2sinsin cos = 2cos = 2 sin +
19. cos + 1 2 = sin + 1 42 + . The members of the family
in the figure correspond to = −5, 0, 5, and 10.
20. ( − 22) = − 2 33 + The members of the family in the
figure correspond to = −5, 0, 2, and 5.
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© Cengage Learning. All Rights Reserved.38 ¤ CHAPTER 5 INTEGRALS
21. −32(2 − 3) = 1 33 − 33 −2 = (9 − 9) − − 8 3 + 6 = 8 3 − 18 3 = − 10 3
22. 12(43 − 32 + 2) = 4 − 3 + 22 1 = (16 − 8 + 4) − (1 − 1 + 1) = 12 − 1 = 11
23. −02 1 24 + 1 43 − = 10 1 5 + 16 1 4 − 1 220 −2 = 0 − 10 1 (−32) + 16 1 (16) − 1 2(4) = − − 16 5 + 1 − 2 = 21 5
24. 03(1 + 62 − 104) = + 23 − 253 0 = (3 + 54 − 486) − 0 = −429
25. 02(2 − 3)(42 + 1) = 02(83 − 122 + 2 − 3) = 24 − 43 + 2 − 32 0 = (32 − 32 + 4 − 6) − 0 = −2
26. −11 (1 − )2 = −11 (1 − 2 + 2) = −11( − 22 + 3) = 1 22 − 2 33 + 1 441 −1
= 1 2 − 2 3 + 1 4 − 1 2 + 2 3 + 1 4 = − 4 3
27. 0(5 + 3 sin) = [5 − 3cos ] 0 = [5 − 3(−1)] − [5(1) − 3(1)] = 5 + 1
28. 12 12 − 43 = 12(−2 − 4−3) = −−11 − 4−−22 2 1 = −1 + 22 2 1 = −12 + 12 − (−1 + 2) = −1
29. 14 4 + 6 √ = 14 √4 + √6 = 14(4−12 + 612) = 812 + 4324 1 = (16 + 32) − (8 + 4) = 36
30. 01 1 +42 = 4arctan1 0 = 4 arctan 1 − 4arctan0 = 44 − 4(0) =
31. 01 √3 + √4 = 01(43 + 54) = 3 773 + 4 9941 0 = 3 7 + 4 9 − 0 = 55 63
32. 14 √−2 = 14 √2 − 2 = 14(−32 − −1) = −2−12 − ln|| 4 1 = −√2 − ln||4 1
= (−1 − ln 4) − (−2 − ln 1) = 1 − ln 4
33. 12 2 − 2 = 1 42 − 2ln||2 1 = (1 − 2ln2) − 14 − 2ln1 = 34 − 2ln2
34. 01(5 − 5) = 522 − ln 5 5 1 0 = 5 2 − ln 5 5 − 0 − ln 5 1 = 52 − ln 5 4
35. 01(10 + 10) = 11 11 + ln 10 10 1 0 = 11 1 + ln 10 10 − 0 + ln 10 1 = 11 1 + ln 10 9
36. 04 sec tan = sec 0 4 = sec 4 − sec 0 = √2 − 1
37. 04 1 + cos cos2 2 = 04cos12 + cos cos2 2 = 04(sec2 + 1)
= tan + 0 4 = tan 4 + 4 − (0 + 0) = 1 + 4
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38. 03 sin + sin sec2 tan2 = 03 sin (1 + tan sec2 2 ) = 03 sinsec sec 2 2 = 03 sin
= −cos 0 3 = − 1 2 − (−1) = 1 2
39. 18 2 + √3 2 = 18 223 + 23 = 18(2−23 + 13) = 2 · 313 + 3 4 438 1 = (12 + 12) − 6 + 3 4 = 69 4
40. −10 10 sinh2+ cosh = −10 10 − −2
2
+
+ −
2
= −10 10 2 = −10 10 2 = 210 −10 = 20 − (−20) = 40
41. 0√32 √1 − 2 = arcsin√ 0 32 = arcsin√32 − arcsin 0 = 3 − 0 = 3
42. 12 ( −21)3 = 12 3 − 322+ 3 − 1 = 12 − 3 + 3 − 12 = 1 2 2 − 3 + 3 ln|| + 12 1
= 2 − 6 + 3 ln 2 + 1 2 − 1 2 − 3 + 0 + 1 = 3 ln 2 − 2
43. 01√3 2 4 − − 1 1 = 01√3 (2 + 1)( 2 −21− 1) = 01√3 2 1+ 1 = arctan1 0√3 = arctan1√3 − arctan 0
=
6
− 0 =
6
44. |2 − 1| = 2−(2−1− 1) if if 2 2 − − 1 1 ≥ 0 0 = 2 1−−21 if if ≥ 1 2 1 2
Thus, 02 |2 − 1| = 012(1 − 2) + 122(2 − 1) = − 21 02 + 2 − 2 12
= ( 1 2 − 1 4) − 0 + (4 − 2) − ( 1 4 − 1 2) = 1 4 + 2 − (− 1 4) = 5 2
45. −21 ( − 2||) = −01[ − 2(−)] + 02[ − 2()] = −01 3 + 02(−) = 3 1 2 20 −1 − 1 2 22 0
= 30 − 1 2 − (2 − 0) = − 7 2 = −35
46. 032 |sin| = 0 sin + 32(−sin) = − cos 0 + cos 3 2 = [1 − (−1)] + [0 − (−1)] = 2 + 1 = 3
47. The graph shows that = 1 − 2 − 54 has -intercepts at
= ≈ −086 and at = ≈ 042. So the area of the region that lies
under the curve and above the -axis is
(1 − 2 − 54) = − 2 − 5
= ( − 2 − 5) − ( − 2 − 5) ≈ 136
48. The graph shows that = (2 + 1)−1 − 4 has -intercepts at
= ≈ −087 and at = ≈ 087. So the area of the region that lies
under the curve and above the -axis is
(2 + 1)−1 − 4 = tan−1 − 1 5 5
= tan−1 − 1 5 5 − tan−1 − 1 5 5
≈ 123
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49. = 022 − 2 = 2 − 1 332 0 = 4 − 8 3 − 0 = 4 3
50. = √4 ⇒ = 4, so = 01 4 = 1 551 0 = 1 5 .
51. If 0() is the rate of change of weight in pounds per year, then () represents the weight in pounds of the child at age . We
know from the Net Change Theorem that 510 0() = (10) − (5), so the integral represents the increase in the child’s
weight (in pounds) between the ages of 5 and 10.
52. () = 0() = () − () by the Net Change Theorem, so it represents the change in the charge from time
= to = .
53. Since () is the rate at which oil leaks, we can write () = − 0(), where () is the volume of oil at time . [Note that the
minus sign is needed because is decreasing, so 0() is negative, but () is positive.] Thus, by the Net Change Theorem,
0120 () = − 0120 0() = −[ (120) − (0)] = (0) − (120), which is the number of gallons of oil that leaked
from the tank in the first two hours (120 minutes).
54. By the Net Change Theorem, 015 0() = (15) − (0) = (15) − 100 represents the increase in the bee population in
15 weeks. So 100 + 015 0() = (15) represents the total bee population after 15 weeks.
55. By the Net Change Theorem, 1000 5000 0() = (5000) − (1000), so it represents the increase in revenue when
production is increased from 1000 units to 5000 units.
56. The slope of the trail is the rate of change of the elevation , so () = 0(). By the Net Change Theorem,
35 () = 35 0() = (5) − (3) is the change in the elevation between = 3 miles and = 5 miles from the
start of the trail.
57. In general, the unit of measurement for () is the product of the unit for () and the unit for . Since () is
measured in newtons and is measured in meters, the units for 0100 () are newton-meters (or joules). (A newton-meter
is abbreviated N·m.)
58. The units for () are pounds per foot and the units for are feet, so the units for are pounds per foot per foot, denoted
(lbft)ft. The unit of measurement for 28 () is the product of pounds per foot and feet; that is, pounds.
59. (a) Displacement = 03(3 − 5) = 3 22 − 53 0 = 27 2 − 15 = − 3 2 m
(b) Distance traveled = 03 |3 − 5| = 053(5 − 3) + 533(3 − 5)
= 5 − 3 225 03 + 3 22 − 53 53 = 25 3 − 3 2 · 25 9 + 27 2 − 15 − 3 2 · 25 9 − 25 3 = 41 6 m
60. (a) Displacement = 24(2 − 2 − 3) = 1 33 − 2 − 34 2 = 64 3 − 16 − 12 − 8 3 − 4 − 6 = 2 3 m
(b) () = 2 − 2 − 3 = ( + 1)( − 3), so () 0 for −1 3, but on the interval [24], () 0 for 2 ≤ 3.
