Section 5.2 The Definite Integral
23. Show that the definite integral is equal to lim
n→∞Rn and then evaluate the limit.
Z 4 0
(x − x2)dx, Rn = 4 n
n
X
i=1
4i n −16i2
n2
Solution:
530 ¤ CHAPTER 5 INTEGRALS 18. 2
0 −2with = 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
() = −2is decreasing on (0 2). We cannot make a similar statement for2
−1−2since is increasing on (−1 0).
19. On [0 1], lim
→∞
=1
1 +
∆ =
1 0
1 + .
20. On [2 5], lim
→∞
=1
1 + 3∆ =5 2 √
1 + 3.
21. On [2 7], lim
→∞
=1
[5(∗)3− 4∗] ∆ =7
2(53− 4) .
22. On [1 3], lim
→∞
=1
∗
(∗)2+ 4∆ =
3 1
2+ 4.
23. For
4
0 ( − 2) , ∆ = 4 − 0
= 4
, and = 0 + ∆ = 4
. Then
4
0 ( − 2) = lim
→∞
=1
4
4
= lim
→∞
=1
4
−
4
2 4
= lim
→∞
4
=1
4
−162
2
= lim
→∞.
lim→∞
4
=1
4
−162
2
= lim
→∞
4
4
=1 − 16
2
=1
2
= lim
→∞
16
2
( + 1)
2 − 64
3
( + 1)(2 + 1) 6
= lim
→∞
8
( + 1) − 32
32( + 1)(2 + 1)
= lim
→∞
8
1 +1
− 32 3
1 + 1
2 + 1
= 8(1) − 32
3(1)(2) = −40 3
24. For 3 1
(3+ 52) , ∆ = 3 − 1
= 2
, and = 1 + ∆ = 1 +2
. Then
3 1
(3+ 52) = lim
→∞
=1
1 +2
2
= lim
→∞
=1
1 +2
3
+ 5
1 + 2
2 2
= lim
→∞
2
=1
1 +6
+122
2 +83
3
+
5 +20
+202
2
= lim
→∞
2
=1
6 +26
+322
2 + 83
3
= lim
→∞
[continued]
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57. Write as a single integral in the form Rb
af (x)dx:
Z 2
−2
f (x)dx + Z 5
2
f (x)dx − Z −1
−2
f (x)dx Solution:
SECTION 5.2 THE DEFINITE INTEGRAL ¤ 501 37. 0
−3
1 +√ 9 − 2
can be interpreted as the area under the graph of
() = 1 +√
9 − 2between = −3 and = 0. This is equal to one-quarter the area of the circle with radius 3, plus the area of the rectangle, so
0
−3
1 +√ 9 − 2
= 14 · 32+ 1 · 3 = 3 +94.
38. 5
−5
−√
25 − 2
=5
−5 −5
−5
√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, 12(5)2 = 252. Thus, the value of the original integral is 0 −252 = −252.
39. 3
−4
12 can be interpreted as the sum of the areas of the two shaded
triangles; that is, 12(4)(2) +12(3)3 2
= 4 +94 = 254.
40. 1
0 |2 − 1| can be interpreted as the sum of the areas of the two shaded triangles; that is, 21
2
1 2
(1) = 12.
41. 1 1
√1 + 4 = 0since 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]
= −38 [given value]
43. 1
0(5 − 62) =1
0 5 − 61
0 2 = 5(1 − 0) − 61 3
= 5 − 2 = 3
44. 3
1 (2− 1) = 23
1 −3
1 1 = 2(3− ) − 1(3 − 1) = 23− 2 − 2 45. 3
1 + 2 =3
1 · 2 = 23
1 = 2(3− ) = 5− 3 46. 2
0 (2 cos − 5) =2
0 2 cos −2
0 5 = 22
0 cos − 52
0
= 2(1) − 5(2)2− 02
2 = 2 − 52 8 47. 2
−2 () +5
2 () −−1
−2 () =5
−2 () +−2
−1 () [by Property 5 and reversing limits]
=5
−1 () [Property 5]
° 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.c
60. FindR5
0 f (x)dx if
3 for x < 3 x for x ≥ 3 Solution:
502 ¤ CHAPTER 5 INTEGRALS 48. 4
2 () +8
4 () =8
2 () , so8
4 () =8
2 () −4
2 () = 73 − 59 = 14.
