Calculus 2nd Edition Rogawski

ANSWERS TO ODD-NUMBERED EXERCISES
A85
Section 8.8 Preliminary Questions
1. T 1 = 6; T 2 = 7
3. The Trapezoidal Rule integrates linear functions exactly, so the
error will be zero.
5. The two graphical interpretations of the Midpoint Rule are the
sum of the areas of the midpoint rectangles and the sum of the areas
of the tangential trapezoids.
Section 8.8 Exercises
1. T 4 = 2.75; M 4 = 2.625
3. T 6 = 64.6875; M 6 ≈ 63.2813
5. T 6 ≈ 1.4054; M 6 ≈ 1.3769
7. T 6 = 1.1703; M 6 = 1.2063
9. T 4 ≈ 0.3846; M 5 ≈ 0.3871
11. T 5 = 1.4807; M 5 = 1.4537
13. S 4 ≈ 5.2522 15. S 6 ≈ 1.1090 17. S 4 ≈ 0.7469
19. S 8 ≈ 2.5450 21. S 1 0 ≈ 0.3466 23. ≈ 2.4674
25. ≈ 1.8769 27. ≈ 608.611
29. (a) Assuming the speed of the tsunami is a continuous function,
at x miles from the shore, the speed is √ 15f (x). Covering an
infinitesimally small distance, dx, the time T required for the tsunami
dx
to cover that distance becomes √ . It follows from this that
15f (x)
T = ∫ M
0
dx √ 15f (x)
.
(b) ≈ 3.347 hours.
31. (a) Since x 3 is concave up on [0, 2], T 6 is too large.
(b) We have f ′ (x) = 3x 2 and f ′′ (x) = 6x. Since |f ′′ (x)| =|6x| is
increasing on [0, 2], its maximum value occurs at x = 2 and we may
take K 2 =|f ′′ (2)| =12. Thus, Error(T 6 ) ≤ 2 9 .
(c) Error(T 6 ) ≈ 0.1111 < 2 9
33. T 1 0 will overestimate the integral. Error(T 10 ) ≤ 0.045.
35. M 1 0 will overestimate the integral. Error(M 10 ) ≤ 0.0113
37. N ≥ 10 3 ; Error ≈ 3.333 × 10 −7
39. N ≥ 750; Error ≈ 2.805 × 10 −7
41. Error(T 10 ) ≤ 0.0225; Error(M 10 ) ≤ 0.01125
43. S 8 ≈ 4.0467; N ≥ 23
45. Error(S 40 ) ≤ 1.017 × 10 −4 .
47. N ≥ 305 49. N ≥ 186
51. (a) The maximum value of |f (4) (x)| on the interval [0, 1] is 24.
(b) N ≥ 20; S 20 ≈ 0.785398; |0.785398 − π 4 | ≈ 1.55 × 10−10 .
53. (a) Notice |f ′′ (x)| =|2 cos(x 2 ) − 4x 2 sin(x 2 )|; proof follows.
(b) When K 2 = 2, Error(M N ) ≤ 1
4N 2 .
(c) N ≥ 16
55. Error(T 4 ) ≈ 0.1039; Error(T 8 ) ≈ 0.0258; Error(T 16 ) ≈ 0.0064;
Error(T 32 ) ≈ 0.0016; Error(T 64 ) ≈ 0.0004. Thes are about twice as
large as the error in M N .
57. S 2 = 1 4 . This is the exact value of the integral.
59. T N = r(b2 − a 2 ∫
)
b
+ s(b − a) = f (x) dx
2
a
61. (a) This result follows because the even-numbered interior
endpoints overlap:
(N−2)/2 ∑
i=0
= b − a
6
S 2j
2 = b − a [(y 0 + 4y 1 + y 2 ) + (y 2 + 4y 3 + y 4 ) +···]
6
[
y0 + 4y 1 + 2y 2 + 4y 3 + 2y 4 +···+4y N−1 + y N
]
= SN .
(b) If f (x) is a quadratic polynomial, then by part (a) we have
∫ b
S N = S2 0 + S2 2 +···+SN−2 2 = f (x) dx.
a
63. Let f (x) = ax 3 + bx 2 + cx + d, with a ̸= 0, be any cubic
polynomial. Then, f (4) (x) = 0, so we can take K 4 = 0. This yields
Error(S N ) ≤ 0
180N 4 = 0. In other words, S N is exact for all cubic
polynomials for all N.
Chapter 8 Review
1. (a) (v) (b) (iv) (c) (iii) (d) (i) (e) (ii)
3.
sin 9 θ
9 − sin11 θ
11 + C.
5.
tan θsec 5 θ
6 − 7 tan θsec3 θ
24 +
tan θ sec θ
16 + 16 1 ln | sec θ + tan θ|+C
7. − √ 1 − x 2 −1 sec−1 x + C 9. 2tan −1√ x + C
11. − tan−1 x
x + ln |x| − 1 2 ln (1 + x 2) + C.
13.
5
32 e4 − 32 1 ≈ 8.50 15. cos 12 6θ
72 − cos10 6θ
60 + C
17. 5ln|x − 1|+ln |x + 1|+C
19.
tan 3 θ
3 + tan θ + C 21. ≈ 1.0794
23. − cos5 θ
5 + 2cos3 θ
3 − cos θ + C 25. − 1 4
27.
2
3 (tan x)3/2 + C
29.
sin 6 θ
6 − sin8 θ
8 + C 31. − 1 3 u3 + C = − 1 3 cot3 x + C
∣ 33. ≈ 0.4202 35.
1 ∣∣
49 ln t+4
t−3 ∣ − 1 7 · 1
t−3 + C
37.
1
2 sec−1 x 2 + C
√a 2
tan
∫
⎧⎪ −1√ x
a + C a > 0
dx ⎨ √ √ ∣
39.
x 3/2 + ax 1/2 = √1
x− −a
ln −a
∣ √ √ ∣∣ + C a < 0
x+ −a
⎪ ⎩
− 2 √ x
+ C a = 0
41. ln |x + 2|+ x+2 5 − 3
(x+2) 2 + C
)
43. − ln |x − 2| − 2 x−2 1 + 1 2
(x ln 2
( + 4 + C
)
45.
1
3 tan−1 x+4
3 + C 47. ln |x + 2|+ x+2 5 − 3
(x+2) 2 + C
(
x
49. −
2 +4 ) 3/2 √
48x 3 + x 2 +4
16x + C 51. − 1 9 e4−3x (3x + 4) + C
53.
1
2 x2 sin x 2 + 1 2 cos x2 + C
∣ 55.
x 2
2 tanh−1 x + x 2 − 1 ∣∣
4 ln 1+x
1−x ∣ + C
( )
57. x ln x 2 + 9 − 2x + 6tan −1 ( x )
3 + C
59.
1
2 sinh 2 61. t + 1 4 coth(1 − 4t) + C 63. π
3
65. tan −1 (tanh x) + C