Solution Manual : Chapter - 05 Integration

186
CHAPTER 5
Integration
EXERCISE SET 5.1
1. Endpoints 0,
1
n
,
2
n
, . . . ,
n − 1
n
, 1; using right endpoints
An =
1
n
+
2
n
+ · · · +
n − 1
n
+ 1
1
n
n 2 5 10 50 100
An 0.853553 0.749739 0.710509 0.676095 0.671463
2. Endpoints 0,
1
n
,
2
n
, . . . ,
n − 1
n
An =
n
n + 1
+
n
n + 2
+
n
n + 3
+ · · · +
n
2n − 1
+
1
2
1
n
n 2 5 10 50 100
An 0.583333 0.645635 0.668771 0.688172 0.690653
3. Endpoints 0,
π
n
,
2π
n
, . . . ,
(n − 1)π
n
, π; using right endpoints
An = [sin(π/n) + sin(2π/n) + · · · + sin(π(n − 1)/n) + sin π]
π
n
n 2 5 10 50 100
An 1.57080 1.93376 1.98352 1.99935 1.99984
4. Endpoints 0,
π
2n
,
2π
2n
, . . . ,
(n − 1)π
2n
,
π
2
; using right endpoints
An = [cos(π/2n) + cos(2π/2n) + · · · + cos((n − 1)π/2n) + cos(π/2)]
π
2n
n 2 5 10 50 100
An 0.555359 0.834683 0.919405 0.984204 0.992120
5. Endpoints 1,
n + 1
n
,
n + 2
n
, . . . ,
2n − 1
n
An =
n
n + 1
+
n
n + 2
+ · · · +
n
2n − 1
+
1
2
1
n
n 2 5 10 50 100
An 0.583333 0.645635 0.668771 0.688172 0.690653
6. Endpoints −
π
2
, −
π
2
+
π
n
, −
π
2
+
2π
n
, . . . , −
π
2
+
(n − 1)π
n
,
π
2
; using right endpoints
An = cos −
π
2
+
π
n
+ cos −
π
2
+
2π
n
+ · · · + cos −
π
2
+
(n − 1)π
n
+ cos
π
2
π
n
n 2 5 10 50 100
An 1.99985 1.93376 1.98352 1.99936 1.99985

Exercise Set 5.2 187
7. Endpoints 0,
1
n
,
2
n
, . . . ,
n − 1
n
An =

 1 −
1
n
2
+ 1 −
2
n
2
+ · · · + 1 −
n − 1
n
2
+ 0

 1
n
n 2 5 10 50 100
An 0.433013 0.659262 0.726130 0.774567 0.780106
8. Endpoints −1, −1 +
2
n
, −1 +
4
n
, . . . , −1 +
2(n − 1)
n
An =

 1 −
n − 2
n
2
+ 1 −
n − 4
n
2
+ · · · + 1 −
n − 2
n
2
+ 0

 2
n
n 2 5 10 50 100
An 1 1.423837 1.518524 1.566097 1.569136
9. 3(x − 1) 10. 5(x − 2) 11. x(x + 2) 12.
3
2
(x − 1)2
13. (x + 3)(x − 1) 14.
3
2
x(x − 2)
15. The area in Exercise 13 is always 3 less than the area in Exercise 11. The regions are identical
except that the area in Exercise 11 has the extra trapezoid with vertices at (0, 0), (1, 0), (0, 2), (1, 4)
(with area 3).
16. (a) The region in question is a trapezoid, and the area of a trapezoid is
1
2
(h1 + h2)w.
(b) From Part (a), A (x) =
1
2
[f(a) + f(x)] + (x − a)
1
2
f (x)
=
1
2
[f(a) + f(x)] + (x − a)
1
2
f(x) − f(a)
x − a
= f(x)
17. B is also the area between the graph of f(x) =
√
x and the interval [0, 1] on the y−axis, so A + B
is the area of the square.
18. If the plane is rotated about the line y = x then A becomes B and vice versa.
EXERCISE SET 5.2
1. (a)
x
√
1 + x2
dx = 1 + x2 + C (b) x2
cos(1 + x3
)dx =
1
3
sin(1 + x3
) + C
2. (a)
d
dx
(sin x − x cos x + C) = cos x − cos x + x sin x = x sin x
(b)
d
dx
x
√
1 − x2
+ C =
√
1 − x2 + x2
/
√
1 − x2
1 − x2
=
1
(1 − x2)3/2
3.
d
dx
x3 + 5 =
3x2
2
√
x3 + 5
so
3x2
2
√
x3 + 5
dx = x3 + 5 + C

188 Chapter 5
4.
d
dx
x
x2 + 3
=
3 − x2
(x2 + 3)2
so
3 − x2
(x2 + 3)2
dx =
x
x2 + 3
+ C
5.
d
dx
sin 2
√
x =
cos (2
√
x)
√
x
so
cos (2
√
x)
√
x
dx = sin 2
√
x + C
6.
d
dx
[sin x − x cos x] = x sin x so x sin x dx = sin x − x cos x + C
7. (a) x9
/9 + C (b)
7
12
x12/7
+ C (c)
2
9
x9/2
+ C
8. (a)
3
5
x5/3
+ C (b) −
1
5
x−5
+ C = −
1
5x5
+ C (c) 8x1/8
+ C
9. (a)
1
2
x−3
dx = −
1
4
x−2
+ C (b) u4
/4 − u2
+ 7u + C
10.
3
5
x5/3
− 5x4/5
+ 4x + C
11. (x−3
+ x1/2
− 3x1/4
+ x2
)dx = −
1
2
x−2
+
2
3
x3/2
−
12
5
x5/4
+
1
3
x3
+ C
12. (7y−3/4
− y1/3
+ 4y1/2
)dy = 28y1/4
−
3
4
y4/3
+
8
3
y3/2
+ C
13. (x + x4
)dx = x2
/2 + x5
/5 + C
14. (4 + 4y2
+ y4
)dy = 4y +
4
3
y3
+
1
5
y5
+ C
15. x1/3
(4 − 4x + x2
)dx = (4x1/3
− 4x4/3
+ x7/3
)dx = 3x4/3
−
12
7
x7/3
+
3
10
x10/3
+ C
16. (2 − x + 2x2
− x3
)dx = 2x −
1
2
x2
+
2
3
x3
−
1
4
x4
+ C
17. (x + 2x−2
− x−4
)dx = x2
/2 − 2/x + 1/(3x3
) + C
18. (t−3
− 2)dt = −
1
2
t−2
− 2t + C
19. −4 cos x + 2 sin x + C 20. 4 tan x − csc x + C
21. (sec2
x + sec x tan x)dx = tan x + sec x + C
22. (sec x tan x + 1)dx = sec x + x + C 23.
sec θ
cos θ
dθ = sec2
θ dθ = tan θ + C
24. sin y dy = − cos y + C 25. sec x tan x dx = sec x + C

26. (φ + 2 csc2
φ)dφ = φ2
/2 − 2 cot φ + C 27. (1 + sin θ)dθ = θ − cos θ + C
28.
2 sin x cos x
cos x
dx = 2 sin x dx = −2 cos x + C
29.
1 − sin x
1 − sin2
x
dx =
1 − sin x
cos2 x
dx = sec2
x − sec x tan x dx = tan x − sec x + C
30.
1
1 + cos 2x
dx =
1
2 cos2 x
dx =
1
2
sec2
x dx =
1
2
tan x + C
31. (a)
2
y
–2 2
x
(b) y
x
4
-1 1
(c) f(x) = x2
/2 − 1
32. (a)
5
6
x
y (b)
1
y
x
1
(c) y =
2
3
x3/2
− 2
33. y
x
-5
5
c/4 c/2
34.
–4
2
y
x
1 2
35. f (x) = m = − sin x so f(x) = (− sin x)dx = cos x + C; f(0) = 2 = 1 + C
so C = 1, f(x) = cos x + 1
36. f (x) = m = (x + 1)2
, so f(x) = (x + 1)2
dx =
1
3
(x + 1)3
+ C;
f(−2) = 8 =
1
3
(−2 + 1)3
+ C = −
1
3
+ C, = 8 +
1
3
=
25
3
, f(x) =
1
3
(x + 1)3
+
25
3
37. (a) y(x) = x1/3
dx =
3
4
x4/3
+ C, y(1) =
3
4
+ C = 2, C =
5
4
; y(x) =
3
4
x4/3
+
5
4
(b) y(t) = (sin t + 1) dt = − cos t + t + C, y
π
3
= −
1
2
+
π
3
+ C =
1
2
, C = 1 −
π
3
;
y(t) = − cos t + t + 1 −
π
3

190 Chapter 5
(c) y(x) = (x1/2
+ x−1/2
)dx =
2
3
x3/2
+ 2x1/2
+ C, y(1) = 0 =
8
3
+ C, C = −
8
3
,
y(x) =
2
3
x3/2
+ 2x1/2
−
8
3
38. (a) y(x) =
1
8
x−3
dx = −
1
16
x−2
+ C, y(1) = 0 = −
1
16
+ C, C =
1
16
; y(x) = −
1
16
x−2
+
1
16
(b) y(t) = (sec2
t − sin t) dt = tan t + cos t + C, y(
π
4
) = 1 = 1 +
√
2
2
+ C, C = −
√
2
2
;
y(t) = tan t + cos t −
√
2
2
(c) y(x) = x7/2
dx =
2
9
x9/2
+ C, y(0) = 0 = C, C = 0; y(x) =
2
9
x9/2
39. f (x) =
2
3
x3/2
+ C1; f(x) =
4
15
x5/2
+ C1x + C2
40. f (x) = x2
/2 + sin x + C1, use f (0) = 2 to get C1 = 2 so f (x) = x2
/2 + sin x + 2,
f(x) = x3
/6 − cos x + 2x + C2, use f(0) = 1 to get C2 = 2 so f(x) = x3
/6 − cos x + 2x + 2
41. dy/dx = 2x + 1, y = (2x + 1)dx = x2
+ x + C; y = 0 when x = −3
so (−3)2
+ (−3) + C = 0, C = −6 thus y = x2
+ x − 6
42. dy/dx = x2
, y = x2
dx = x3
/3 + C; y = 2 when x = −1 so (−1)3
/3 + C = 2, C = 7/3
thus y = x3
/3 + 7/3
43. dy/dx = 6xdx = 3x2
+ C1. The slope of the tangent line is −3 so dy/dx = −3 when x = 1.
Thus 3(1)2
+ C1 = −3, C1 = −6 so dy/dx = 3x2
− 6, y = (3x2
− 6)dx = x3
− 6x + C2. If x = 1,
then y = 5 − 3(1) = 2 so (1)2
− 6(1) + C2 = 2, C2 = 7 thus y = x3
− 6x + 7.
44. dT/dx = C1, T = C1x + C2; T = 25 when x = 0 so C2 = 25, T = C1x + 25. T = 85 when x = 50
so 50C1 + 25 = 85, C1 = 1.2, T = 1.2x + 25
45. (a) F (x) = G (x) = 3x + 4
(b) F(0) = 16/6 = 8/3, G(0) = 0, so F(0) − G(0) = 8/3
(c) F(x) = (9x2
+ 24x + 16)/6 = 3x2
/2 + 4x + 8/3 = G(x) + 8/3
46. (a) F (x) = G (x) = 10x/(x2
+ 5)2
(b) F(0) = 0, G(0) = −1, so F(0) − G(0) = 1
(c) F(x) =
x2
x2 + 5
=
(x2
+ 5) − 5
x2 + 5
= 1 −
5
x2 + 5
= G(x) + 1
47. (sec2
x − 1)dx = tan x − x + C 48. (csc2
x − 1)dx = − cot x − x + C
49. (a)
1
2
(1 − cos x)dx =
1
2
(x − sin x) + C (b)
1
2
(1 + cos x) dx =
1
2
(x + sin x) + C