Distance traveled = 24 2 − 2 − 3 = 23 −(2 − 2 − 3) + 34(2 − 2 − 3)
= − 1 33 + 2 + 33 2 + 1 33 − 2 − 34 3
= (−9 + 9 + 9) − − 8 3 + 4 + 6 + 64 3 − 16 − 12 − (9 − 9 − 9) = 4 m
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61. (a) 0() = () = + 4 ⇒ () = 1 2 2 + 4 + ⇒ (0) = = 5 ⇒ () = 1 2 2 + 4 + 5 ms
(b) Distance traveled = 010 |()| = 010 1 2 2 + 4 + 5 = 010 1 2 2 + 4 + 5 = 1 6 3 + 22 + 510 0
=
500
3 + 200 + 50 = 416 2 3 m
62. (a) 0() = () = 2 + 3 ⇒ () = 2 + 3 + ⇒ (0) = = −4 ⇒ () = 2 + 3 − 4
(b) Distance traveled = 03 2 + 3 − 4 = 03 |( + 4)( − 1)| = 01−2 − 3 + 4 + 132 + 3 − 4
= − 1 3 3 − 3 2 2 + 41 0 + 1 3 3 + 3 2 2 − 43 1
= − 1 3 − 3 2 + 4 + 9 + 27 2 − 12 − 1 3 + 3 2 − 4 = 89 6 m
63. Since 0() = (), = 04 () = 049 + 2√ = 9 + 4 3 324 0 = 36 + 32 3 − 0 = 140 3 = 46 2 3 kg.
64. By the Net Change Theorem, the amount of water that flows from the tank during the first 10 minutes is
010 () = 010(200 − 4) = 200 − 2210 0 = (2000 − 200) − 0 = 1800 liters.
65. Let be the position of the car. We know from Equation 2 that (100) − (0) = 0100 (). We use the Midpoint Rule for
0 ≤ ≤ 100 with = 5. Note that the length of each of the five time intervals is 20 seconds = 3600 20 hour = 180 1 hour.
So the distance traveled is
0100 () ≈ 180 1 [(10) + (30) + (50) + (70) + (90)] = 180 1 (38 + 58 + 51 + 53 + 47) = 247 180 ≈ 14 miles.
66. (a) By the Net Change Theorem, the total amount spewed into the atmosphere is (6) − (0) = 06 () = (6) since
(0) = 0. The rate () is positive, so is an increasing function. Thus, an upper estimate for (6) is 6 and a lower
estimate for (6) is 6. ∆ = −
=
6 − 0
6
= 1.
6 =
6
=1
()∆ = 10 + 24 + 36 + 46 + 54 + 60 = 230 tonnes.
6 =
6
=1
(−1)∆ = 6 + (0) − (6) = 230 + 2 − 60 = 172 tonnes.
(b) ∆ = −
=
6 − 0
3
= 2. (6) ≈ 3 = 2[(1) + (3) + (5)] = 2(10 + 36 + 54) = 2(100) = 200 tonnes.
67. From the Net Change Theorem, the increase in cost if the production level is raised from 2000 yards to 4000 yards is
(4000) − (2000) = 2000 4000 0().
2000 4000 0() = 2000 4000(3 − 001 + 00000062) = 3 − 00052 + 000000234000 2000
= 60,000 − 2,000 = $58,000
68. By the Net Change Theorem, the amount of water after four days is
25,000 + 04 () ≈ 25,000 + 4 = 25,000 + 4 −4 0 [(05) + (15) + (25) + (35)]
≈ 25,000 + [1500 + 1770 + 740 + (−690)] = 28,320 liters
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69. To use the Midpoint Rule, we’ll use the midpoint of each of three 2-second intervals.
(6) − (0) = 06 () ≈ [(1) + (3) + (5)] 6 −3 0 ≈ (06 + 10 + 93)(2) = 398 fts
70. Use the midpoint of each of four 2-day intervals. Let = 0 correspond to July 18 and note that the inflow rate, (), is in ft3s.
Amount of water = 08 () ≈ [(1) + (3) + (5) + (7)] 8 −4 0 ≈ [6401 + 4249 + 3821 + 2628](2) = 34,198.
Now multiply by the number of seconds in a day, 24 · 602, to get 2,954,707,200 ft3.
71. Let () denote the bacteria population at time (in hours). By the Net Change Theorem,
(1) − (0) = 01 0() = 01(1000 · 2) = 1000 ln 2 2 1 0 = 1000 ln 2 (21 − 20) = 1000 ln 2 ≈ 1443
Thus, the population after one hour is 4000 + 1443 = 5443.
72. Let () denote the number of megabits transmitted at time (in hours) [note that () is measured in megabits/second]. By
the Net Change Theorem and the Midpoint Rule,
(8) − (0) = 08 3600() ≈ 3600 · 8−4 0[(1) + (3) + (5) + (7)]
≈ 7200(032 + 050 + 056 + 083) = 7200(221) = 15,912 megabits
73. Power is the rate of change of energy with respect to time; that is, () = 0(). By the Net Change Theorem and the
Midpoint Rule,
(24) − (0) = 024 () ≈ 2412 − 0[(1) + (3) + (5) + · · · + (21) + (23)]
≈ 2(16,900 + 16,400 + 17,000 + 19,800 + 20,700 + 21,200
+ 20,500 + 20,500 + 21,700 + 22,300 + 21,700 + 18,900)
= 2(237,600) = 475,200
Thus, the energy used on that day was approximately 4.75 × 105 megawatt-hours.
74. (a) From Exercise 4.1.74(a), () = 0001463 − 0115532 + 2498169 − 2126872.
(b) (125) − (0) = 0125 () = 00003654 − 0038513 + 124908452 − 2126872125 0 ≈ 206,407 ft
5.5 The Substitution Rule
1. Let = 2. Then = 2 and = 1
2 , so cos 2 = cos 1 2 = 1 2 sin + = 1 2 sin 2 + .
2. Let = −2. Then = −2 and = − 1
2 , so −2 = − 1 2 = − 1 2 + = − 1 2 −2 + .
3. Let = 3 + 1. Then = 32 and 2 = 1
3 , so
23 + 1 = √ 1 3 = 13 3322 + = 13 · 2332 + = 2 9(3 + 1)32 + .
4. Let = sin. Then = cos , so sin2 cos = 2 = 1 3 3 + = 1 3 sin3 + .
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© Cengage Learning. All Rights Reserved.SECTION 5.5 THE SUBSTITUTION RULE ¤ 43
5. Let = 4 − 5. Then = 43 and 3 = 1
4 , so
4−3 5 = 114 = 1 4 ln|| + = 1 4 ln 4 − 5 + .
6. Let = 2 + 1. Then = 2 and = 1
2 , so √2 + 1 = √ 1 2 = 1 2 · 2 332 + = 1 3(2 + 1)32 + .
7. Let = 1 − 2. Then = −2 and = − 1
2 , so
√1 − 2 = √ − 1 2 = − 1 2 · 2 332 + = − 1 3(1 − 2)32 + .
8. Let = 3. Then = 32 and 2 = 1
3 , so 23 = 1 3 = 1 3 + = 1 33 + .
9. Let = 1 − 2. Then = −2 and = − 1
2 , so
(1 − 2)9 = 9 − 1 2 = − 1 2 · 10 1 10 + = − 20 1 (1 − 2)10 + .
10. Let = 1 + cos. Then = −sin and sin = −, so
sin √1 + cos = √(−) = − 2 332 + = − 2 3(1 + cos)32 + .
11. Let =
2 . Then = 2 and = 2 , so cos 2 = cos 2 = 2 sin + = 2 sin 2 + .
12. Let = 2. Then = 2 and = 1
2 , so sec2 2 = sec2 1 2 = 1 2 tan + = 1 2 tan 2 + .
13. Let = 5 − 3. Then = −3 and = − 1
3 , so
5 −3 = 1 − 1 3 = − 1 3 ln|| + = − 1 3 ln|5 − 3| + .
14. Let = 4 − 3. Then = −32 and 2 = − 1 3 , so
2(4 − 3)23 = 23 − 1 3 = − 1 3 · 3 553 + = − 1 5(4 − 3)53 + .
15. Let = cos. Then = −sin and sin = −, so
cos3 sin = 3(−) = − 1 44 + = − 1 4 cos4 + .
16. Let = −5. Then = −5 and = − 1
5 , so −5 = − 1 5 = − 1 5 + = − 1 5−5 + .
17. Let = 1 − . Then = − and = −, so
(1 −)2 = 12 (−) = − −2 = −(−−1) + = 1 + = 1 −1 + .
18. Let = √. Then = 1
2√ and 2 = √1 , so
sin√√ = sin(2) = −2cos + = −2cos √ + .
19. Let = 3 + 3. Then = (3 + 32) = 3( + 2), so
√3 ++ 2 3 = 1 3 1 2 = 13 −12 = 1 3 · 212 + = 2 33 + 3 + .
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20. Let = 3 + 1. Then = 32 and 1
3 = 2 + , so
3+ 1 2 = 1 1 3 = 1 3 ln|| + = 1 3 ln|3 + 1| + .
21. Let = ln. Then =
, so (ln)2 = 2 = 1 33 + = 1 3(ln)3 + .
22. Let = cos. Then = −sin and − = sin , so
sin sin(cos) = sin(−) = (−cos )(−1) + = cos(cos) + .
23. Let = tan. Then = sec2 , so sec2 tan3 = 3 = 1 44 + = 1 4 tan4 + .
24. Let = + 2. Then = and = − 2, so
√ + 2 = ( − 2)√ = (32 − 212) = 2 552 − 2 · 2 332 + = 2 5( + 2)52 − 4 3( + 2)32 + .