49. 9
0[2 () + 3()] = 29
0 () + 39
0 () = 2(37) + 3(16) = 122 50. If () =
3 for 3
for ≥ 3, then5
0 () 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,5
0 () = 5(3) +12(2)(2) = 17.
51. 3
0 () 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 is8
3 () , followed by8
4 () , which has a value about 1 unit less than8
3 () . Still positive, but with a smaller value than8
4 () , is8
0 () . Ordering these quantities from smallest to largest gives us
3
0 () 0(1) 8
0 () 8
4 () 8
3 () or B E A D C 52. (0) =0
2 () = −2
0 () , 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) =2
2 () = 0is the only non-negative value, choice C is the largest.
53. =2
−4[ () + 2 + 5] =2
−4 () + 22
−4 +2
−45 = 1+ 22+ 3
1= −3 [area below -axis] + 3 − 3 = −3
2= −12(4)(4) [area of triangle, see figure] +12(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) ≤2
0 () ≤ (2 − 0) ⇒ 2 ≤2
0 () ≤ 2.
55. 2− 4 + 4 = ( − 2)2 ≥ 0 on [0 4], so4
0(2− 4 + 4) ≥ 0 [Property 6].
56. 2≤ on [0 1] so√
1 + 2≤√
1 + on [0 1]. Hence,1 0
√1 + 2 ≤1 0
√1 + [Property 7].
57. If −1 ≤ ≤ 1, then 0 ≤ 2 ≤ 1 and 1 ≤ 1 + 2 ≤ 2, so 1 ≤√
1 + 2 ≤√ 2and 1[1 − (−1)] ≤1
−1
√1 + 2 ≤√
2 [1 − (−1)] [Property 8]; that is, 2 ≤1
−1
√1 + 2 ≤ 2√2.
58. If
6 ≤ ≤ 3, then 1
2 ≤ sin ≤
√3 2
sin is increasing on
63, so 1
2
3−
6
≤
3
6 sin ≤
√3 2
3−
6
[Property 8]; that is, 12 ≤
3
6 sin ≤
√3 12 .
° 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.c
68. Use the properties of integrals to verify the inequality without evaluating the integrals.
π 12 ≤
Z π3
π 6
sin xdx ≤
√3π 12 Solution:
502 ¤ CHAPTER 5 INTEGRALS 48. 4
2 () +8
4 () =8
2 () , so8
4 () =8
2 () −4
2 () = 73 − 59 = 14.
49. 9
0[2 () + 3()] = 29
0 () + 39
0 () = 2(37) + 3(16) = 122 50. If () =
3 for 3
for ≥ 3, then5
0 () 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,5
0 () = 5(3) +12(2)(2) = 17.
51. 3
0 () 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 is8
3 () , followed by8
4 () , which has a value about 1 unit less than8
3 () . Still positive, but with a smaller value than8
4 () , is8
0 () . Ordering these quantities from smallest to largest gives us
3
0 () 0(1) 8
0 () 8
4 () 8
3 () or B E A D C 52. (0) =0
2 () = −2
0 () , 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) =2
2 () = 0is the only non-negative value, choice C is the largest.
53. =2
−4[ () + 2 + 5] =2
−4 () + 22
−4 +2
−45 = 1+ 22+ 3
1= −3 [area below -axis] + 3 − 3 = −3
2= −12(4)(4) [area of triangle, see figure] + 12(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) ≤2
0 () ≤ (2 − 0) ⇒ 2 ≤2
0 () ≤ 2.
55. 2− 4 + 4 = ( − 2)2≥ 0 on [0 4], so4
0(2− 4 + 4) ≥ 0 [Property 6].
56. 2≤ on [0 1] so√
1 + 2≤√
1 + on [0 1]. Hence,1 0
√1 + 2 ≤1 0
√1 + [Property 7].
57. If −1 ≤ ≤ 1, then 0 ≤ 2 ≤ 1 and 1 ≤ 1 + 2 ≤ 2, so 1 ≤√
1 + 2 ≤√ 2and 1[1 − (−1)] ≤1
−1
√1 + 2 ≤√
2 [1 − (−1)] [Property 8]; that is, 2 ≤1
−1
√1 + 2 ≤ 2√2.
58. If
6 ≤ ≤ 3, then 1
2 ≤ sin ≤
√3 2
sin is increasing on
63, so 1
2
3 −
6
≤
3
6 sin ≤
√3 2
3 −
6
[Property 8]; that is, 12 ≤
3
6 sin ≤
√3 12 .
° 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.c
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