50. (a) F (x) = G (x) = f(x), where f(x) =
1, x > 0
−1, x < 0
(b) G(x) − F(x) =
2, x > 0
3, x < 0
so G(x) = F(x) plus a constant
(c) no, because (−∞, 0) ∪ (0, +∞) is not an interval
51. v =
1087
2
√
273
T−1/2
dT =
1087
√
273
T1/2
+C, v(273) = 1087 = 1087+C so C = 0, v =
1087
√
273
T1/2
ft/s
EXERCISE SET 5.3
1. (a) u23
du = u24
/24 + C = (x2
+ 1)24
/24 + C
(b) − u3
du = −u4
/4 + C = −(cos4
x)/4 + C
(c) 2 sin u du = −2 cos u + C = −2 cos
√
x + C
(d)
3
8
u−1/2
du =
3
4
u1/2
+ C =
3
4
4x2 + 5 + C
2. (a)
1
4
sec2
u du =
1
4
tan u + C =
1
4
tan(4x + 1) + C
(b)
1
4
u1/2
du =
1
6
u3/2
+ C =
1
6
(1 + 2y2
)3/2
+ C
(c)
1
π
u1/2
du =
2
3π
u3/2
+ C =
2
3π
sin3/2
(πθ) + C
(d) u4/5
du =
5
9
u9/5
+ C =
5
9
(x2
+ 7x + 3)9/5
+ C
3. (a) − u du = −
1
2
u2
+ C = −
1
2
cot2
x + C
(b) u9
du =
1
10
u10
+ C =
1
10
(1 + sin t)10
+ C
(c)
1
2
cos u du =
1
2
sin u + C =
1
2
sin 2x + C
(d)
1
2
sec2
u du =
1
2
tan u + C =
1
2
tan x2
+ C
4. (a) (u − 1)2
u1/2
du = (u5/2
− 2u3/2
+ u1/2
)du =
2
7
u7/2
−
4
5
u5/2
+
2
3
u3/2
+ C
=
2
7
(1 + x)7/2
−
4
5
(1 + x)5/2
+
2
3
(1 + x)3/2
+ C
(b) csc2
u du = − cot u + C = − cot(sin x) + C
(c) sin u du = − cos u + C = − cos(x − π) + C
(d)
du
u2
= −
1
u
+ C = −
1
x5 + 1
+ C

192 Chapter 5
5. u = 2 − x2
, du = −2x dx; −
1
2
u3
du = −u4
/8 + C = −(2 − x2
)4
/8 + C
6. u = 3x − 1, du = 3dx;
1
3
u5
du =
1
18
u6
+ C =
1
18
(3x − 1)6
+ C
7. u = 8x, du = 8dx;
1
8
cos u du =
1
8
sin u + C =
1
8
sin 8x + C
8. u = 3x, du = 3dx;
1
3
sin u du = −
1
3
cos u + C = −
1
3
cos 3x + C
9. u = 4x, du = 4dx;
1
4
sec u tan u du =
1
4
sec u + C =
1
4
sec 4x + C
10. u = 5x, du = 5dx;
1
5
sec2
u du =
1
5
tan u + C =
1
5
tan 5x + C
11. u = 7t2
+ 12, du = 14t dt;
1
14
u1/2
du =
1
21
u3/2
+ C =
1
21
(7t2
+ 12)3/2
+ C
12. u = 4 − 5x2
, du = −10x dx; −
1
10
u−1/2
du = −
1
5
u1/2
+ C = −
1
5
4 − 5x2 + C
13. u = x3
+ 1, du = 3x2
dx;
1
3
u−1/2
du =
2
3
u1/2
+ C =
2
3
x3 + 1 + C
14. u = 1 − 3x, du = −3dx; −
1
3
u−2
du =
1
3
u−1
+ C =
1
3
(1 − 3x)−1
+ C
15. u = 4x2
+ 1, du = 8x dx;
1
8
u−3
du = −
1
16
u−2
+ C = −
1
16
(4x2
+ 1)−2
+ C
16. u = 3x2
, du = 6x dx;
1
6
cos u du =
1
6
sin u + C =
1
6
sin(3x2
) + C
17. u = 5/x, du = −(5/x2
)dx; −
1
5
sin u du =
1
5
cos u + C =
1
5
cos(5/x) + C
18. u =
√
x, du =
1
2
√
x
dx; 2 sec2
u du = 2 tan u + C = 2 tan
√
x + C
19. u = x3
, du = 3x2
dx;
1
3
sec2
u du =
1
3
tan u + C =
1
3
tan(x3
) + C
20. u = cos 2t, du = −2 sin 2t dt; −
1
2
u3
du = −
1
8
u4
+ C = −
1
8
cos4
2t + C
21. u = sin 3t, du = 3 cos 3t dt;
1
3
u5
du =
1
18
u6
+ C =
1
18
sin6
3t + C
22. u = 5 + cos 2θ, du = −2 sin 2θ dθ; −
1
2
u−3
du =
1
4
u−2
+ C =
1
4
(5 + cos 2θ)−2
+ C

23. u = 2 − sin 4θ, du = −4 cos 4θ dθ; −
1
4
u1/2
du = −
1
6
u3/2
+ C = −
1
6
(2 − sin 4θ)3/2
+ C
24. u = tan 5x, du = 5 sec2
5x dx;
1
5
u3
du =
1
20
u4
+ C =
1
20
tan4
5x + C
25. u = sec 2x, du = 2 sec 2x tan 2x dx;
1
2
u2
du =
1
6
u3
+ C =
1
6
sec3
2x + C
26. u = sin θ, du = cos θ dθ; sin u du = − cos u + C = − cos(sin θ) + C
27. u = x − 3, x = u + 3, dx = du
(u + 3)u1/2
du = (u3/2
+ 3u1/2
)du =
2
5
u5/2
+ 2u3/2
+ C =
2
5
(x − 3)5/2
+ 2(x − 3)3/2
+ C
28. u = y + 1, y = u − 1, dy = du
u − 1
u1/2
du = (u1/2
− u−1/2
)du =
2
3
u3/2
− 2u1/2
+ C =
2
3
(y + 1)3/2
− 2(y + 1)1/2
+ C
29. sin2
2θ sin 2θ dθ = (1 − cos2
2θ) sin 2θ dθ; u = cos 2θ, du = −2 sin 2θ dθ,
−
1
2
(1 − u2
)du = −
1
2
u +
1
6
u3
+ C = −
1
2
cos 2θ +
1
6
cos3
2θ + C
30. sec2
3θ = tan2
3θ + 1, u = 3θ, du = 3dθ
sec4
3θ dθ =
1
3
(tan2
u + 1) sec2
u du =
1
9
tan3
u +
1
3
tan u + C =
1
9
tan3
3θ +
1
3
tan 3θ + C
31. u = a + bx, du = bdx,
(a + bx)n
dx =
1
b
un
du =
(a + bx)n+1
b(n + 1)
+ C
32. u = a + bx, du = b dx, dx =
1
b
du
1
b
u1/n
du =
n
b(n + 1)
u(n+1)/n
+ C =
n
b(n + 1)
(a + bx)(n+1)/n
+ C
33. u = sin(a + bx), du = b cos(a + bx)dx
1
b
un
du =
1
b(n + 1)
un+1
+ C =
1
b(n + 1)
sinn+1
(a + bx) + C
35. (a) with u = sin x, du = cos x dx; u du =
1
2
u2
+ C1 =
1
2
sin2
x + C1;
with u = cos x, du = − sin x dx; − u du = −
1
2
u2
+ C2 = −
1
2
cos2
x + C2
(b) because they diﬀer by a constant:
1
2
sin2
x + C1 − −
1
2
cos2
x + C2 =
1
2
(sin2
x + cos2
x) + C1 − C2 = 1/2 + C1 − C2

194 Chapter 5
36. (a) First method: (25x2
− 10x + 1)dx =
25
3
x3
− 5x2
+ x + C1;
second method:
1
5
u2
du =
1
15
u3
+ C2 =
1
15
(5x − 1)3
+ C2
(b)
1
15
(5x − 1)3
+ C2 =
1
15
(125x3
− 75x2
+ 15x − 1) + C2 =
25
3
x3
− 5x2
+ x −
1
15
+ C2;
the answers diﬀer by a constant.
37. y(x) =
√
3x + 1dx =
2
9
(3x + 1)3/2
+ C,
y(1) =
16
9
+ C = 5, C =
29
9
so y(x) =
2
9
(3x + 1)3/2
+
29
9
38. y(x) = (6 − 5 sin 2x)dx = 6x +
5
2
cos 2x + C,
y(0) =
5
2
+ C = 3, C =
1
2
so y(x) = 6x +
5
2
cos 2x +
1
2
39. f (x) = m =
√
3x + 1, f(x) = (3x + 1)1/2
dx =
2
9
(3x + 1)3/2
+ C; f(0) = 1 =
2
9
+ C, C =
7
9
,
so f(x) =
2
9
(3x + 1)3/2
+
7
9
40. 5
0
-5 5
41. p(t) = (4 + 0.15t)3/2
dt =
8
3
(4 + 0.15t)5/2
+ C; p(0) = 100,000 =
8
3
45/2
+ C =
256
3
+ C,
C = 100,000−
256
3
≈ 99,915, p(t) ≈
8
3
(4+0.15t)5/2
+99,915, p(5) ≈
8
3
(4.75)5/2
+99,915 ≈ 100,046
EXERCISE SET 5.4
1. (a) 1 + 8 + 27 = 36 (b) 5 + 8 + 11 + 14 + 17 = 55
(c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (d) 1 + 1 + 1 + 1 + 1 + 1 = 6
(e) 1 − 2 + 4 − 8 + 16 = 11 (f) 0 + 0 + 0 + 0 + 0 + 0 = 0
2. (a) 1 + 0 − 3 + 0 = −2 (b) 1 − 1 + 1 − 1 + 1 − 1 = 0
(c) π2
+ π2
+ · · · + π2
= 14π2
(14 terms)
(d) 24
+ 25
+ 26
= 112
(e)
√
1 +
√
2 +
√
3 +
√
4 +
√
5 +
√
6
(f) 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 = 1

3.
10
k=1
k 4.
20
k=1
3k 5.
10
k=1
2k
6.
8
k=1
(2k − 1) 7.
6
k=1
(−1)k+1
(2k − 1) 8.
5
k=1
(−1)k+1 1
k
9. (a)
50
k=1
12k (b)
50
k=1
(2k − 1)
10. (a)
5
k=1
(−1)k+1
ak (b)
5
k=0
(−1)k+1
bk (c)
n
k=0
akxk
(d)
5
k=0
a5−k
bk
11.
1
2
(100)(100 + 1) = 5050 12. 7
100
k=1
k +
100
k=1
1 =
7
2
(100)(101) + 100 = 35,450
13.
1
6
(20)(21)(41) = 2870 14.
20
k=1
k2
−
3
k=1
k2
= 2870 − 14 = 2856
15.
30
k=1
k(k2
− 4) =
30
k=1
(k3
− 4k) =
30
k=1
k3
− 4
30
k=1
k =
1
4
(30)2
(31)2
− 4 ·
1
2
(30)(31) = 214,365
16.
6
k=1
k −
6
k=1
k3
=
1
2
(6)(7) −
1
4
(6)2
(7)2
= −420
17.
n
k=1
3k
n
=
3
n
n
k=1
k =
3
n
·
1
2
n(n + 1) =
3
2
(n + 1)
18.
n−1
k=1
k2
n
=
1
n
n−1
k=1
k2
=
1
n
·
1
6
(n − 1)(n)(2n − 1) =
1
6
(n − 1)(2n − 1)
19.
n−1
k=1
k3
n2
=
1
n2
n−1
k=1
k3
=
1
n2
·
1
4
(n − 1)2
n2
=
1
4
(n − 1)2
20.
n
k=1
5
n
−
2k
n
=
5
n
n
k=1
1 −
2
n
n
k=1
k =
5
n
(n) −
2
n
·
1
2
n(n + 1) = 4 − n
22.
n(n + 1)
2
= 465, n2
+ n − 930 = 0, (n + 31)(n − 30) = 0, n = 30.
23.
1 + 2 + 3 + · · · + n
n2
=
n
k=1
k
n2
=
1
n2
n
k=1
k =
1
n2
·
1
2
n(n + 1) =
n + 1
2n
; lim
n→+∞
n + 1
2n
=
1
2