25. Let = 1 + . Then = , so √1 + = √ = 2 332 + = 2 3(1 + )32 + .
Or: Let = √1 + . Then 2 = 1 + and 2 = , so
√1 + = · 2 = 2 33 + = 2 3(1 + )32 + .
26. Let = + . Then = and = (1), so
+ = (1) = 1 1 = 1 ln|| + = 1 ln| + | + .
27. Let = 3 + 3. Then = (32 + 3) and 1 3 = (2 + 1), so
(2 + 1)(3 + 3)4 = 4 ( 1 3 ) = 1 3 · 1 55 + = 15 1 (3 + 3)5 + .
28. Let = cos. Then = −sin and sin = −, so cos sin = (−) = − + = −cos + .
29. Let = 5. Then = 5 ln5 and 5 = 1
ln5 , so
5 sin(5) = sin ln5 1 = −ln5 1 cos + = −ln5 1 cos(5) + .
30. Let = tan. Then = sec2 , so
tan sec22 = 12 = −2 = −1−1 + = −tan 1 + = −cot + .
Or: tan sec22 = cos12 · cos sin22 = csc2 = −cot +
31. Let = arctan. Then = 1
2 + 1 , so (arctan 2 + 1)2 = 2 = 1 33 + = 1 3(arctan)3 + .
32. Let = 2 + 4. Then = 2 and = 1
2 , so
2+ 4 = 1 12 = 1 2 ln|| + = 1 2 ln 2 + 4 + = 12 ln(2 + 4) + [since 2 + 4 0].
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33. Let = 1 + 5. Then = 5 and = 1
5 , so
cos(1 + 5) = cos 1 5 = 1 5 sin + = 1 5 sin(1 + 5) + .
34. Let =
. Then = −
2 and 12 = −1 , so
cos( 2 ) = cos−1 = −1 sin + = −1 sin +
35. Let = cot. Then = −csc2 and csc2 = −, so
√cot csc2 = √(−) = −3322 + = − 2 3(cot)32 + .
36. Let = 2 + 3. Then = 2 ln 2 and 2 = 1
ln 2 , so
22+ 3 = 1ln 2 1 = ln 2 1 ln|| + = ln 2 1 ln(2 + 3) + .
37. Let = sinh. Then = cosh , so sinh2 cosh = 2 = 1 33 + = 1 3 sinh3 + .
38. Let = 1 + tan. Then = sec2 , so
cos2 √ 1 + tan = √sec 1 + tan 2 = √ = −12 = 1122 + = 2√1 + tan + .
39. 1 + cos sin 22 = 2 1 + cos sincos 2 = 2. Let = cos. Then = −sin , so
2 = −2 1 + 2 = −2 · 1 2 ln(1 + 2) + = −ln(1 + 2) + = −ln(1 + cos2 ) + .
Or: Let = 1 + cos2 .
40. Let = cos. Then = −sin and sin = −, so
1 + cos sin2 = 1 + − 2 = −tan−1 + = −tan−1(cos) + .
41. cot = cos sin . Let = sin. Then = cos , so cot = 1 = ln|| + = ln|sin| + .
42. Let = ln. Then = 1
, so cos(ln ) = cos = sin + = sin(ln) + .
43. Let = sin−1 . Then = √1 1− 2 , so √1 − 2 sin−1 = 1 = ln|| + = ln sin−1 + .
44. Let = 2. Then = 2 , so 1 +4 = 1 + 1 2 2 = 1 2 tan−1 + = 1 2 tan−1(2) + .
45. Let = 1 + 2. Then = 2 , so
1 + 1 +2 = 1 +12 + 1 +2 = tan−1 + 1 2 = tan−1 + 1 2 ln|| +
= tan−1 + 1
2 ln 1 + 2 + = tan−1 + 1 2 ln1 + 2 + [since 1 + 2 0].
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46. Let = 2 + . Then = , = − 2, and 2 = ( − 2)2, so
2√2 + = ( − 2)2√ = (2 − 4 + 4)12 = (52 − 432 + 412)
=
2 7
72 − 8
552 + 8 332 + = 2 7(2 + )72 − 8 5(2 + )52 + 8 3(2 + )32 +
47. Let = 2 + 5. Then = 2 and = 1
2( − 5), so
(2 + 5)8 = 1 2( − 5)8 ( 1 2 ) = 1 4 (9 − 58)
=
1 4
( 10 1 10 − 5 99) + = 40 1 (2 + 5)10 − 36 5 (2 + 5)9 +
48. Let = 2 + 1 [so 2 = − 1]. Then = 2 and = 1 2 , so
3 √2 + 1 = 2 √2 + 1 = ( − 1)√ 1 2 = 1 2 (32 − 12)
=
1 2
2 552 − 2 332 + = 1 5(2 + 1)52 − 1 3(2 + 1)32 + .
Or: Let = √2 + 1. Then 2 = 2 + 1 ⇒ 2 = 2 ⇒ = , so
3 √2 + 1 = 2 √2 + 1 = (2 − 1) · = (4 − 2)
=
1 5
5 − 1
33 + = 1 5(2 + 1)52 − 1 3(2 + 1)32 + .
Note: This answer can be written as 1
15 √2 + 1(34 + 2 − 2) + .
49. () = (2 − 1)3. = 2 − 1 ⇒ = 2 so
(2 − 1)3 = 3 1 2 = 1 84 + = 1 8(2 − 1)4 +
Where is positive (negative), is increasing (decreasing). Where
changes from negative to positive (positive to negative), has a local
minimum (maximum).
50. () = tan2 sec2 . = tan ⇒ = sec2 , so
tan2 sec2 = 2 = 1 33 + = 1 3 tan3 +
Note that is positive and is increasing. At = 0, = 0 and has a
horizontal tangent.
51. () = cos sin. = cos ⇒ = −sin , so
cos sin = (−) = − + = −cos +
Note that at = , changes from positive to negative and has a local
maximum. Also, both and are periodic with period 2, so at = 0 and
at = 2, changes from negative to positive and has a local minimum.
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52. () = sincos4. = cos ⇒ = −sin , so
sincos4 = 4 (−) = − 1 5 5 + = − 1 5 cos5 +
Note that at = , changes from positive to negative and has a local
maximum. Also, both and are periodic with period 2, so at = 0 and
at = 2, changes from negative to positive and has a local minimum.
53. Let =
2 , so = 2 . When = 0, = 0; when = 1, = 2 . Thus,
01 cos(2) = 02 cos 2 = 2 [sin] 0 2 = 2 sin 2 − sin 0 = 2 (1 − 0) = 2
54. Let = 3 − 1, so = 3. When = 0, = −1; when = 1, = 2. Thus,
01(3 − 1)50 = −21 50 1 3 = 1 3 51 1 512 −1 = 153 1 251 − (−1)51 = 153 1 (251 + 1)
55. Let = 1 + 7, so = 7. When = 0, = 1; when = 1, = 8. Thus,
01 √3 1 + 7 = 18 13( 1 7 ) = 1 7 3 4 438 1 = 28 3 (843 − 143) = 28 3 (16 − 1) = 45 28
56. Let = 5 + 1, so = 5. When = 0, = 1; when = 3, = 16. Thus,
03 5 + 1 = 116 115 = 1 5 ln|| 16 1 = 15(ln 16 − ln 1) = 1 5 ln 16.
57. Let = cos, so = −sin . When = 0, = 1; when = 6 , = √32. Thus,
06 cos sin2 = 1√32 12 (−) = 1 √ 1 32 = √23 − 1.
58. Let = 1
2 , so = 1 2 . When = 3 , = 6 ; when = 23 , = 3 . Thus,
2 3 3 csc2 1 2 = 63 csc2 (2) = 2−cot 3 6 = −2cot 3 − cot 6
= −2√13 − √3 = −2 1 3√3 − √3 = 4 3√3
59. Let = 1, so = −12 . When = 1, = 1; when = 2, = 1 2 . Thus,
12 1 2 = 112 (−) = − 1 12 = −(12 − ) = − √.
60. Let = −2, so = −2 . When = 0, = 0; when = 1, = −1. Thus,
01 −2 = 0−1 − 1 2 = − 1 2 − 0 1 = − 1 2 −1 − 0 = 1 2(1 − 1).
61. − 44(3 + 4 tan) = 0 by Theorem 7(b), since () = 3 + 4 tan is an odd function.
62. Let = sin, so = cos . When = 0, = 0; when = 2 , = 1. Thus,
02 cos sin(sin) = 01 sin = −cos1 0 = −(cos 1 − 1) = 1 − cos 1.
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© Cengage Learning. All Rights Reserved.48 ¤ CHAPTER 5 INTEGRALS
63. Let = 1 + 2, so = 2. When = 0, = 1; when = 13, = 27. Thus,
013 3 (1 + 2 )2 = 127 −23 1 2 = 1 2 · 31327 1 = 3 2(3 − 1) = 3.
64. Assume 0. Let = 2 − 2, so = −2 . When = 0, = 2; when = , = 0. Thus,
0 √2 − 2 = 02 12− 1 2 = 1 2 02 12 = 1 2 · 2 332 02 = 1 33.