196 Chapter 5
24.
12
+ 22
+ 32
+ · · · + n2
n3
=
n
k=1
k2
n3
=
1
n3
n
k=1
k2
=
1
n3
·
1
6
n(n + 1)(2n + 1) =
(n + 1)(2n + 1)
6n2
;
lim
n→+∞
(n + 1)(2n + 1)
6n2
= lim
n→+∞
1
6
(1 + 1/n)(2 + 1/n) =
1
3
25.
n
k=1
5k
n2
=
5
n2
n
k=1
k =
5
n2
·
1
2
n(n + 1) =
5(n + 1)
2n
; lim
n→+∞
5(n + 1)
2n
=
5
2
26.
n−1
k=1
2k2
n3
=
2
n3
n−1
k=1
k2
=
2
n3
·
1
6
(n − 1)(n)(2n − 1) =
(n − 1)(2n − 1)
3n2
;
lim
n→+∞
(n − 1)(2n − 1)
3n2
= lim
n→+∞
1
3
(1 − 1/n)(2 − 1/n) =
2
3
27. (a)
5
j=0
2j
(b)
6
j=1
2j−1
(c)
7
j=2
2j−2
28. (a)
5
k=1
(k + 4)2k+8
(b)
13
k=9
(k − 4)2k
29. Endpoints 2, 3, 4, 5, 6; ∆x = 1;
(a) Left endpoints:
4
k=1
f(x∗
k)∆x = 7 + 10 + 13 + 16 = 46
(b) Midpoints:
4
k=1
f(x∗
k)∆x = 8.5 + 11.5 + 14.5 + 17.5 = 52
(c) Right endpoints:
4
k=1
f(x∗
k)∆x = 10 + 13 + 16 + 19 = 58
30. Endpoints 1, 3, 5, 7, 9, ∆x = 2;
(a) Left endpoints:
4
k=1
f(x∗
k)∆x = 1 +
1
3
+
1
5
+
1
7
2 =
352
105
(b) Midpoints:
4
k=1
f(x∗
k)∆x =
1
2
+
1
4
+
1
6
+
1
8
2 =
25
12
4
k=1
f(x∗
k)∆x =
1
3
+
1
5
+
1
7
+
1
9
2 =
496
315
31. Endpoints: 0, π/4, π/2, 3π/4, π; ∆x = π/4
(a) Left endpoints:
4
k=1
f(x∗
k)∆x = 1 +
√
2/2 + 0 −
√
2/2 (π/4) = π/4
(b) Midpoints:
4
k=1
f(x∗
k)∆x = [cos(π/8) + cos(3π/8) + cos(5π/8) + cos(7π/8)] (π/4)
= [cos(π/8) + cos(3π/8) − cos(3π/8) − cos(π/8)] (π/4) = 0
4
k=1
f(x∗
k)∆x =
√
2/2 + 0 −
√
2/2 − 1 (π/4) = −π/4

32. Endpoints −1, 0, 1, 2, 3; ∆x = 1
(a)
4
k=1
f(x∗
k)∆x = −3 + 0 + 1 + 0 = −2
(b)
4
k=1
f(x∗
k)∆x = −
5
4
+
3
4
+
3
4
+
15
4
= 4
(c)
4
k=1
f(x∗
k)∆x = 0 + 1 + 0 − 3 = −2
33. (a) 0.718771403, 0.705803382, 0.698172179
(b) 0.668771403, 0.680803382, 0.688172179
(c) 0.692835360, 0.693069098, 0.693134682
34. (a) 0.761923639, 0.712712753, 0.684701150
(b) 0.584145862, 0.623823864, 0.649145594
(c) 0.663501867, 0.665867079, 0.666538346
35. (a) 4.884074734, 5.115572731, 5.248762738
(b) 5.684074734, 5.515572731, 5.408762738
(c) 5.34707029, 5.338362719, 5.334644416
36. (a) 0.919403170, 0.960215997, 0.984209789
(b) 1.076482803, 1.038755813, 1.015625715
(c) 1.001028824, 1.000257067, 1.000041125
37. ∆x =
3
n
, x∗
k = 1 +
3
n
k; f(x∗
k)∆x =
1
2
x∗
k∆x =
1
2
1 +
3
n
k
3
n
=
3
2
1
n
+
3
n2
k
n
k=1
f(x∗
k)∆x =
3
2
n
k=1
1
n
+
n
k=1
3
n2
k =
3
2
1 +
3
n2
·
1
2
n(n + 1) =
3
2
1 +
3
2
n + 1
n
A = lim
n→+∞
3
2
1 +
3
2
1 +
1
n
=
3
2
1 +
3
2
=
15
4
38. ∆x =
5
n
, x∗
k = 0 + k
5
n
; f(x∗
k)∆x = (5 − x∗
k)∆x = 5 −
5
n
k
5
n
=
25
n
−
25
n2
k
n
k=1
f(x∗
k)∆x =
n
k=1
25
n
−
25
n2
n
k=1
k = 25 −
25
n2
·
1
2
n(n + 1) = 25 −
25
2
n + 1
n
A = lim
n→+∞
25 −
25
2
1 +
1
n
= 25 −
25
2
=
25
2
39. ∆x =
3
n
, x∗
k = 0 + k
3
n
; f(x∗
k)∆x = 9 − 9
k2
n2
3
n
n
k=1
f(x∗
k)∆x =
n
k=1
(9 − 9
k2
n2
)
3
n
=
27
n
n
k=1
1 −
k2
n2
= 27 −
27
n3
n
k=1
k2
A = lim
n→+∞
27 −
27
n3
n
k=1
k2
= 27 − 27
1
3
= 18

198 Chapter 5
40. ∆x =
3
n
, x∗
k = k
3
n
f(x∗
k)∆x = 4 −
1
4
(x∗
k)2
∆x = 4 −
1
4
9k2
n2
3
n
=
12
n
−
27k2
4n3
n
k=1
f(x∗
k)∆x =
n
k=1
12
n
−
27
4n3
n
k=1
k2
= 12 −
27
4n3
·
1
6
n(n + 1)(2n + 1) = 12 −
9
8
(n + 1)(2n + 1)
n2
A = lim
n→+∞
12 −
9
8
1 +
1
n
2 +
1
n
= 12 −
9
8
(1)(2) = 39/4
41. ∆x =
4
n
, x∗
k = 2 + k
4
n
f(x∗
k)∆x = (x∗
k)3
∆x = 2 +
4
n
k
3
4
n
=
32
n
1 +
2
n
k
3
=
32
n
1 +
6
n
k +
12
n2
k2
+
8
n3
k3
n
k=1
f(x∗
k)∆x =
32
n
n
k=1
1 +
6
n
n
k=1
k +
12
n2
n
k=1
k2
+
8
n3
n
k=1
k3
=
32
n
n +
6
n
·
1
2
n(n + 1) +
12
n2
·
1
6
n(n + 1)(2n + 1) +
8
n3
·
1
4
n2
(n + 1)2
= 32 1 + 3
n + 1
n
+ 2
(n + 1)(2n + 1)
n2
+ 2
(n + 1)2
n2
A = lim
n→+∞
32 1 + 3 1 +
1
n
+ 2 1 +
1
n
2 +
1
n
+ 2 1 +
1
n
2
= 32[1 + 3(1) + 2(1)(2) + 2(1)2
] = 320
42. ∆x =
2
n
, x∗
k = −3 + k
2
n
; f(x∗
k)∆x = [1 − (x∗
k)3
]∆x = 1 − −3 +
2
n
k
3
2
n
=
2
n
28 −
54
n
k +
36
n2
k2
−
8
n3
k3
n
k=1
f(x∗
k)∆x =
2
n
28n − 27(n + 1) + 6
(n + 1)(2n + 1)
n
− 2
(n + 1)2
n
A = lim
n→+∞
2 28 − 27 1 +
1
n
+ 6 1 +
1
n
2 +
1
n
− 2 1 +
1
n
2
= 2(28 − 27 + 12 − 2) = 22
43. ∆x =
3
n
, x∗
k = 1 + (k − 1)
3
n
f(x∗
k)∆x =
1
2
x∗
k∆x =
1
2
1 + (k − 1)
3
n
3
n
=
1
2
3
n
+ (k − 1)
9
n2
n
k=1
f(x∗
k)∆x =
1
2
n
k=1
3
n
+
9
n2
n
k=1
(k − 1) =
1
2
3 +
9
n2
·
1
2
(n − 1)n =
3
2
+
9
4
n − 1
n
A = lim
n→+∞
3
2
+
9
4
1 −
1
n
=
3
2
+
9
4
=
15
4

44. ∆x =
5
n
, x∗
k =
5
n
(k − 1)
f(x∗
k)∆x = (5 − x∗
k)∆x = 5 −
5
n
(k − 1)
5
n
=
25
n
−
25
n2
(k − 1)
n
k=1
f(x∗
k)∆x =
25
n
n
k=1
1 −
25
n2
n
k=1
(k − 1) = 25 −
25
2
n − 1
n
A = lim
n→+∞
25 −
25
2
1 −
1
n
= 25 −
25
2
=
25
2
45. ∆x =
3
n
, x∗
k = 0 + (k − 1)
3
n
; f(x∗
k)∆x = (9 − 9
(k − 1)2
n2
)
3
n
n
k=1
f(x∗
k)∆x =
n
k=1
9 − 9
(k − 1)2
n2
3
n
=
27
n
n
k=1
1 −
(k − 1)2
n2
= 27 −
27
n3
n
k=1
k2
+
54
n3
n
k=1
k −
27
n2
A = lim
n→+∞
= 27 − 27
1
3
+ 0 + 0 = 18
46. ∆x =
3
n
, x∗
k = (k − 1)
3
n
f(x∗
k)∆x = 4 −
1
4
(x∗
k)2
∆x = 4 −
1
4
9(k − 1)2
n2
3
n
=
12
n
−
27k2
4n3
+
27k
2n3
−
27
4n3
n
k=1
f(x∗
k)∆x =
n
k=1
12
n
−
27
4n3
n
k=1
k2
+
27
2n3
n
k=1
k −
27
4n3
n
k=1
1
= 12 −
27
4n3
·
1
6
n(n + 1)(2n + 1) +
27
2n3
n(n + 1)
2
−
27
4n2
= 12 −
9
8
(n + 1)(2n + 1)
n2
+
27
4n
+
27
4n2
−
27
4n2
A = lim
n→+∞
12 −
9
8
1 +
1
n
2 +
1
n
+ 0 + 0 − 0 = 12 −
9
8
(1)(2) = 39/4
47. ∆x =
1
n
, x∗
k =
2k − 1
2n
f(x∗
k)∆x =
(2k − 1)2
(2n)2
1
n
=
k2
n3
−
k
n3
+
1
4n3
n
k=1
f(x∗
k)∆x =
1
n3
n
k=1
k2
−
1
n3
n
k=1
k +
1
4n3
n
k=1
1
Using Theorem 5.4.4,
A = lim
n→+∞
n
k=1
f(x∗
k)∆x =
1
3
+ 0 + 0 =
1
3
48. ∆x =
2
n
, x∗
k = −1 +
2k − 1
n
f(x∗
k)∆x = −1 +
2k − 1
n
2
2
n
=
8k2
n3
−
8k
n3
+
2
n3
−
2
n