65. Let = 2 + 2, so = 2 and = 1 2 . When = 0, = 2; when = , = 22. Thus,
0 2 + 2 = 222 12 1 2 = 1 2 2 3322 22 = 1 3322 22 = 1 3(22)32 − (2)32 = 1 32√2 − 13
66. − 33 4 sin = 0 by Theorem 7(b), since () = 4 sin is an odd function.
67. Let = − 1, so + 1 = and = . When = 1, = 0; when = 2, = 1. Thus,
12 √ − 1 = 01( + 1)√ = 01(32 + 12) = 2 552 + 2 3321 0 = 2 5 + 2 3 = 16 15.
68. Let = 1 + 2, so = 1 2( − 1) and = 2. When = 0, = 1; when = 4, = 9. Thus,
04 √1 + 2 = 19 1 2(√− 1) 2 = 1 4 19(12 − −12) = 1 4 2 332 − 2129 1 = 1 4 · 2 332 − 3129 1
=
1 6
[(27 − 9) − (1 − 3)] = 20 6 = 10 3
69. Let = ln, so =
. When = , = 1; when = 4; = 4. Thus,
4 √ln = 14 −12 = 2124 1 = 2(2 − 1) = 2.
70. Let = ( − 1)2, so = 2( − 1). When = 0, = 1; when = 2, = 1. Thus,
02( − 1)(−1)2 = 11 1 2 = 0 since the limits are equal.
71. Let = + , so = ( + 1) . When = 0, = 1; when = 1, = + 1. Thus,
01 + 1 + = 1+1 1 = ln|| 1+1 = ln| + 1| − ln|1| = ln( + 1).
72. Let = 2
− , so = 2 . When = 0 = −; when = 2 , = − . Thus,
02 sin2 − = −− sin 2 = 2 −cos −− = −2[cos( − ) − cos(−)]
= −
2
(−cos − cos) = −
2
(−2cos) =
cos
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73. Let = 1 + √, so = 1
2√ ⇒ 2√ = ⇒ 2( − 1) = . When = 0, = 1; when = 1,
= 2. Thus,
01 (1 + √)4 = 12 14 · [2( − 1)] = 212 13 − 14 = 2−212 + 313 2 1
= 2− 1 8 + 24 1 − − 1 2 + 1 3 = 2 12 1 = 1 6
74. If () = sin √3 , then (−) = sin √3 − = sin(−√3 ) = −sin √3 = −(), so is an odd function. Now
= −32 sin √3 = −22 sin √3 + 23 sin √3 = 1 + 2. 1 = 0 by Theorem 7(b). To estimate 2, note that
2 ≤ ≤ 3 ⇒ √3 2 ≤ √3 ≤ √3 3 [≈ 144] ⇒ 0 ≤ √3 ≤ 2 [≈ 157] ⇒ sin 0 ≤ sin √3 ≤ sin 2 [since sine is
increasing on this interval] ⇒ 0 ≤ sin √3 ≤ 1. By comparison property 8, 0(3 − 2) ≤ 2 ≤ 1(3 − 2) ⇒
0 ≤ 2 ≤ 1 ⇒ 0 ≤ ≤ 1.
75. From the graph, it appears that the area under the curve is about
1 + a little more than 1 2 · 1 · 07, or about 14. The exact area is given by
= 01 √2 + 1. Let = 2 + 1, so = 2. The limits change to
2 · 0 + 1 = 1 and 2 · 1 + 1 = 3, and
= 13 √ 1 2 = 1 2 2 3323 1 = 1 33√3 − 1 = √3 − 1 3 ≈ 1399.
76. From the graph, it appears that the area under the curve is almost 1 2 · · 26,
or about 4. The exact area is given by
= 0(2 sin − sin 2) = −2cos 0 − 0 sin 2
= −2(−1 − 1) − 0 = 4
Note: 0 sin2 = 0 since it is clear from the graph of = sin2 that 2 sin2 = − 02 sin2 .
77. First write the integral as a sum of two integrals:
= −22( + 3)√4 − 2 = 1 + 2 = −22 √4 − 2 + −22 3√4 − 2 . 1 = 0 by Theorem 7(b), since
() = √4 − 2 is an odd function and we are integrating from = −2 to = 2. We interpret 2 as three times the area of
a semicircle with radius 2, so = 0 + 3 · 1 2 · 22 = 6.
78. Let = 2. Then = 2 and the limits are unchanged (02 = 0 and 12 = 1), so
= 01 √1 − 4 = 1 2 01 √1 − 2 . But this integral can be interpreted as the area of a quarter-circle with radius 1.
So = 1
2 ·
1 4
· 12 = 1 8.
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79. First Figure Let = √, so = 2 and = 2 . When = 0, = 0; when = 1, = 1. Thus,
1 = 01 √ = 01 (2 ) = 201 .
Second Figure 2 = 01 2 = 201 .
Third Figure Let = sin, so = cos . When = 0, = 0; when = 2 , = 1. Thus,
3 = 02 sin sin 2 = 02 sin (2 sin cos) = 01 (2 ) = 201 .
Since 1 = 2 = 3, all three areas are equal.
80. Let =
12
. Then =
12
and
024() = 02485 − 018 cos12 = 02(85 − 018 cos)12 = 12 85 − 018 sin2 0
=
12
[(85 · 2 − 0) − (0 − 0)] = 2040 kcal
81. The rate is measured in liters per minute. Integrating from = 0 minutes to = 60 minutes will give us the total amount of oil
that leaks out (in liters) during the first hour.
060 () = 060 100−001 [ = −001, = −001]
= 1000−06(−100) = −10,000− 0 06 = −10,000(−06 − 1) ≈ 45119 ≈ 4512 liters
82. Let () = with = 450268 and = 112567, and () = population after hours. Since () = 0(),
03 () = (3) − (0) is the total change in the population after three hours. Since we start with 400 bacteria, the
population will be
(3) = 400 + 03 () = 400 + 03 = 400 + 3 0 = 400 + 3 − 1
≈ 400 + 11,313 = 11,713 bacteria
83. The volume of inhaled air in the lungs at time is
() = 0 () = 0 1 2 sin25 = 025 1 2 sin 25 substitute = 25 , = 25
=
5
4 −cos2 05 = 45 −cos25 + 1 = 45 1 − cos25 liters
84. The rate is measured in kilograms per year. Integrating from = 0 years (2000) to = 20 years (2020) will give us the net
change in biomass from 2000 to 2020.
020 (1 + 5 60,000−−0066)2 = 61+5−12 60,000 2 − 1 3 = 1 + 5 = −3−0−606
= 20,0001+5 6 −12 = 1 + 5 20,000 −12 − 20,6000 ≈ 16,666
Thus, the predicted biomass for the year 2020 is approximately 25,000 + 16,666 = 41,666 kg.
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85. 030 () = 030 0− = 0 1−30 (−) == −− −
= 0− 1−30 = 0(−−30 + 1)
The integral 030 () represents the total amount of urea removed from the blood in the first 30 minutes of dialysis.
86. Number of calculators = (4) − (2) = 24 50001 − 100( + 10)−2
= 5000 + 100( + 10)−14 2 = 50004 + 100 14 − 2 + 100 12 ≈ 4048
87. Let = 2. Then = 2, so 02 (2) = 04 () 1 2 = 1 2 04 () = 1 2(10) = 5.
88. Let = 2. Then = 2 , so 03 (2) = 09 () 1 2 = 1 2 09 () = 1 2(4) = 2.
89. Let = −. Then = −, so
(−) = −− ()(−) = −− () = −− ()
From the diagram, we see that the equality follows from the fact that we are
reflecting the graph of , and the limits of integration, about the -axis.
90. Let = + . Then = , so
( + ) = ++ () = ++ ()
From the diagram, we see that the equality follows from the fact that we are
translating the graph of , and the limits of integration, by a distance .
91. Let = 1 − . Then = 1 − and = −, so
01 (1 − ) = 10 (1 − ) (−) = 01 (1 − ) = 01 (1 − ) .
92. Let = − . Then = −. When = , = 0 and when = 0, = . So
0 (sin) = − 0( − )(sin( − )) = 0( − )(sin)
= 0 (sin) − 0 (sin) = 0 (sin) − 0 (sin) ⇒
20 (sin) = 0 (sin) ⇒ 0 (sin) = 2 0 (sin).
93. sin
1 + cos2 = ·
sin
2 − sin2 = (sin), where () = 2 − 2 . By Exercise 92,
0 1 + cos sin2 = 0 (sin) = 2 0 (sin) = 2 0 1 + cos sin2
Let = cos. Then = −sin . When = , = −1 and when = 0, = 1. So
2
0 1 + cos sin2 = −2 1−1 1 +2 = 2 −11 1 +2 = 2 tan−1 1 −1
=
2
[tan−1 1 − tan−1(−1)] =
2 4 − −4 = 42
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© Cengage Learning. All Rights Reserved.52 ¤ CHAPTER 5 INTEGRALS
94. (a) 02 (cos) = 02 sin 2 − = 2 − , = −
= 0 2 (sin)(−) = 02 (sin) = 02 (sin)
Continuity of is needed in order to apply the substitution rule for definite integrals.