200 Chapter 5
n
k=1
f(x∗
kk)∆x =
8
n3
n
k=1
k2
−
8
n3
n
k=1
k +
2
n2
− 2
A = lim
n→+∞
n
k=1
f(x∗
k)∆x =
8
3
+ 0 + 0 − 2 =
2
3
49. ∆x =
2
n
, x∗
k = −1 +
2k
n
f(x∗
k)∆x = −1 +
2k
n
2
n
= −
2
n
+ 4
k
n2
n
k=1
f(x∗
k)∆x = −2 +
4
n2
n
k=1
k = −2 +
4
n2
n(n + 1)
2
= −2 + 2 +
2
n
A = lim
n→+∞
n
k=1
f(x∗
k)∆x = 0
The area below the x-axis cancels the area above the x-axis.
50. ∆x =
3
n
, x∗
k = −1 +
3k
n
f(x∗
k)∆x = −1 +
3k
n
3
n
= −
3
n
+
9
n2
k
n
k=1
f(x∗
k)∆x = −3 +
9
n2
n(n + 1)
2
A = lim
n→+∞
n
k=1
f(x∗
k)∆x = −3 +
9
2
+ 0 =
3
2
The area below the x-axis cancels the area above the x-axis that lies to the right of the line x = 1;
the remaining area is a trapezoid of width 1 and heights 1, 2, hence its area is
1 + 2
2
=
3
2
51. ∆x =
2
n
, x∗
k =
2k
n
f(x∗
k) =
2k
n
2
− 1
2
n
=
8k2
n3
−
2
n
n
k=1
f(x∗
k)∆x =
8
n3
n
k=1
k2
−
2
n
n
k=1
1 =
8
n3
n(n + 1)(2n + 1)
6
− 2
A = lim
n→+∞
n
k=1
f(x∗
k)∆x =
16
6
− 2 =
2
3
52. ∆x =
2
n
, x∗
k = −1 +
2k
n
f(x∗
k)∆x = −1 +
2k
n
3
2
n
= −
2
n
+ 12
k
n2
− 24
k2
n3
+ 16
k3
n4
n
k=1
f(x∗
k)∆x = −2 +
12
n2
n(n + 1)
2
−
24
n3
n(n + 1)(2n + 1)
6
+
16
n4
n(n + 1)
2
2
A = lim
n→+∞
n
k=1
f(x∗
k) = −2 +
12
2
−
48
6
+
16
22
= 0

53. ∆x =
b − a
n
, x∗
k = a +
b − a
n
(k − 1)
f(x∗
k)∆x = mx∗
k∆x = m a +
b − a
n
(k − 1)
b − a
n
= m(b − a)
a
n
+
b − a
n2
(k − 1)
n
k=1
f(x∗
k)∆x = m(b − a) a +
b − a
2
·
n − 1
n
A = lim
n→+∞
m(b − a) a +
b − a
2
1 −
1
n
= m(b − a)
b + a
2
=
1
2
m(b2
− a2
)
54. ∆x =
b − a
n
, x∗
k = a +
k
n
(b − a)
f(x∗
k)∆x =
ma
n
(b − a) +
mk
n2
(b − a)2
n
k=1
f(x∗
k)∆x = ma(b − a) +
m
n2
(b − a)2 n(n + 1)
2
A = lim
n→+∞
n
k=1
f(x∗
k)∆x = ma(b − a) +
m
2
(b − a)2
= m(b − a)
a + b
2
55. (a) With x∗
k as the right endpoint, ∆x =
b
n
, x∗
k =
b
n
k
f(x∗
k)∆x = (x∗
k)3
∆x =
b4
n4
k3
,
n
k=1
f(x∗
k)∆x =
b4
n4
n
k=1
k3
=
b4
4
(n + 1)2
n2
A = lim
n→+∞
b4
4
1 +
1
n
2
= b4
/4
(b) ∆x =
b − a
n
, x∗
k = a +
b − a
n
k
f(x∗
k)∆x = (x∗
k)3
∆x = a +
b − a
n
k
3
b − a
n
=
b − a
n
a3
+
3a2
(b − a)
n
k +
3a(b − a)2
n2
k2
+
(b − a)3
n3
k3
n
k=1
f(x∗
k)∆x = (b − a) a3
+
3
2
a2
(b − a)
n + 1
n
+
1
2
a(b − a)2 (n + 1)(2n + 1)
n2
+
1
4
(b − a)3 (n + 1)2
n2
A = lim
n→+∞
n
k=1
f(x∗
k)∆x
= (b − a) a3
+
3
2
a2
(b − a) + a(b − a)2
+
1
4
(b − a)3
=
1
4
(b4
− a4
).
56. Let A be the area of the region under the curve and above the interval 0 ≤ x ≤ 1 on the x-axis,
and let B be the area of the region between the curve and the interval 0 ≤ y ≤ 1 on the y-axis.
Together A and B form the square of side 1, so A + B = 1.
But B can also be considered as the area between the curve x = y2
and the interval 0 ≤ y ≤ 1 on
the y-axis. By Exercise 47 above, B =
1
3
,so A = 1 −
1
3
=
2
3
.

202 Chapter 5
57. If n = 2m then 2m + 2(m − 1) + · · · + 2 · 2 + 2 = 2
m
k=1
k = 2 ·
m(m + 1)
2
= m(m + 1) =
n2
+ 2n
4
;
if n = 2m + 1 then (2m + 1) + (2m − 1) + · · · + 5 + 3 + 1 =
m+1
k=1
(2k − 1)
= 2
m+1
k=1
k −
m+1
k=1
1 = 2 ·
(m + 1)(m + 2)
2
− (m + 1) = (m + 1)2
=
n2
+ 2n + 1
4
58. 50·30+49·29+· · ·+22·2+21·1 =
30
k=1
k(k+20) =
30
k=1
k2
+20
30
k=1
k =
30 · 31 · 61
6
+20
30 · 31
2
= 18,755
59. both are valid 60. none is valid
61.
n
k=1
(ak − bk) = (a1 − b1) + (a2 − b2) + · · · + (an − bn)
= (a1 + a2 + · · · + an) − (b1 + b2 + · · · + bn) =
n
k=1
ak −
n
k=1
bk
62.
n
k=1
(k + 1)4
− k4
= (n + 1)4
− 1 (telescoping sum), expand the
quantity in brackets to get
n
k=1
(4k3
+ 6k2
+ 4k + 1) = (n + 1)4
− 1,
4
n
k=1
k3
+ 6
n
k=1
k2
+ 4
n
k=1
k +
n
k=1
1 = (n + 1)4
− 1
n
k=1
k3
=
1
4
(n + 1)4
− 1 − 6
n
k=1
k2
− 4
n
k=1
k −
n
k=1
1
=
1
4
[(n + 1)4
− 1 − n(n + 1)(2n + 1) − 2n(n + 1) − n]
=
1
4
(n + 1)[(n + 1)3
− n(2n + 1) − 2n − 1]
=
1
4
(n + 1)(n3
+ n2
) =
1
4
n2
(n + 1)2
63. (a)
n
k=1
1 means add 1 to itself n times, which gives the result.
(b)
1
n2
n
k=1
k =
1
n2
n(n + 1)
2
=
1
2
+
1
2n
, so lim
n→+∞
1
n2
n
k=1
k =
1
2
(c)
1
n3
n
k=1
k2
=
1
n3
n(n + 1)(2n + 1)
6
=
2
6
+
3
6n
+
1
6n2
, so lim
n→+∞
1
n3
n
k=1
k2
=
1
3
(d)
1
n4
n
k=1
k3
=
1
n4
n(n + 1)
2
2
=
1
4
+
1
2n
+
1
4n2
, so lim
n→+∞
1
n4
n
k=1
k3
=
1
4

EXERCISE SET 5.5
1. (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/6 (b) 2
2. (a) (
√
2/2)(π/2) + (−1)(3π/4) + (0)(π/2) + (
√
2/2)(π/4) = 3(
√
2 − 2)π/8
(b) 3π/4
3. (a) (−9/4)(1) + (3)(2) + (63/16)(1) + (−5)(3) = −117/16
(b) 3
4. (a) (−8)(2) + (0)(1) + (0)(1) + (8)(2) = 0 (b) 2
5.
2
−1
x2
dx 6.
2
1
x3
dx
7.
3
−3
4x(1 − 3x)dx 8.
π/2
0
sin2
x dx
9. (a) lim
max ∆xk→0
n
k=1
2x∗
k∆xk; a = 1, b = 2 (b) lim
max ∆xk→0
n
k=1
x∗
k
x∗
k + 1
∆xk; a = 0, b = 1
10. (a) lim
max ∆xk→0
n
k=1
x∗
k ∆xk, a = 1, b = 2
(b) lim
max ∆xk→0
n
k=1
(1 + cos x∗
k) ∆xk, a = −π/2, b = π/2
11. (a) A =
1
2
(3)(3) = 9/2
3
x
y
A
(b) −A = −
1
2
(1)(1 + 2) = −3/2
-2 -1 x
y
A
(c) −A1 + A2 = −
1
2
+ 8 = 15/2
-1
4
x
y
A1
A2
(d) −A1 + A2 = 0
-5
5
x
y
A1
A2

204 Chapter 5
12. (a) A =
1
2
(1)(2) = 1
2
1
x
y
A
(b) A =
1
2
(2)(3/2 + 1/2) = 2
-1 1
1
x
y
A
(c) −A = −
1
2
(1/2)(1) = −1/4
2
1
x
y
A
(d) A1 − A2 = 1 − 1/4 = 3/4
2
1
x
y
A2
A1
13. (a) A = 2(5) = 10
y
x
1
2
5
A
(b) 0; A1 = A2 by symmetry
6
c x
y
A1
A2
(c) A1 + A2 =
1
2
(5)(5/2) +
1
2
(1)(1/2)
= 13/2
-1
5
2
x
y
3
2
A1
A2
(d)
1
2
[π(1)2
] = π/2
y
x
1
-1 1
A
14. (a) A = (6)(5) = 30
-10 -5
6
x
y
A
(b) −A1 + A2 = 0 because
A1 = A2 by symmetry
$
4
x
y
A1
A2
(c) A1 + A2 =
1
2
(2)(2) +
1
2
(1)(1) = 5/2
2
2
x
y
A1
A2
(d)
1
4
π(2)2
= π
y
x
2
2
A

15. (a) 0.8 (b) −2.6 (c) −1.8 (d) −0.3
16. (a)
1
0
f(x)dx =
1
0
2xdx = x2
1
0
= 1
(b)
1
−1
f(x)dx =
1
−1
2xdx = x2
1
−1
= 12
− (−1)2
= 0
(c)
10
1
f(x)dx =
10
1
2dx = 2x
10
1
= 18
(d)
5
1/2
f(x)dx =
1
1/2
2xdx +
5
1
2dx = x2
1
1/2
+ 2x