(b) In part (a), take () = 2, so 02 cos2 = 02 sin2 . Now
02 cos2 + 02 sin2 = 02(cos2 + sin2 ) = 02 1 = 0 2 = 2 ,
so 202 cos2 = 2 ⇒ 02 cos2 = 4 = 02 sin2 .
5 Review
1. True by Property 2 of the Integral in Section 5.2.
2. False. Try = 0, = 2, () = () = 1 as a counterexample.
3. True by Property 3 of the Integral in Section 5.2.
4. False. You can’t take a variable outside the integral sign. For example, using () = 1 on [01],
01 () = 01 = 1 2 21 0 = 1 2 (a constant) while 01 1 = []1 0 = · 1 = (a variable).
5. False. For example, let () = 2. Then 01 √2 = 01 = 1 2, but 01 2 = 1 3 = √13.
6. True by the Net Change Theorem.
7. True by Comparison Property 7 of the Integral in Section 5.2.
8. False. For example, let = 0, = 1, () = 3, () = . () () for each in (0 1), but 0() = 0 1 = 0()
for ∈ (01).
9. True. The integrand is an odd function that is continuous on [−1 1].
10. True. −552 + + = −552 + + −55
= 2052 + + 0 [because 2 + is even and is odd]
11. False. For example, the function = || is continuous on R, but has no derivative at = 0.
12. True by FTC1.
13. True by Property 5 of Integrals.
14. False. For example, 01 − 1 2 = 1 2 2 − 1 2 1 0 = 1 2 − 1 2 − (0 − 0) = 0, but () = − 1 2 6= 0.
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© Cengage Learning. All Rights Reserved.CHAPTER 5 REVIEW ¤ 53
15. False. () is a constant, so () = 0, not () [unless () = 0]. Compare the given statement
carefully with FTC1, in which the upper limit in the integral is .
16. False. See the paragraph before Note 4 and Figure 4 in Section 5.2, and notice that = − 3 0 for 1 ≤ 2.
17. False. The function () = 14 is not bounded on the interval [−21]. It has an infinite discontinuity at = 0, so it is
not integrable on the interval. (If the integral were to exist, a positive value would be expected, by Comparison
Property 6 of Integrals.)
18. False. For example, if () = 10 if if 0−≤1 ≤ ≤ 1 0 then has a jump discontinuity at 0, but −11 () exists and is
equal to 1.
1. (a) 6 =
6
=1
(−1)∆ [∆ = 6 −6 0 = 1]
= (0) · 1 + (1) · 1 + (2) · 1 + (3) · 1 + (4) · 1 + (5) · 1
≈ 2 + 35 + 4 + 2 + (−1) + (−25) = 8
The Riemann sum represents the sum of the areas of the four rectangles
above the -axis minus the sum of the areas of the two rectangles below the
-axis.
(b) 6 =
6
=1
()∆ [∆ = 6 −6 0 = 1]
= (1) · 1 + (2) · 1 + (3) · 1 + (4) · 1 + (5) · 1 + (6) · 1
= (05) + (15) + (25) + (35) + (45) + (55)
≈ 3 + 39 + 34 + 03 + (−2) + (−29) = 57
The Riemann sum represents the sum of the areas of the four rectangles
above the -axis minus the sum of the areas of the two rectangles below the
-axis.
2. (a) () = 2 − and ∆ = 2 −4 0 = 05 ⇒
4 = 05(05) + 05(1) + 05(15) + 05(2)
= 05(−025 + 0 + 075 + 2) = 125
The Riemann sum represents the sum of the areas of the two rectangles
above the x-axis minus the area of the rectangle below the x-axis. (The
second rectangle vanishes.)
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(b) 022 − = lim
→∞
=1
() ∆ [∆ = 2 and = 2]
= lim
→∞
=1422 − 22 = lim →∞ 2 42 =1 2 − 2 =1
= lim
→∞83 · ( + 1)(2 6 + 1) − 42 · (2+ 1) = lim →∞4 3 · + 1 · 2+ 1 − 2 · + 1
= lim
→∞4 31 + 1 2 + 1 − 21 + 1 = 4 3 · 1 · 2 − 2 · 1 = 2 3
(c) 022 − = 1 33 − 1 222 0 = 8 3 − 2 = 2 3
(d) 022 − = 1 − 2, where 1 and 2 are the areas shown in the
diagram.
3. 01 + √1 − 2 = 01 + 01 √1 − 2 = 1 + 2.
1 can be interpreted as the area of the triangle shown in the figure
and 2 can be interpreted as the area of the quarter-circle.
Area = 1
2(1)(1) + 1 4()(1)2 = 1 2 + 4 .
4. On [0 ], lim
→∞
=1
sin ∆ = 0 sin = [−cos] 0 = −(−1) − (−1) = 2.
5. 06 () = 04 () + 46 () ⇒ 10 = 7 + 46 () ⇒ 46 () = 10 − 7 = 3
6. (a) 15( + 25) = lim
→∞
=1
()∆ ∆ = 5 − 1 = 4 , = 1 + 4
= lim
→∞
=1 1 + 4 + 21 + 45 · 4 CAS = lim →∞ 13054 + 312633+ 20802 − 256 · 4
= 5220
(b) 15( + 25) = 1 22 + 2 665 1 = 25 2 + 15,3625 − 1 2 + 1 3 = 12 + 5208 = 5220
7. First note that either or must be the graph of 0 (), since 00 () = 0, and (0) 6= 0. Now notice that 0 when
is increasing, and that 0 when is increasing. It follows that is the graph of (), is the graph of 0(), and is the
graph of 0 ().
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8. (a) By the Net Change Theorem (FTC2), 01 arctan = arctan 1 0 = 4 − 1
(b)
01 arctan = 0 since this is the derivative of a constant.
(c) By FTC1,
0 arctan = arctan .
9. (4) = 04 () = 01 () + 12 () + 23 () + 34 ()
= −
1 2
· 1 · 2 area of triangle, below -axis + 1 2 · 1 · 2 + 1 · 2 + 1 2 · 1 · 2 = 3
By FTC1, 0() = (), so 0(4) = (4) = 0.
10. () = 0 () ⇒ 0() = () [by FTC1] ⇒ 00() = 0(), so 00(4) = 0(4) = −2, which is the slope of
the line segment at = 4.
11. 1283 + 32 = 8 · 1 44 + 3 · 1 332 1 = 24 + 32 1 = 2 · 24 + 23 − (2 + 1) = 40 − 3 = 37
12. 0 4 − 8 + 7 = 1 55 − 42 + 7 0 = 1 5 5 − 4 2 + 7 − 0 = 1 5 5 − 4 2 + 7
13. 011 − 9 = − 10 1 101 0 = 1 − 10 1 − 0 = 10 9
14. Let = 1 − , so = − and = −. When = 0, = 1; when = 1, = 0. Thus,
01(1 − )9 = 10 9(−) = 01 9 = 10 1 101 0 = 10 1 (1 − 0) = 10 1 .
15. 19 √ − 22 = 19(−12 − 2) = 212 − 29 1 = (6 − 81) − (2 − 1) = −76
16. 01 (√4 + 1)2 = 01(12 + 214 + 1) = 2 332 + 8 554 + 1 0 = 2 3 + 8 5 + 1 − 0 = 49 15
17. Let = 2 + 1, so = 2 and = 1 2 . When = 0, = 1; when = 1, = 2. Thus,
01 (2 + 1)5 = 12 5 1 2 = 1 2 1 662 1 = 12 1 (64 − 1) = 63 12 = 21 4 .
18. Let = 1 + 3, so = 32 and 2 = 1 3 . When = 0, = 1; when = 2, = 9. Thus,
02 21 + 3 = 19 12 1 3 = 1 3 2 3329 1 = 2 9(27 − 1) = 52 9 .
19. 15 ( −4)2 does not exist because the function () = ( −14)2 has an infinite discontinuity at = 4;
that is, is discontinuous on the interval [15].
20. Let = 3, so = 3 . When = 0, = 1; when = 1, = 3. Thus,
01 sin(3) = 03 sin31 = 31 −cos3 0 = −31(−1 − 1) = 32 .
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21. Let = 3, so = 32 . When = 0, = 0; when = 1, = 1. Thus,
01 2 cos(3) = 01 cos 1 3 = 1 3 sin1 0 = 1 3(sin 1 − 0) = 1 3 sin 1.
22. −11 1 + sin2 = 0 by Theorem 5.5.7(b), since () = 1 + sin2 is an odd function.
23. − 44 2 + cos 4 tan = 0 by Theorem 5.5.7(b), since () = 2 + cos 4 tan is an odd function.
24. Let = , so = When = 0 = 1; when = 1 = . Thus,
01 1 +2 = 1 1 +12 = arctan 1 = arctan − arctan 1 = arctan − 4 .
25. 1 − 2 = 1 − 12 = 12 − 2 + 1 = −1 − 2ln|| + +
26. 110 2− 4 does not exist because the function () = 2− 4 has an infinite discontinuity at = 2;
that is, is discontinuous on the interval [110].
27. Let = 2 + 4. Then = (2 + 4) = 2( + 2), so
√2+ 2 + 4 = −12 1 2 = 1 2 · 212 + = √ + = 2 + 4 + .