5
1
= 12
−(1/2)2
+2·5−2·1 = 3/4+8 = 35/4
17.
2
−1
f(x)dx + 2
2
−1
g(x)dx = 5 + 2(−3) = −1
18. 3
4
1
f(x)dx −
4
1
g(x)dx = 3(2) − 10 = −4
19.
5
1
f(x)dx =
5
0
f(x)dx −
1
0
f(x)dx = 1 − (−2) = 3
20.
−2
3
f(x)dx = −
3
−2
f(x)dx = −
1
−2
f(x)dx +
3
1
f(x)dx = −(2 − 6) = 4
21. (a)
1
0
xdx + 2
1
0
1 − x2dx = 1/2 + 2(π/4) = (1 + π)/2
(b) 4
3
−1
dx − 5
3
−1
xdx = 4 · 4 − 5(−1/2 + (3 · 3)/2) = −4
22. (a)
0
−3
2dx +
0
−3
9 − x2dx = 2 · 3 + (π(3)2
)/4 = 6 + 9π/4
(b)
2
−2
dx − 3
2
−2
|x|dx = 4 · 1 − 3(2)(2 · 2)/2 = −8
23. (a)
√
x > 0, 1 − x < 0 on [2, 3] so the integral is negative
(b) x2
> 0, 3 − cos x > 0 for all x so the integral is positive
24. (a) x4
> 0,
√
3 − x > 0 on [−3, −1] so the integral is positive
(b) x3
− 9 < 0, |x| + 1 > 0 on [−2, 2] so the integral is negative
25.
10
0
25 − (x − 5)2dx = π(5)2
/2 = 25π/2 26.
3
0
9 − (x − 3)2dx = π(3)2
/4 = 9π/4
27.
1
0
(3x + 1)dx = 5/2 28.
2
−2
4 − x2dx = π(2)2
/2 = 2π

206 Chapter 5
29. (a) f is continuous on [−1, 1] so f is integrable there by Part (a) of Theorem 5.5.8
(b) |f(x)| ≤ 1 so f is bounded on [−1, 1], and f has one point of discontinuity, so by Part (b) of
Theorem 5.5.8 f is integrable on [−1, 1]
(c) f is not bounded on [-1,1] because lim
x→0
f(x) = +∞, so f is not integrable on [0,1]
(d) f(x) is discontinuous at the point x = 0 because lim
x→0
sin
1
x
does not exist. f is continuous
elsewhere. −1 ≤ f(x) ≤ 1 for x in [−1, 1] so f is bounded there. By Part (b), Theorem 5.5.8,
f is integrable on [−1, 1].
30. Each subinterval of a partition of [a, b] contains both rational and irrational numbers. If all x∗
k are
chosen to be rational then
n
k=1
f(x∗
k)∆xk =
n
k=1
(1)∆xk =
n
k=1
∆xk = b − a so lim
max ∆xk→0
n
k=1
f(x∗
k)∆xk = b − a.
If all x∗
k are irrational then lim
max ∆xk→0
n
k=1
f(x∗
k)∆xk = 0. Thus f is not integrable on [a, b] because
the preceding limits are not equal.
31. (a) Let Sn =
n
k=1
f(x∗
k)∆xk and S =
b
a
f(x)dx then
n
k=1
cf(x∗
k)∆xk = cSn and we want to prove
that lim
max ∆xk→0
cSn = cS. If c = 0 the result follows immediately, so suppose that c = 0 then
for any > 0, |cSn − cS| = |c||Sn − S| < if |Sn − S| < /|c|. But because f is integrable
on [a, b], there is a number δ > 0 such that |Sn − S| < /|c| whenever max ∆xk < δ so
|cSn − cS| < and hence lim
max ∆xk→0
cSn = cS.
(b) Let Rn =
n
k=1
f(x∗
k)∆xk, Sn =
n
k=1
g(x∗
k)∆xk, Tn =
n
k=1
[f(x∗
k) + g(x∗
k)]∆xk, R =
b
a
f(x)dx,
and S =
b
a
g(x)dx then Tn = Rn + Sn and we want to prove that lim
max ∆xk→0
Tn = R + S.
|Tn − (R + S)| = |(Rn − R) + (Sn − S)| ≤ |Rn − R| + |Sn − S|
so for any > 0 |Tn − (R + S)| < if |Rn − R| + |Sn − S| < .
Because f and g are integrable on [a, b], there are numbers δ1 and δ2 such that
|Rn − R| < /2 for max ∆xk < δ1 and |Sn − S| < /2 for max ∆xk < δ2.
If δ = min(δ1, δ2) then |Rn − R| < /2 and |Sn − S| < /2 for max ∆xk < δ thus
|Rn − R| + |Sn − S| < and so |Tn − (R + S)| < for max ∆xk < δ which shows that
lim
max ∆xk→0
Tn = R + S.
32. For the smallest, ﬁnd x∗
k so that f(x∗
k) is minimum on each subinterval: x∗
1 = 1, x∗
2 = 3/2, x∗
3 = 3
so (2)(1) + (7/4)(2) + (4)(1) = 9.5. For the largest, ﬁnd x∗
k so that f(x∗
k) is maximum on each
subinterval: x∗
1 = 0, x∗
2 = 3, x∗
3 = 4 so (4)(1) + (4)(2) + (8)(1) = 20.
33. ∆xk =
4k2
n2
−
4(k − 1)2
n2
=
4
n2
(2k − 1), x∗
k =
4k2
n2
,
f(x∗
k) =
2k
n
, f(x∗
k)∆xk =
8k
n3
(2k − 1) =
8
n3
(2k2
− k),
n
k=1
f(x∗
k)∆xk =
8
n3
n
k=1
(2k2
− k) =
8
n3
1
3
n(n + 1)(2n + 1) −
1
2
n(n + 1) =
4
3
(n + 1)(4n − 1)
n2
,
lim
n→+∞
n
k=1
f(x∗
k)∆xk = lim
n→+∞
4
3
1 +
1
n
4 −
1
n
=
16
3
.

34. For any partition of [a, b] use the right endpoints to form the sum
n
k=1
f(x∗
k)∆xk. Since f(x∗
k) = 0
for each k, the sum is zero and so is
b
a
f(x) dx = lim
n→+∞
n
k=1
f(x∗
k)∆xk.
35. With f(x) = g(x) then f(x) − g(x) = 0 for a < x ≤ b. By Theorem 5.5.4(b)
b
a
f(x) dx =
b
a
[(f(x) − g(x) + g(x)]dx =
b
a
[f(x) − g(x)]dx +
b
a
g(x)dx.
But the first term on the right hand side is zero (from Exercise 34), so
b
a
f(x) dx =
b
a
g(x) dx
36. Choose any large positive integer N and any partition of [0, a]. Then choose x∗
1 in the first interval
so small that f(x∗
1)∆x1 > N. For example choose x∗
1 < ∆x1/N. Then with this partition and
choice of x∗
1,
n
k=1
f(x∗
k)∆xk > f(x∗
1)∆x1 > N. This shows that the sum is dependent on partition
and/or points, so Definition 5.5.1 is not satisfied.
EXERCISE SET 5.6
1. (a)
2
0
(2 − x)dx = (2x − x2
/2)
2
0
= 4 − 4/2 = 2
(b)
1
−1
2dx = 2x
1
−1
= 2(1) − 2(−1) = 4
(c)
3
1
(x + 1)dx = (x2
/2 + x)
3
1
= 9/2 + 3 − (1/2 + 1) = 6
2. (a)
5
0
xdx = x2
/2
5
0
= 25/2 (b)
9
3
5dx = 5x
9
3
= 5(9) − 5(3) = 30
(c)
2
−1
(x + 3)dx = (x2
/2 + 3x)
2
−1
= 4/2 + 6 − (1/2 − 3) = 21/2
3.
3
2
x3
dx = x4
/4
3
2
= 81/4 − 16/4 = 65/4 4.
1
−1
x4
dx = x5
/5
1
−1
= 1/5 − (−1)/5 = 2/5
5.
9
1
√
xdx =
2
3
x3/2
9
1
=
2
3
(27 − 1) = 52/3 6.
4
1
x−3/5
dx =
5
2
x2/5
4
1
=
5
2
(42/5
− 1)
7.
1
3
x3
− 2x2
+ 7x
0
−3
= 48 8.
1
2
x2
+
1
5
x5
2
−1
= 81/10
9.
3
1
x−2
dx = −
1
x
3
1
= 2/3 10.
2
1
x−6
dx = −
1
5x5
2
1
= 31/160

208 Chapter 5
11.
4
5
x5/2
9
4
= 844/5 12. 3x5/3
+
4
x
8
1
= 179/2
13. − cos θ
π/2
−π/2
= 0 14. tan θ
π/4
0
= 1
15. sin x
π/4
−π/4
=
√
2 16.
1
2
x2
− sec x
1
0
= 3/2 − sec(1)
17. 6
√
t −
10
3
t3/2
+
2
√
t
4
1
= −55/3 18. 8
√
y +
4
3
y3/2
−
2
3y3/2
9
4
= 10819/324
19.
1
2
x2
− 2 cot x
π/2
π/6
= π2
/9 + 2
√
3 20. a1/2
x −
2
3
x3/2
4a
a
= −
5
3
a3/2
21. (a)
3/2
0
(3 − 2x)dx +
2
3/2
(2x − 3)dx = (3x − x2
)
3/2
0
+ (x2
− 3x)
2
3/2
= 9/4 + 1/4 = 5/2
(b)
π/2
0
cos x dx +
3π/4
π/2
(− cos x)dx = sin x
π/2
0
− sin x
3π/4
π/2
= 2 −
√
2/2
22. (a)
0
−1
√
2 − x dx +
2
0
√
2 + x dx = −
2
3
(2 − x)3/2
0
−1
+
2
3
(2 + x)3/2
2
0
= −
2
3
(2
√
2 − 3
√
3) +
2
3
(8 − 2
√
2) =
2
3
(8 − 4
√
2 + 3
√
3)
(b)
π/6
0
(1/2 − sin x) dx +
π/2
π/6
(sin x − 1/2) dx
= (x/2 + cos x)
π/6
0
− (cos x + x/2)
π/2
π/6
= (π/12 +
√
3/2) − 1 − π/4 + (
√
3/2 + π/12) =
√
3 − π/12 − 1
23. (a) 17/6 (b) F(x) =



1
2
x2
, x ≤ 1
1
3
x3
+
1
6
, x > 1
24. (a)
1
0
√
x dx +
4
1
1
x2
dx =
2
3
x3/2
1
0
−
1
x
4
1
= 17/12
(b) F(x) =



2
3
x3/2
, x < 1
−
1
x
+
5
3
, x ≥ 1
25. 0.665867079;
3
1
1
x2
dx = −
1
x
3
1
= 2/3 26. 1.000257067;
π/2
0
sin x dx = − cos x
π/2
0
= 1

27. 3.106017890;
1
−1
sec2
x dx = tan x
1
−1
= 2 tan 1 ≈ 3.114815450
29. A =
3
0
(x2
+ 1)dx =
1
3
x3
+ x
3
0
= 12
30. A =
2
1
(−x2
+ 3x − 2)dx = −
1
3
x3
+
3
2
x2
− 2x
2
1
= 1/6
31. A =
2π/3
0
3 sin x dx = −3 cos x
2π/3
0
= 9/2 32. A = −
−1
−2
x3
dx = −
1
4
x4
−1
−2
= 15/4
33. A1 =
−2
−3
(x2
− 3x − 10)dx =
1
3
x3
−
3
2
x2
− 10x
−2
−3
= 23/6,
A2 = −
5
−2
(x2
− 3x − 10)dx = 343/6,
A3 =
8
5
(x2
− 3x − 10)dx = 243/6, A = A1 + A2 + A3 = 203/2 -3
-2
85
10
x
y
A1
A2
A3
34. (a) the area is positive
(b)
5
−2
1
100
x3
−
1
20
x2
−
1
25
x +
1
5
dx =
1
400
x4
−
1
60
x3
−
1
50
x2
+
1
5
x
5
−2
=
343
1200
35. (a) the area between the curve and the x-axis breaks into equal parts, one above and one below
the x-axis, so the integral is zero
(b)
1
−1
x3
dx =
1
4
x4
1
−1
=
1
4
(14
− (−1)4
) = 0;
π/2
−π/2
sin xdx = − cos x
π/2
−π/2
= − cos(π/2) + cos(−π/2) = 0 + 0 = 0
(c) The area on the left side of the y-axis is equal to the area on the right side, so
a
−a
f(x)dx = 2
a
0
f(x)dx
(d)
1
−1
x2
dx =
1
3
x3
1
−1
=
1
3
(13
− (−1)3
) =
2
3
= 2
1
0
x2
dx;
π/2
−π/2
cos xdx = sin x
π/2
−π/2
= sin(π/2) − sin(−π/2) = 1 + 1 = 2 = 2
π/2
0
cos xdx
36. The numerator is an odd function and the denominator is an even function, so the integrand is an
odd function and the integral is zero.
37. (a) x3
+ 1 (b) F(x) =
1
4
t4
+ t
x
1
=
1
4
x4
+ x −
5
4
; F (x) = x3
+ 1
38. (a) cos 2x (b) F(x) =
1
2
sin 2t
x
π/4
=
1
2
sin 2x −
1
2
, F (x) = cos 2x

210 Chapter 5
39. (a) sin
√
x (b)
√
1 + cos2 x 40. (a)
1
1 +
√
x
(b)
1
1 + x + x2
41. −
x
cos x
42. |u|
43. F (x) =
√
3x2 + 1, F (x) =
3x
√
3x2 + 1
(a) 0 (b)
√
13 (c) 6/
√
13
44. F (x) =
cos x
x2 + 3
, F (x) =
−(x2
+ 3) sin x − 2x cos x
(x2 + 3)2
(a) 0 (b) 1/3 (c) 0
45. (a) F (x) =
x − 3
x2 + 7
= 0 when x = 3, which is a relative minimum, and hence the absolute
minimum, by the ﬁrst derivative test.
(b) increasing on [3, +∞), decreasing on (−∞, 3]
(c) F (x) =
7 + 6x − x2
(x2 + 7)2
=
(7 − x)(1 + x)
(x2 + 7)2
; concave up on (−1, 7), concave down on (−∞, −1)
and on (7, +∞)
46. F
t
2
3
-20 -10 20
47. (a) (0, +∞) because f is continuous there and 1 is in (0, +∞)
(b) at x = 1 because F(1) = 0
48. (a) (−3, 3) because f is continuous there and 1 is in (−3, 3)
(b) at x = 1 because F(1) = 0
49. (a) fave =
1
9
9
0
x1/2
dx = 2;
√
x∗ = 2, x∗
= 4
(b) fave =
1
3
2
−1
(3x2
+ 2x + 1) dx =
1
3
(x3
+ x2
+ x)
2
−1
= 5; 3x∗2
+ 2x∗
+ 1 = 5,
with solutions x∗
= −(1/3)(1 ±
√
13), but only x∗
= −(1/3)(1 −
√
13) lies in the interval
[−1, 2].
50. (a) fave =
1
2π
π
−π
sin x dx = 0; sin x∗
= 0, x∗
= −π, 0, π
(b) fave =
1
2
3
1
1
x2
dx =
1
3
;
1
(x∗)2
=
1
3
, x∗
=
√
3

51.
√
2 ≤
√
x3 + 2 ≤
√
29, so 3
√
2 ≤
3
0
x3 + 2dx ≤ 3
√
29
52. Let f(x) = x sin x, f(0) = f(1) = 0, f (x) = sin x + x cos x = 0 when x = − tan x, x ≈ 2.0288,
so f has an absolute maximum at x ≈ 2.0288; f(2.0288) ≈ 1.8197, so 0 ≤ x sin x ≤ 1.82 and
0 ≤
π
0
x sin xdx ≤ 1.82π = 5.72
53. (a) cF(x)
b
a
= cF(b) − cF(a) = c[F(b) − F(a)] = c F(x)
b
a
(b) F(x) + G(x)
b
a
= [F(b) + G(b)] − [F(a) + G(a)]
= [F(b) − F(a)] + [G(b) − G(a)] = F(x)
b
a
+ G(x)
b
a
(c) F(x) − G(x)
b
a
= [F(b) − G(b)] − [F(a) − G(a)]
= [F(b) − F(a)] − [G(b) − G(a)] = F(x)
b
a
− G(x)
b
a
54. Let f be continuous on a closed interval [a, b] and let F be an antiderivative of f on [a, b]. By
Theorem 4.8.2,
F(b) − F(a)
b − a
= F (x∗
) for some x∗
in (a, b). By Theorem 5.6.1,
b
a
f(x) dx = F(b) − F(a), i.e.
b
a
f(x) dx = F (x∗
)(b − a) = f(x∗
)(b − a).
EXERCISE SET 5.7
1. (a) the increase in height in inches, during the ﬁrst ten years
(b) the change in the radius in centimeters, during the time interval t = 1 to t = 2 seconds
(c) the change in the speed of sound in ft/s, during an increase in temperature from t = 32◦
F
to t = 100◦
F
(d) the displacement of the particle in cm, during the time interval t = t1 to t = t2 seconds
2. (a)
1
0
V (t)dt gal
(b) the change f(x1) − f(x2) in the values of f over the interval
3. (a) displ = s(3) − s(0)
=
3
0
v(t)dt =
2
0
(1 − t)dt +
3
2
(t − 3)dt = (t − t2
/2)
2
0
+ (t2
/2 − 3t)
3
2
= −1/2;
dist =
3
0
|v(t)|dt = (t − t2
/2)
1
0
+ (t2
/2 − t)
2
1
− (t2
/2 − 3t)
3
2
= 3/2
(b) displ = s(3) − s(0)
=
3
0
v(t)dt =
1
0
tdt +
2
1
dt +
3
2
(5 − 2t)dt = t2
/2
1
0
+ t
2
1
+ (5t − t2
)
3
2
= 3/2;
dist =
1
0
tdt +
2
1
dt +
5/2
2
(5 − 2t)dt +
3
5/2
(2t − 5)dt
= t2
/2
1
0
+ t
2
1
+ (5t − t2
)
5/2
2
+ (t2
− 5t)
3
5/2
= 2

212 Chapter 5
4. v
t
-1
1
2 4 8 10
5. (a) v(t) = 20 +
t
0
a(u)du; add areas of the small blocks to get
v(4) ≈ 20 + 1.4 + 3.0 + 4.7 + 6.2 = 35.3 m/s
(b) v(6) = v(4) +
6
4
a(u)du ≈ 35.3 + 7.5 + 8.6 = 51.4 m/s
6. a > 0 and therefore (Theorem 5.5.6(a)) v > 0, so the particle is always speeding up for 0 < t < 10
7. (a) s(t) = (t3
− 2t2
+ 1)dt =
1
4
t4
−
2
3
t3
+ t + C,
s(0) =
1
4
(0)4
−
2
3
(0)3
+ 0 + C = 1, C = 1, s(t) =
1
4
t4
−
2
3
t3
+ t + 1
(b) v(t) = 4 cos 2t dt = 2 sin 2t + C1, v(0) = 2 sin 0 + C1 = −1, C1 = −1,
v(t) = 2 sin 2t − 1, s(t) = (2 sin 2t − 1)dt = − cos 2t − t + C2,
s(0) = − cos 0 − 0 + C2 = −3, C2 = −2, s(t) = − cos 2t − t − 2
8. (a) s(t) = (1 + sin t)dt = t − cos t + C, s(0) = 0 − cos 0 + C = −3, C = −2, s(t) = t − cos t − 2
(b) v(t) = (t2
− 3t + 1)dt =
1
3
t3
−
3
2
t2
+ t + C1,
v(0) =
1
3
(0)3
−
3
2
(0)2
+ 0 + C1 = 0, C1 = 0, v(t) =
1
3
t3
−
3
2
t2
+ t,
s(t) =
1
3
t3
−
3
2
t2
+ t dt =
1
12
t4
−
1
2
t3
+
1
2
t2
+ C2,
s(0) =
1
12
(0)4
−
1
2
(0)3
+
1
2
(0)2
+ C2 = 0, C2 = 0, s(t) =
1
12
t4
−
1
2
t3
+
1
2
t2
9. (a) s(t) = (2t − 3)dt = t2
− 3t + C, s(1) = (1)2
− 3(1) + C = 5, C = 7, s(t) = t2
− 3t + 7
(b) v(t) = cos tdt = sin t + C1, v(π/2) = 2 = 1 + C1, C1 = 1, v(t) = sin t + 1,
s(t) = (sin t + 1)dt = − cos t + t + C2, s(π/2) = 0 = π/2 + C2, C2 = −π/2,
s(t) = − cos t + t − π/2
10. (a) s(t) = t2/3
dt =
3
5
t5/3
+ C, s(8) = 0 =
3
5
32 + C, C = −
96
5
, s(t) =
3
5
t5/3
−
96
5
(b) v(t) =
√
tdt =
2
3
t3/2
+ C1, v(4) = 1 =
2
3
8 + C1, C1 = −
13
3
, v(t) =
2
3
t3/2
−
13
3
,

s(t) =
2
3
t3/2
−
13
3
dt =
4
15
t5/2
−
13
3
t + C2, s(4) = −5 =
4
15
32 −
13
3
4 + C2 = −
44
5
+ C2,
C2 =
19
5
, s(t) =
4
15
t5/2
−
13
3
t +
19
5
11. (a) displacement = s(π/2) − s(0) =
π/2
0
sin tdt = − cos t
π/2
0
= 1 m
distance =
π/2
0
| sin t|dt = 1 m
(b) displacement = s(2π) − s(π/2) =
2π
π/2
cos tdt = sin t
2π
π/2
= −1 m
distance =
2π
π/2
| cos t|dt = −
3π/2
π/2
cos tdt +
2π
3π/2
cos tdt = 3 m
12. (a) displacement = s(6) − s(0) =
6
0
(2t − 4)dt = (t2
− 4t)
6
0
= 12 m
distance =
6
0
|2t−4|dt =
2
0
(4−2t)dt+
6
2
(2t−4)dt = (4t−t2
)
2
0
+(t2
−4t)
6
2
= 20 m
(b) displacement =
5
0
|t − 3|dt =
3
0
−(t − 3)dt +
5
3
(t − 3)dt = 13/2 m
distance =
5
0
|t − 3|dt = 13/2 m
13. (a) v(t) = t3
− 3t2
+ 2t = t(t − 1)(t − 2)
displacement =
3
0
(t3
− 3t2
+ 2t)dt = 9/4 m
distance =
3
0
|v(t)|dt =
1
0
v(t)dt +
2
1
−v(t)dt +
3
2
v(t)dt = 11/4 m
(b) displacement =
3
0
(
√
t − 2)dt = 2
√
3 − 6 m
distance =
3
0
|v(t)|dt = −
3
0
v(t)dt = 6 − 2
√
3 m
14. (a) displacement =
3
1
(
1
2
−
1
t2
)dt = 1/3 m
distance =
3
1
|v(t)|dt = −
√
2
1
v(t)dt +
3
√
2
v(t)dt = 10/3 − 2
√
2 m
(b) displacement =
9
4
3t−1/2
dt = 6 m
distance =
9
4
|v(t)|dt =
9
4
v(t)dt = 6 m
15. v(t) = −2t + 3
displacement =
4
1
(−2t + 3)dt = −6 m
distance =
4
1
| − 2t + 3|dt =
3/2
1
(−2t + 3)dt +
4
3/2
(2t − 3)dt = 13/2 m