28. Let = 1 + cot. Then = −csc2 , so 1 + cot csc2 = 1 (−) = −ln|| + = −ln|1 + cot| + .
29. Let = sin. Then = cos , so sin cos = 1 = 1 · 1 22 + = 21(sin)2 + .
30. Let = cos. Then = −sin , so sin cos(cos) = − cos = −sin + = −sin(cos) + .
31. Let = √. Then =
2√, so √√ = 2 = 2 + = 2√ + .
32. Let = ln. Then = 1
, so sin(ln ) = sin = −cos + = −cos(ln) + .
33. Let = ln(cos). Then = −sin
cos
= −tan , so
tan ln(cos) = − = − 1 22 + = − 1 2[ln(cos )]2 + .
34. Let = 2. Then = 2 , so √1− 4 = 12 √1 − 2 = 1 2 sin−1 + = 1 2 sin−12 + .
35. Let = 1 + 4. Then = 43 , so 1 +34 = 14 1 = 1 4 ln|| + = 1 4 ln1 + 4 + .
36. Let = 1 + 4. Then = 4, so sinh(1 + 4) = 1 4 sinh = 1 4 cosh + = 1 4 cosh(1 + 4) + .
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37. Let = 1 + sec. Then = sec tan , so
sec 1 + sec tan = 1 + sec 1 (sec tan ) = 1 = ln|| + = ln|1 + sec| + .
38. Let = 1 + tan, so = sec2 . When = 0, = 1; when = 4 , = 2. Thus,
04(1 + tan)3 sec2 = 12 3 = 1 442 1 = 1 424 − 14 = 1 4(16 − 1) = 15 4 .
39. Since 2 − 4 0 for 0 ≤ 2 and 2 − 4 0 for 2 ≤ 3, we have 2 − 4 = −(2 − 4) = 4 − 2 for 0 ≤ 2 and
2 − 4 = 2 − 4 for 2 ≤ 3. Thus,
03 2 − 4 = 02(4 − 2) + 23(2 − 4) = 4 − 33 2 0 + 33 − 43 2
= 8 − 8 3 − 0 + (9 − 12) − 8 3 − 8 = 16 3 − 3 + 16 3 = 32 3 − 9 3 = 23 3
40. Since √ − 1 0 for 0 ≤ 1 and √ − 1 0 for 1 ≤ 4, we have √ − 1 = − √ − 1 = 1 − √
for 0 ≤ 1 and
√ − 1 = √ − 1 for 1 ≤ 4. Thus,
04 √ − 1 = 01 1 − √ + 14 √ − 1 = − 2 3321 0 + 2 332 − 4 1
= 1 − 2 3 − 0 + 16 3 − 4 − 2 3 − 1 = 1 3 + 16 3 − 4 + 1 3 = 6 − 4 = 2
41. Let = 1 + sin. Then = cos , so
√cos 1 + sin = −12 = 212 + = 2√1 + sin + .
42. Let = 2 + 1. Then 2 = − 1 and = 1
2 , so
√23+ 1 = (√−1) 1 2 = 12 (12 − −12)
=
1 2
2 332 − 212 + = 1 3(2 + 1)32 − (2 + 1)12 +
=
1 3
(2 + 1)12 (2 + 1) − 3 + = 1 3√2 + 1 (2 − 2) +
43. From the graph, it appears that the area under the curve = √ between = 0
and = 4 is somewhat less than half the area of an 8 × 4 rectangle, so perhaps
about 13 or 14. To find the exact value, we evaluate
04 √ = 04 32 = 2 5524 0 = 2 5(4)52 = 64 5 = 128.
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44. From the graph, it seems as though 02 cos2 sin is equal to 0. To evaluate
the integral, let = cos ⇒ = −sin . Thus, = 11 2 (−) = 0.
45. () = 0 1 +23 ⇒ 0() = 0 1 +23 = 1 +23
46. () = 1√ + sin = − 1 √ + sin ⇒ 0() = − 1√ + sin = −√ + sin
47. Let = 4. Then
= 43. Also, = , so
0() =
04 cos(2) = 0 cos(2) · = cos(2) = 43 cos(8).
48. Let = sin Then
= cos Also, = so
0() =
1sin 1 1 +− 2 4 = 1 1 1 +− 2 4 · = 11 + − 2 4 · = 11 + sin − sin2 4 · cos = 1 + sin cos3 4
49. = √ = √1 + 1 = − 1√ + 1 ⇒
= −
1√ + 1 . Let = √. Then
1√ = 1 = 1 = · 2√1 = √√ · 2√1 = 2√ ,
so
= −
√
2
+
=
2 − √
2
.
50. = 23+1 sin4 = 20 sin4 + 03+1 sin4 = 03+1 sin4 − 02 sin4 ⇒
0 = sin(3 + 1)4 · (3 + 1) − sin(2)4 · (2) = 3 sin(3 + 1)4 − 2sin(2)4
51. If 1 ≤ ≤ 3, then √12 + 3 ≤ √2 + 3 ≤ √32 + 3 ⇒ 2 ≤ √2 + 3 ≤ 2√3, so
2(3 − 1) ≤ 13 √2 + 3 ≤ 2√3(3 − 1); that is, 4 ≤ 13 √2 + 3 ≤ 4√3.
52. If 3 ≤ ≤ 5, then 4 ≤ + 1 ≤ 6 and 1
6
≤ 1
+ 1
≤ 1
4, so
1 6
(5 − 3) ≤ 35 + 1 1 ≤ 14(5 − 3);
that is, 1
3
≤ 35 + 1 1 ≤ 12.
53. 0 ≤ ≤ 1 ⇒ 0 ≤ cos ≤ 1 ⇒ 2 cos ≤ 2 ⇒ 01 2 cos ≤ 01 2 = 1 3 31 0 = 1 3 [Property 7].
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54. On the interval 4 2 , is increasing and sin is decreasing, so sin is decreasing. Therefore, the largest value of sin on
4 2 is sin( 44) = √ 242 = 2√ 2. By Property 8 with = 2√ 2 we get 42 sin ≤ 2√ 2 2 − 4 = √22.
55. cos ≤ 1 ⇒ cos ≤ ⇒ 01 cos ≤ 01 = []1 0 = − 1
56. For 0 ≤ ≤ 1, 0 ≤ sin−1 ≤ 2 , so 01 sin−1 ≤ 01 2 = 4 21 0 = 4 .
57. ∆ = (3 − 0)6 = 1 2, so the endpoints are 0, 1 2, 1, 3 2, 2, 5 2, and 3, and the midpoints are 1 4, 3 4, 5 4, 7 4, 9 4, and 11 4 .
The Midpoint Rule gives
03 sin(3) ≈ 6
=1
()∆ = 1 2 sin 1 4 3 + sin 3 4 3 + sin 5 4 3 + sin 7 4 3 + sin 9 4 3 + sin 11 4 3 ≈ 0280981.
58. (a) Displacement = 052 − = 1 33 − 1 225 0 = 125 3 − 25 2 = 175 6 = 2916 meters
(b) Distance traveled = 05 2 − = 05 |( − 1)| = 01 − 2 + 152 −
= 1 22 − 1 331 0 + 1 33 − 1 225 1 = 1 2 − 1 3 − 0 + 125 3 − 25 2 − 1 3 − 1 2 = 177 6 = 295 meters
59. Note that () = 0(), where () = the number of barrels of oil consumed up to time . So, by the Net Change Theorem,
08 () = (8) − (0) represents the number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2008.
60. Distance covered = 050 () ≈ 5 = 505− 0[(05) + (15) + (25) + (35) + (45)]
= 1(467 + 886 + 1022 + 1067 + 1081) = 4523 m
61. We use the Midpoint Rule with = 6 and ∆ = 246− 0 = 4. The increase in the bee population was
024 () ≈ 6 = 4[(2) + (6) + (10) + (14) + (18) + (22)]
≈ 4[50 + 1000 + 7000 + 8550 + 1350 + 150] = 4(18,100) = 72,400
62. 1 = 1
2 = 1 2(2)(2) = 2, 2 = 1 2 = 1 2(1)(1) = 1 2, and since
= −√1 − 2 for 0 ≤ ≤ 1 represents a quarter-circle with radius 1,
3 = 1 42 = 1 4(1)2 = 4 . So
−13 () = 1 − 2 − 3 = 2 − 1 2 − 4 = 1 4(6 − )
63. Let = 2 sin. Then = 2 cos and when = 0, = 0; when = 2 , = 2. Thus,
02 (2 sin)cos = 02 () 1 2 = 1 2 02 () = 1 2 02 () = 1 2(6) = 3.
64. (a) is increasing on those intervals where 0 is positive. By the Fundamental Theorem of Calculus,
0() =
0 cos 2 2 = cos 2 2. This is positive when 2 2 is in the interval 2 − 1 2 2 + 1 2 ,
any integer. This implies that 2 − 1 2 2 2 2 + 1 2 ⇔ 0 ≤ || 1 or √4 − 1 || √4 + 1,
any positive integer. So is increasing on the intervals (−11), √3 √5, −√5 −√3, √73, −3 −√7, .
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(b) is concave upward on those intervals where 00 0. We differentiate 0 to find 00: 0() = cos 2 2 ⇒
00() = −sin 2 2 2 · 2 = −sin 2 2. For 0, this is positive where (2 − 1) 2 2 2, any
positive integer ⇔ 2(2 − 1) 2√, any positive integer. Since there is a factor of − in 00, the intervals
of upward concavity for 0 are −2(2 + 1) −2√ , any nonnegative integer. That is, is concave upward on
−√20, √22, −√6 −2, √62√2, .
(c) From the graphs, we can determine
that 0 cos 2 2 = 07 at
≈ 076 and ≈ 122.
(d) The graphs of () and () have similar shapes, except that ’s flattens out
near the origin, while ’s does not. Note that for 0, is increasing
where is concave up, and is decreasing where is concave down.
Similarly, is increasing where is concave down, and is decreasing
where is concave up. For 0, these relationships are reversed; that is,
is increasing where is concave down, and is increasing where is
concave up. See Example 5.3.3 and Exercise 5.3.65 for a discussion of ().
65. Area under the curve = sinh between = 0 and = 1 is equal to 1 ⇒
01 sinh = 1 ⇒ 1 cosh1 0 = 1 ⇒ 1(cosh − 1) = 1 ⇒
cosh − 1 = ⇒ cosh = + 1. From the graph, we get = 0 and
≈ 16161, but = 0 isn’t a solution for this problem since the curve
= sinh becomes = 0 and the area under it is 0. Thus, ≈ 16161.
66. Both numerator and denominator approach 0 as → 0, so we use l’Hospital’s Rule. (Note that we are differentiating with
respect to , since that is the quantity which is changing.) We also use FTC1:
lim
→0
( ) = lim
→0
0 −(−)2(4)
√4
H
= lim
→0
−(−)2(4)
√4 =
−2(4)
√4
67. Using FTC1, we differentiate both sides of the given equation, 1 () = ( − 1)2 + 1 −(), and get
() = 2 + 2( − 1)2 + −() ⇒ ()(1 − −) = 2 + 2( − 1)2 ⇒ () = 2(2 − 1)
1 − −
.
68. The second derivative is the derivative of the first derivative, so we’ll apply the Net Change Theorem with = 0
1200() = 12(0)0() = 0(2) − 0(1) = 5 − 2 = 3. The other information is unnecessary.
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69. Let = () and = 0(). So 2 () 0() = 2(()) = 2 ( ( )) = [()]2 − [()]2.
70. Let () = 2 1 + 3 . Then 0(2) = lim →0 (2 + ) − (2) = lim →0 1 22+ 1 + 3 , and 0() = √1 + 3, so
lim
→0
1
22+ 1 + 3 = 0(2) = 1 + 23 = √9 = 3.
71. Let = 1 − . Then = −, so 01 (1 − ) = 10 ()(−) = 01 () = 01 ().
72. lim
→∞
1
1 9 + 2 9 + 3 9 + · · · + 9 = lim →∞ 1 − 0 =1 9 = 01 9 = 10 10 1 0 = 10 1
The limit is based on Riemann sums using right endpoints and subintervals of equal length.
73. The shaded region has area 01 () = 1 3. The integral 01 −1()
gives the area of the unshaded region, which we know to be 1 − 1 3 = 2 3.
So 01 −1() = 2 3.
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1. Differentiating both sides of the equation sin = 02 () (using FTC1 and the Chain Rule for the right side) gives
sin + cos = 2(2). Letting = 2 so that (2) = (4), we obtain sin 2 + 2 cos 2 = 4(4), so
(4) = 1 4(0 + 2 · 1) = 2 .
2. The area under the curve = + 1 from = to = + 15 is given by () = +15 + 1
To find the minimum value of we’ll differentiate using FTC1 and set the derivative equal to 0.
0() =
+15 + 1
=
1 + 1 + 1+15 + 1
= −
1 + 1 + 1+15 + 1
= − + 1 + + 15 + + 1 1 5 = 15 + + 1 1 5 − 1
0() = 0 ⇔ 15 + 1
+ 15
−
1
= 0 ⇔ 15( + 15) + − ( + 15) = 0 ⇔
152 + 225 − 15 = 0 multiply by 4 3 ⇔ 22 + 3 − 2 = 0 ⇔ (2 − 1)( + 2) = 0 ⇔ = 1 2 or
= −2 Since 0, = 1 2 00() = − 1
( + 15)2 +
1 2
0, so
1 2 = 122 + 1 = 1 22 + ln|| 2 12 = (2 + ln 2) − 1 8 − ln 2 = 15 8 + 2 ln 2 is the minimum value of .
3. For = 04 (−2)4 , let = − 2 so that = + 2 and = . Then
= −22( + 2)4 = −22 4 + −22 24 = 0 [by 5.5.7(b)] +204 (−2)4 = 2.
4. (a) From the graph of () = 2 − 2
3 , it appears that the areas
are equal; that is, the area enclosed is independent of .
(b) We first find the x-intercepts of the curve, to determine the limits of integration: = 0 ⇔ 2 − 2 = 0 ⇔ = 0
or = 2. Now we integrate the function between these limits to find the enclosed area:
= 02 2−3 2 = 13 2 − 1 332 0 = 13 (2)2 − 1 3(2)3 = 13 43 − 8 33 = 4 3, a constant.
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(c) The vertices of the family of parabolas seem to determine a
branch of a hyperbola.
(d) For a particular , the vertex is the point where the maximum occurs. We have seen that the x-intercepts are 0 and 2, so by
symmetry, the maximum occurs at = , and its value is 2() − 2
3 =
1
. So we are interested in the curve consisting of
all points of the form 1 , 0. This is the part of the hyperbola = 1 lying in the first quadrant.
5. () = 0() √1 + 1 3 , where () = 0cos [1 + sin(2)]. Using FTC1 and the Chain Rule (twice) we have
0() = 1 + [ 1()]3 0() = 1 + [ 1()]3 [1 + sin(cos2 )](−sin). Now 2 = 00 [1 + sin(2)] = 0, so
0 2 = √1 + 0 1 (1 + sin 0)(−1) = 1 · 1 · (−1) = −1.
6. If () = 0 2 sin(2) = 2 0 sin(2), then 0() = 2 sin(2) + 2 0 sin(2), by the Product Rule and FTC1.
7. By l’Hospital’s Rule and the Fundamental Theorem, using the notation exp() = ,
lim
→0
0(1 − tan 2)1
H
= lim
→0
(1 − tan 2)1
1
= explim →0 ln(1 −tan 2)
H
= explim →0 −1 2sec − tan 2 2 2 = exp−12−· 102 = −2
8. The area () = 0 sin(2), and the area () = 1 2 sin(2). Since lim
→0+
() = 0 = lim
→0+
(), we can use
l’Hospital’s Rule:
lim
→0+
()
()
H
= lim
→0+
sin(2)
1 2
sin(2) + 1 2 [2cos(2)] [by FTC1 and the Product Rule]
H
= lim
→0+
2cos(2)
cos(2) − 23 sin(2) + 2cos(2) = lim →0+ 3cos(2cos( 2) − 222)sin(2) = 3 −2 0 = 23
9. () = 2 + − 2 = (− + 2)( + 1) = 0 ⇔ = 2 or = −1. () ≥ 0 for ∈ [−12] and () 0 everywhere
else. The integral (2 + − 2) has a maximum on the interval where the integrand is positive, which is [−12]. So
= −1, = 2. (Any larger interval gives a smaller integral since () 0 outside [−12]. Any smaller interval also gives a
smaller integral since () ≥ 0 in [−1 2].)
10. This sum can be interpreted as a Riemann sum, with the right endpoints of the subintervals as sample
points and with = 0, = 10,000, and () = √. So we approximate
10,000
=1
√ ≈ lim
→∞
10,000
=1
10,000 = 010000 √ = 2 3 3210 0 ,000 = 2 3(1,000,000) ≈ 666,667.
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Alternate method: We can use graphical methods as follows:
From the figure we see that −1 √ √ +1 √ , so
010,000 √
10000
=1
√ 110001 √ . Since
√ = 2 332 + , we get 010,000 √ = 666,6666 and
110001 √ = 2 3[(10,001)32 − 1] ≈ 666,766.
Hence, 666,6666
10000
=1
√ 666,766. We can estimate the sum by averaging these bounds:
10,000
=1
≈
666,6666 + 666,766
2
≈ 666,716. The actual value is about 666,71646.
11. (a) We can split the integral 0 [[]] into the sum
=1 −1 [[]] . But on each of the intervals [ − 1 ) of integration,
[[]] is a constant function, namely − 1. So the ith integral in the sum is equal to ( − 1)[ − ( − 1)] = ( − 1). So the
original integral is equal to
=1
( − 1) =
−1
=1
= ( − 1)
2
.
(b) We can write [[]] = 0 [[]] − 0 [[]] .
Now 0 [[]] = 0[[]] [[]] + [[]] [[]]. The first of these integrals is equal to 1 2([[]] − 1)[[]],
by part (a), and since [[]] = [[]] on [[[]] ], the second integral is just [[]]( − [[]]). So
0 [[]] = 1 2([[]] − 1)[[]] + [[]]( − [[]]) = 1 2 [[]](2 − [[]] − 1) and similarly 0 [[]] = 1 2 [[]] (2 − [[]] − 1).