214 Chapter 5
16. v(t) =
1
2
t2
− 2t
displacement =
5
1
1
2
t2
− 2t dt = −10/3 m
distance =
5
1
1
2
t2
− 2t dt =
4
1
−
1
2
t2
− 2t dt +
5
4
1
2
t2
− 2t dt = 17/3 m
17. v(t) =
2
5
√
5t + 1 +
8
5
displacement =
3
0
2
5
√
5t + 1 +
8
5
dt =
4
75
(5t + 1)3/2
+
8
5
t
3
0
= 204/25 m
distance =
3
0
|v(t)|dt =
3
0
v(t)dt = 204/25 m
18. v(t) = − cos t + 2
displacement =
π/2
π/4
(− cos t + 2)dt = (π +
√
2 − 2)/2 m
distance =
π/2
π/4
| − cos t + 2|dt =
π/2
π/4
(− cos t + 2)dt = (π +
√
2 − 2)/2 m
19. (a) s = sin
1
2
πt dt = −
2
π
cos
1
2
πt + C
s = 0 when t = 0 which gives C =
2
π
so s = −
2
π
cos
1
2
πt +
2
π
.
a =
dv
dt
=
π
2
cos
1
2
πt. When t = 1 : s = 2/π, v = 1, |v| = 1, a = 0.
(b) v = −3 t dt = −
3
2
t2
+ C1, v = 0 when t = 0 which gives C1 = 0 so v = −
3
2
t2
s = −
3
2
t2
dt = −
1
2
t3
+ C2, s = 1 when t = 0 which gives C2 = 1 so s = −
1
2
t3
+ 1.
When t = 1 : s = 1/2, v = −3/2, |v| = 3/2, a = −3.
20. (a) negative, because v is decreasing
(b) speeding up when av > 0, so 2 < t < 5; slowing down when 1 < t < 2
(c) negative, because the area between the graph of v(t) and the t-axis appears to be greater
where v < 0 compared to where v > 0
21. A = A1 + A2 =
1
0
(1 − x2
)dx +
3
1
(x2
− 1)dx = 2/3 + 20/3 = 22/3
22. A = A1 + A2 =
π
0
sin xdx −
3π/2
π
sin xdx = 2 + 1 = 3
23. A = A1 + A2 =
0
−1
1 −
√
x + 1 dx +
1
0
[
√
x + 1 − 1] dx
= x −
2
3
(x + 1)3/2
0
−1
+
2
3
(x + 1)3/2
− x
1
0
= −
2
3
+ 1 +
4
√
2
3
− 1 −
2
3
= 4
√
2 − 1
3

24. A = A1 + A2 =
1
1/2
1 − x2
x2
dx +
2
1
x2
− 1
x2
dx = −
1
x
− x
1
1/2
+ x +
1
x
2
1
= −2 + 2 +
1
2
+ 2 +
1
2
− 2 = 1
25. s(t) =
20
3
t3
− 50t2
+ 50t + s0, s(0) = 0 gives s0 = 0, so s(t) =
20
3
t3
− 50t2
+ 50t, a(t) = 40t − 100
150
-100
0 6
150
-100
0 6
50
0
0 6
26. v(t) = 2t2
− 30t + v0, v(0) = 3 = v0, so v(t) = 2t2
− 30t + 3, s(t) =
2
3
t3
− 15t2
+ 3t + s0,
s(0) = −5 = s0, so s(t) =
2
3
t3
− 15t2
+ 3t − 5
1200
-1200
0 25
s(t)
500
-200
0 25
v(t)
70
-30
0 25
a(t)
–1
0
1
2
3
y
1 2 4 5
x
27. (a) From the graph the velocity is at ﬁrst positive, but then turns
negative, then positive again. The displacement, which is the
cumulative area from x = 0 to x = 5, starts positive, turns
negative, and then turns positive again.
(b) displ = 5/2 − sin 5 + 5 cos 5
28. (a) If t0 < 1 then the area between the velocity curve and the
t-axis, between t = 0 and t = t0, will always be positive,
so the displacement will be positive.
1
v
0.4 0.8
t
(b) displ =
π2
+ 4
2π2

216 Chapter 5
29. (a) a(t) =
0, t < 4
−10, t > 4
a t
-10
-5
2 4 12
(b) v(t) =
25, t < 4
65 − 10t, t > 4
v
t
-40
-20
20
2 4 6 8 10 12
(c) x(t) =
25t, t < 4
65t − 5t2
− 80, t > 4
, so x(8) = 120, x(12) = −20
(d) x(6.5) = 131.25
30. (a) From (9) t =
v − v0
a
; from that and (8)
s − s0 = v0
v − v0
a
+
1
2
a
(v − v0)2
a2
; multiply through by a to get
a(s − s0) = v0(v − v0) +
1
2
(v − v0)2
= (v − v0) v0 +
1
2
(v − v0) =
1
2
(v2
− v2
0). Thus
a =
v2
− v2
0
2(s − s0)
.
(b) Put the last result of Part (a) into the ﬁrst equation of Part (a) to obtain
t =
v − v0
a
= (v − v0)
2(s − s0)
v2 − v2
0
=
2(s − s0)
v + v0
.
(c) From (9) v0 = v − at; use this in (8) to get
s − s0 = (v − at)t +
1
2
at2
= vt −
1
2
at2
This expression contains no v0 terms and so diﬀers from (8).
31. (a) a = −1 mi/h/s = −22/15 ft/s2
(b) a = 30 km/h/min = 1/7200 km/s2
32. Take t = 0 when deceleration begins, then a = −10 so v = −10t + C1, but v = 88 when t = 0
which gives C1 = 88 thus v = −10t + 88, t ≥ 0
(a) v = 45 mi/h = 66 ft/s, 66 = −10t + 88, t = 2.2 s
(b) v = 0 (the car is stopped) when t = 8.8 s
s = v dt = (−10t + 88)dt = −5t2
+ 88t + C2, and taking s = 0 when t = 0, C2 = 0 so
s = −5t2
+ 88t. At t = 8.8, s = 387.2. The car travels 387.2 ft before coming to a stop.
33. a = a0 ft/s2
, v = a0t + v0 = a0t + 132 ft/s, s = a0t2
/2 + 132t + s0 = a0t2
/2 + 132t ft; s = 200 ft
when v = 88 ft/s. Solve 88 = a0t + 132 and 200 = a0t2
/2 + 132t to get a0 = −
121
5
when t =
20
11
,
so s = −12.1t2
+ 132t, v = −
121
5
t + 132.
(a) a0 = −
121
5
ft/s2
(b) v = 55 mi/h =
242
3
ft/s when t =
70
33
s
(c) v = 0 when t =
60
11
s

34. dv/dt = 3, v = 3t + C1, but v = v0 when t = 0 so C1 = v0, v = 3t + v0. From ds/dt = v = 3t + v0
we get s = 3t2
/2 + v0t + C2 and, with s = 0 when t = 0, C2 = 0 so s = 3t2
/2 + v0t. s = 40 when
t = 4 thus 40 = 3(4)2
/2 + v0(4), v0 = 4 m/s
35. Suppose s = s0 = 0, v = v0 = 0 at t = t0 = 0; s = s1 = 120, v = v1 at t = t1; and s = s2,
v = v2 = 12 at t = t2. From Exercise 30(a),
2.6 = a =
v2
1 − v2
0
2(s1 − s0)
, v2
1 = 2as1 = 5.2(120) = 624. Applying the formula again,
−1.5 = a =
v2
2 − v2
1
2(s2 − s1)
, v2
2 = v2
1 − 3(s2 − s1), so
s2 = s1 − (v2
2 − v2
1)/3 = 120 − (144 − 624)/3 = 280 m.
36. a(t) =
4, t < 2
0, t > 2
, so, with v0 = 0, v(t) =
4t, t < 2
8, t > 2
and,
since s0 = 0, s(t) =
2t2
, t < 2
8t − 8, t > 2
s = 100 when 8t − 8 = 100, t = 108/8 = 13.5 s
37. The truck’s velocity is vT = 50 and its position is sT = 50t+5000. The car’s acceleration is aC = 2,
so vC = 2t, sC = t2
(initial position and initial velocity of the car are both zero). sT = sC when
50t+5000 = t2
, t2
−50t−5000 = (t+50)(t−100) = 0, t = 100 s and sC = sT = t2
= 10, 000 ft.
38. Let t = 0 correspond to the time when the leader is 100 m from the ﬁnish line; let s = 0 cor-
respond to the ﬁnish line. Then vC = 12, sC = 12t − 115; aL = 0.5 for t > 0, vL = 0.5t + 8,
sL = 0.25t2
+ 8t − 100. sC = 0 at t = 115/12 ≈ 9.58 s, and sL = 0 at t = −16 + 4
√
41 ≈ 9.61,
so the challenger wins.
39. s = 0 and v = 112 when t = 0 so v(t) = −32t + 112, s(t) = −16t2
+ 112t
(a) v(3) = 16 ft/s, v(5) = −48 ft/s
(b) v = 0 when the projectile is at its maximum height so −32t + 112 = 0, t = 7/2 s,
s(7/2) = −16(7/2)2
+ 112(7/2) = 196 ft.
(c) s = 0 when it reaches the ground so −16t2
+ 112t = 0, −16t(t − 7) = 0, t = 0, 7 of which
t = 7 is when it is at ground level on its way down. v(7) = −112, |v| = 112 ft/s.
40. s = 112 when t = 0 so s(t) = −16t2
+ v0t + 112. But s = 0 when t = 2 thus
−16(2)2
+ v0(2) + 112 = 0, v0 = −24 ft/s.
41. (a) s(t) = 0 when it hits the ground, s(t) = −16t2
+ 16t = −16t(t − 1) = 0 when t = 1 s.
(b) The projectile moves upward until it gets to its highest point where v(t) = 0,
v(t) = −32t + 16 = 0 when t = 1/2 s.
42. (a) s(t) = 0 when the rock hits the ground, s(t) = −16t2
+ 555 = 0 when t =
√
555/4 s
(b) v(t) = −32t, v(
√
555/4) = −8
√
555, the speed at impact is 8
√
555 ft/s
43. (a) s(t) = 0 when the package hits the ground,
s(t) = −16t2
+ 20t + 200 = 0 when t = (5 + 5
√
33)/8 s
(b) v(t) = −32t + 20, v[(5 + 5
√
33)/8] = −20
√
33, the speed at impact is 20
√
33 ft/s
44. (a) s(t) = 0 when the stone hits the ground,
s(t) = −16t2
− 96t + 112 = −16(t2
+ 6t − 7) = −16(t + 7)(t − 1) = 0 when t = 1 s
(b) v(t) = −32t − 96, v(1) = −128, the speed at impact is 128 ft/s