Therefore, [[]] = 1 2 [[]] (2 − [[]] − 1) − 1 2 [[]] (2 − [[]] − 1).
12. By FTC1,
01sin 1 + 4 = 1sin 1 + 4 . Again using FTC1,
2
2 01sin 1 + 4 = 1sin 1 + 4 = 1 + sin4 cos.
13. Let () = 0 () = + 2 2 + 3 3 + 44 0 = + 2 2 + 3 3 + 44. Then (0) = 0, and (1) = 0 by the
given condition, +
2
+
3
+
4
= 0. Also, 0() = () = + + 2 + 3 by FTC1. By Rolle’s Theorem, applied to
on [01], there is a number in (01) such that 0() = 0, that is, such that () = 0. Thus, the equation () = 0 has a
root between 0 and 1.
More generally, if () = 0 + 1 + 22 + · · · + and if 0 + 1
2
+
2
3
+ · · · +
+ 1
= 0 then the equation
() = 0 has a root between 0 and 1 The proof is the same as before:
Let () = 0 () = 0 + 21 2 + 32 3 + · · · + + 1 . Then (0) = (1) = 0 and 0() = (). By
Rolle’s Theorem applied to on [0 1], there is a number in (01) such that 0() = 0, that is, such that () = 0.
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14. Let be the distance between the center of the disk and the surface of the liquid.
The wetted circular region has area 2 − 2 while the unexposed wetted region
(shaded in the diagram) has area 2 √2 − 2 , so the exposed wetted region
has area () = 2 − 2 − 2 √2 − 2 , 0 ≤ ≤ . By FTC1, we have
0() = −2 + 2√2 − 2.
Now 0() 0 ⇒ −2 + 2√2 − 2 0 ⇒ √2 − 2 ⇒ 2 − 2 22 ⇒
2 22 + 2 ⇒ 2 2(2 + 1) ⇒ 2 2
2 + 1 ⇒
√2 + 1, and we’ll call this value *.
Since 0() 0 for 0 * and 0() 0 for * , we have an absolute maximum when = *.
15. Note that
0 0 () = 0 () by FTC1, while
0 ()( − ) = 0 () − 0 ()
= 0 () + () − () = 0 ()
Hence, 0 ()( − ) = 0 0 () + . Setting = 0 gives = 0.
16. The parabola = 4 − 2 and the line = + 2 intersect when
4 − 2 = + 2 ⇔ 2 + − 2 = 0 ⇔ ( + 2)( − 1) = 0 ⇔
= −2 or 1. So the point A is (−20) and B is (13). The slope of the line
= + 2 is 1 and the slope of the parabola = 4 − 2 at -coordinate is
−2. These slopes are equal when = − 1 2, so the point C is − 1 2 15 4 .
The area 1 of the parabolic segment is the area under the parabola from
= −2 to = 1, minus the area under the line = + 2 from −2 to 1. Thus,
1 = −12(4 − 2) − −12( + 2) = 4 − 1 331 −2 − 1 22 + 21 −2
= 4 − 1 3 − −8 + 8 3 − 1 2 + 2 − (2 − 4) = 9 − 9 2 = 9 2
The area 2 of the inscribed triangle is the area under the line segment AC plus the area under the line segment CB minus
the area under the line segment AB. The line through A and C has slope 154 − 0
−12 + 2 =
5 2
and equation − 0 = 5 2( + 2), or
= 5 2 + 5. The line through C and B has slope 3 − 154
1 + 12 = −
1 2
and equation − 3 = − 1 2( − 1), or = − 1 2 + 7 2.
Thus,
2 = −−212 5 2 + 5 + −112 − 1 2 + 7 2 − −12( + 2) = 5 42 + 5− −1 22 + − 1 42 + 7 21 −12 − 9 2
= 16 5 − 5 2 − (5 − 10) + − 1 4 + 7 2 − − 16 1 − 7 4 − 9 2 = 45 16 + 81 16 − 72 16 = 54 16 = 27 8
Archimedes’ result states that 1 = 4
32 which is verified in this case since 4 3 · 27 8 = 9 2.
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17. Let be the nonzero -intercept so that the parabola has equation () = ( − ),
or = 2 − , where 0. The area under the parabola is
= 0 ( − ) = 0 (2 − ) = 1 3 3 − 1 2 2 0
= 1 3 3 − 1 2 3 = − 1 6 3
The point ( ) is on the parabola, so () = ⇒ = ( − ) ⇒
=
( − ). Substituting for in gives () = −6 · −3 ⇒
0() = −
6
·
( − )(32) − 3(−1)
( − )2 = −
6
·
2[3( − ) + ]
( − )2 = −
2(3 − 2)
6( − )2
Now 0 = 0 ⇒ = 3
2 . Since 0() 0 for 3 2 and 0() 0 for 3 2 , so has an absolute
minimum when = 3
2 . Substituting for in gives us =
− 3 2 = −
2
2 , so () = −22 − 3 2 , or
() = −2
2 2 + 3 . Note that the vertex of the parabola is 3 4 9 8 and the minimal area under the parabola
is 3 2 = 9 8 .
18. We restrict our attention to the triangle shown. A point in this triangle is
closer to the side shown than to any other side, so if we find the area of
the region consisting of all points in the triangle that are closer to the
center than to that side, we can multiply this area by 4 to find the total
area. We find the equation of the set of points which are equidistant
from the center and the side: the distance of the point ( ) from the
side is 1 − , and its distance from the center is 2 + 2.
So the distances are equal if 2 + 2 = 1 − ⇔ 2 + 2 = 1 − 2 + 2 ⇔ = 1 2(1 − 2). Note that the area
we are interested in is equal to the area of a triangle plus a crescent-shaped area. To find these areas, we have to find the
-coordinate of the horizontal line separating them. From the diagram, 1 − = √2 ⇔ = 1
1 + √2 = √2 − 1.
We calculate the areas in terms of , and substitute afterward.
The area of the triangle is 1 2(2)() = 2, and the area of the crescent-shaped section is
− 1 2(1 − 2) − = 20 1 2 − − 1 2 2 = 2 1 2 − − 1 6 3 0 = − 22 − 1 3 3.
So the area of the whole region is
4 − 22 − 1 3 3 + 2 = 41 − − 1 3 2 = 4√2 − 11 − √2 − 1 − 1 3 √2 − 12
= 4√2 − 11 − 1 3 √2 = 4 3 4√2 − 5
°c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
NOT FOR SALE
INSTRUCTOR USE ONLY
© Cengage Learning. All Rights Reserved.70 ¤ CHAPTER 5 PROBLEMS PLUS
19. lim
→∞√ √1 + 1 + √ √1 + 2 + · · · + √ √1 +
= lim
→∞
1
+ 1 + + 2 + · · · + +
= lim
→∞
1
1 + 1 1 + 1 + 2 1 + · · · + √1 + 1 1
= lim
→∞
1
=1
where () = √1 + 1
= 01 √1 + 1 = 2√1 + 1 0 = 2√2 − 1
20. Note that the graphs of ( − )2 and [( − ) − 2]2 intersect when | − | = | − − 2| ⇔
− = − − 2 ⇔ = + 1. The integration will proceed differently depending on the value of .
Case 1: −2 ≤ −1
In this case, () = ( − − 2)2 for ∈ [01], so
() = 01( − − 2)2 = 1 3( − − 2)31 0 = 1 3(− − 1)3 − (− − 2)3
=
1 3
(32 + 9 + 7) = 2 + 3 + 7 3 = + 3 22 + 12 1
This is a parabola; its maximum value for
−2 ≤ −1 is (−2) = 1 3, and its minimum
value is − 3 2 = 12 1 .
Case 2: −1 ≤ 0
In this case, () =
( − )2 if 0 ≤ ≤ + 1
( − − 2)2 if + 1 ≤ 1
Therefore,
() = 01 () = 0+1( − )2 + 1+1( − − 2)2
=
1 3
( − )3 0+1 + 1 3( − − 2)31 +1 = 1 31 + 3 + (− − 1)3 − (−1)
= −2 − + 1
3 = − + 1 22 + 12 7
Again, this is a parabola, whose maximum value
for −1 ≤ 0 is − 1 2 = 12 7 , and whose
minimum value on this -interval is (−1) = 1 3.
°c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
NOT FOR SALE
INSTRUCTOR USE ONLY
© Cengage Learning. All Rights Reserved.CHAPTER 5 PROBLEMS PLUS ¤ 71
Case 3: 0 ≤ ≤ 2
In this case, () = ( − )2 for ∈ [01], so
() = 01( − )2 = 1 3 ( − )31 0 = 1 3 (1 − )3 − (−)3
= 2 − + 1
3 = − 1 2 2 + 12 1
This parabola has a maximum value of (2) = 7 3
and a minimum value of 1 2 = 12 1 .
We conclude that () has an absolute maximum value of (2) = 7 3, and absolute minimum values of − 3 2 = 1 2 = 12 1 .
°c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
NOT FOR SALE
INSTRUCTOR USE ONLY
© Cengage Learning. All Rights Reserved.72 ¤ CHAPTER 5 PROBLEMS PLUS
°c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
NOT FOR SALE
INSTRUCTOR USE ONLY
© Cengage Learning. All Rights Reserved.
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