218 Chapter 5
45. s(t) = −4.9t2
+ 49t + 150 and v(t) = −9.8t + 49
(a) the projectile reaches its maximum height when v(t) = 0, −9.8t + 49 = 0, t = 5 s
(b) s(5) = −4.9(5)2
+ 49(5) + 150 = 272.5 m
(c) the projectile reaches its starting point when s(t) = 150, −4.9t2
+ 49t + 150 = 150,
−4.9t(t − 10) = 0, t = 10 s
(d) v(10) = −9.8(10) + 49 = −49 m/s
(e) s(t) = 0 when the projectile hits the ground, −4.9t2
+ 49t + 150 = 0 when (use the quadratic
formula) t ≈ 12.46 s
(f) v(12.46) = −9.8(12.46) + 49 ≈ −73.1, the speed at impact is about 73.1 m/s
46. take s = 0 at the water level and let h be the height of the bridge, then s = h and v = 0 when
t = 0 so s(t) = −16t2
+ h
(a) s = 0 when t = 4 thus −16(4)2
+ h = 0, h = 256 ft
(b) First, find how long it takes for the stone to hit the water (find t for s = 0) : −16t2
+ h = 0,
t =
√
h/4. Next, find how long it takes the sound to travel to the bridge: this time is h/1080
because the speed is constant at 1080 ft/s. Finally, use the fact that the total of these two
times must be 4 s:
h
1080
+
√
h
4
= 4, h + 270
√
h = 4320, h + 270
√
h − 4320 = 0, and by
the quadratic formula
√
h =
−270 ± (270)2 + 4(4320)
2
, reject the negative value to get
√
h ≈ 15.15, h ≈ 229.5 ft.
47. g = 9.8/6 = 4.9/3 m/s2
, so v = −(4.9/3)t, s = −(4.9/6)t2
+ 5, s = 0 when t = 30/4.9 and
v = −(4.9/3) 30/4.9 ≈ −4.04, so the speed of the module upon landing is 4.04 m/s
48. s(t) = −1
2 gt2
+ v0t; s = 1000 when v = 0, so 0 = v = −gt + v0, t = v0/g,
1000 = s(v0/g) = −1
2 g(v0/g)2
+ v0(v0/g) = 1
2 v2
0/g, so v2
0 = 2000g, v0 =
√
2000g.
The initial velocity on the Earth would have to be
√
6 times faster than that on the Moon.
49. fave =
1
3 − 1
3
1
3x dx =
3
4
x2
3
1
= 6 50. fave =
1
2 − (−1)
2
−1
x2
dx =
1
9
x3
2
−1
= 1
51. fave =
1
π − 0
π
0
sin x dx = −
1
π
cos x
π
0
= 2/π
52. fave =
1
π − 0
π
0
cos x dx =
1
π
sin x
π
0
= 0
53. (a) fave =
1
2 − 0
2
0
x2
dx = 4/3 (b) (x∗
)2
= 4/3, x∗
= ±2/
√
3,
but only 2/
√
3 is in [0, 2]
(c)
2
4
x
y
2
3

54. (a) fave =
1
4 − 0
4
0
2x dx = 4 (b) 2x∗
= 4, x∗
= 2
(c)
2 4
4
8
x
y
55. (a) vave =
1
4 − 1
4
1
(3t3
+ 2)dt =
1
3
789
4
=
263
4
(b) vave =
s(4) − s(1)
4 − 1
=
100 − 7
3
= 31
56. (a) aave =
1
5 − 0
5
0
(t + 1)dt = 7/2
(b) aave =
v(π/4) − v(0)
π/4 − 0
=
√
2/2 − 1
π/4
= (2
√
2 − 4)/π
57. time to fill tank = (volume of tank)/(rate of filling) = [π(3)2
5]/(1) = 45π, weight of water in tank
at time t = (62.4) (rate of filling)(time) = 62.4t,
weightave =
1
45π
45π
0
62.4t dt = 1404π lb
58. (a) If x is the distance from the cooler end, then the temperature is T(x) = (15 + 1.5x)◦
C, and
Tave =
1
10 − 0
10
0
(15 + 1.5x)dx = 22.5◦
C
(b) By the Mean-Value Theorem for Integrals there exists x∗
in [0, 10] such that
f(x∗
) =
1
10 − 0
10
0
(15 + 1.5x)dx = 22.5, 15 + 1.5x∗
= 22.5, x∗
= 5
59. (a) amount of water = (rate of flow)(time) = 4t gal, total amount = 4(30) = 120 gal
(b) amount of water =
60
0
(4 + t/10)dt = 420 gal
(c) amount of water =
120
0
(10 +
√
t)dt = 1200 + 160
√
30 ≈ 2076.36 gal
60. (a) The maximum value of R occurs at 4:30 P.M. when t = 0.
(b)
60
0
100(1 − 0.0001t2
)dt = 5280 cars
61. (a)
b
a
[f(x) − fave] dx =
b
a
f(x)dx −
b
a
favedx =
b
a
f(x)dx − fave(b − a) = 0
because fave(b − a) =
b
a
f(x)dx

220 Chapter 5
(b) no, because if
b
a
[f(x) − c]dx = 0 then
b
a
f(x)dx − c(b − a) = 0 so
c =
1
b − a
b
a
f(x)dx = fave is the only value
EXERCISE SET 5.8
1. (a)
3
1
u7
du (b) −
1
2
4
7
u1/2
du (c)
1
π
π
−π
sin u du (d)
0
−3
(u + 5)u20
du
2. (a)
1
2
7
−3
u8
du (b)
5/2
3/2
1
√
u
du
(c)
1
0
u2
du (d)
1
2
4
3
(u − 3)u1/2
du
3. u = 2x + 1,
1
2
3
1
u4
du =
1
10
u5
3
1
= 121/5, or
1
10
(2x + 1)5
1
0
= 121/5
4. u = 4x − 2,
1
4
6
2
u3
du =
1
16
u4
6
2
= 80, or
1
16
(4x − 2)4
2
1
= 80
5. u = 1 − 2x, −
1
2
1
3
u3
du = −
1
8
u4
1
3
= 10, or −
1
8
(1 − 2x)4
0
−1
= 10
6. u = 4 − 3x, −
1
3
−2
1
u8
du = −
1
27
u9
−2
1
= 19, or −
1
27
(4 − 3x)9
2
1
= 19
7. u = 1 + x,
9
1
(u − 1)u1/2
du =
9
1
(u3/2
− u1/2
)du =
2
5
u5/2
−
2
3
u3/2
9
1
= 1192/15,
or
2
5
(1 + x)5/2
−
2
3
(1 + x)3/2
8
0
= 1192/15
8. u = 4 − x,
4
9
(u − 4)u1/2
du =
4
9
(u3/2
− 4u1/2
)du =
2
5
u5/2
−
8
3
u3/2
4
9
= −506/15
or
2
5
(4 − x)5/2
−
8
3
(4 − x)3/2
0
−5
= −506/15
9. u = x/2, 8
π/4
0
sin u du = −8 cos u
π/4
0
= 8 − 4
√
2, or − 8 cos(x/2)
π/2
0
= 8 − 4
√
2
10. u = 3x,
2
3
π/2
0
cos u du =
2
3
sin u
π/2
0
= 2/3, or
2
3
sin 3x
π/6
0
= 2/3
11. u = x2
+ 2,
1
2
3
6
u−3
du = −
1
4u2
3
6
= −1/48, or −
1
4
1
(x2 + 2)2
−1
−2
= −1/48

12. u =
1
4
x −
1
4
, 4
π/4
−π/4
sec2
u du = 4 tan u
π/4
−π/4
= 8, or 4 tan
1
4
x −
1
4
1+π
1−π
= 8
13.
1
3
5
0
25 − u2 du =
1
3
1
4
π(5)2
=
25
12
π 14.
1
2
4
0
16 − u2 du =
1
2
1
4
π(4)2
= 2π
15. −
1
2
0
1
1 − u2 du =
1
2
1
0
1 − u2 du =
1
2
·
1
4
[π(1)2
] = π/8
16.
2
−2
4 − u2 du =
1
2
[π(2)2
] = 2π
17.
1
0
sin πxdx = −
1
π
cos πx
1
0
= −
1
π
(−1 − 1) = 2/π
18. A =
π/8
0
3 cos 2x dx =
3
2
sin 2x
π/8
0
= 3
√
2/4
19.
7
3
(x + 5)−2
dx = −(x + 5)−1
7
3
= −
1
12
+
1
8
=
1
24
20. A =
1
0
dx
(3x + 1)2
= −
1
3(3x + 1)
1
0
=
1
4
21.
1
2 − 0
2
0
x
(5x2 + 1)2
dx = −
1
2
1
10
1
5x2 + 1
2
0
=
1
21
22. fave =
1
1/4 − (−1/4)
1/4
−1/4
sec2
πxdx =
2
π
tan πx
1/4
−1/4
=
4
π
23.
2
3
(3x + 1)1/2
1
0
= 2/3 24.
2
15
(5x − 1)3/2
2
1
= 38/15
25.
2
3
(x3
+ 9)1/2
1
−1
=
2
3
(
√
10 − 2
√
2) 26.
1
10
(t3
+ 1)20
0
−1
= 1/10
27. u = x2
+ 4x + 7,
1
2
28
12
u−1/2
du = u1/2
28
12
=
√
28 −
√
12 = 2(
√
7 −
√
3)
28.
2
1
1
(x − 3)2
dx = −
1
x − 3
2
1
= 1/2
29.
1
2
sin2
x
π/4
−3π/4
= 0 30.
2
3
(tan x)3/2
π/4
0
= 2/3 31.
5
2
sin(x2
)
√
π
0
= 0
32. u =
√
x, 2
2π
π
sin u du = −2 cos u
2π
π
= −4

222 Chapter 5
33. u = 3θ,
1
3
π/3
π/4
sec2
u du =
1
3
tan u
π/3
π/4
= (
√
3 − 1)/3
34. u = sin 3θ,
1
3
−1
0
u2
du =
1
9
u3
−1
0
= −1/9
35. u = 4 − 3y, y =
1
3
(4 − u), dy = −
1
3
du
−
1
27
1
4
16 − 8u + u2
u1/2
du =
1
27
4
1
(16u−1/2
− 8u1/2
+ u3/2
)du
=
1
27
32u1/2
−
16
3
u3/2
+
2
5
u5/2
4
1
= 106/405
36. u = 5 + x,
9
4
u − 5
√
u
du =
9
4
(u1/2
− 5u−1/2
)du =
2
3
u3/2
− 10u1/2
9
4
= 8/3
37. (b)
π/6
0
sin4
x(1 − sin2
x) cos x dx =
1
5
sin5
x −
1
7
sin7
x
π/6
0
=
1
160
−
1
896
=
23
4480
38. (b)
π/4
−π/4
tan2
x(sec2
x − 1) dx =
1
3
tan3
x
π/4
−π/4
−
π/4
−π/4
(sec2
x − 1) dx
=
2
3
+ (− tan x + x)
π/4
−π/4
=
2
3
− 2 +
π
2
= −
4
3
+
π
2
39. (a) u = 3x + 1,
1
3
4
1
f(u)du = 5/3 (b) u = 3x,
1
3
9
0
f(u)du = 5/3
(c) u = x2
, 1/2
0
4
f(u)du = −1/2
4
0
f(u)du = −1/2
40. u = 1 − x,
1
0
xm
(1 − x)n
dx = −
0
1
(1 − u)m
un
du =
1
0
un
(1 − u)m
du =
1
0
xn
(1 − x)m
dx
41. sin x = cos(π/2 − x),
π/2
0
sinn
x dx =
π/2
0
cosn
(π/2 − x)dx = −
0
π/2
cosn
u du (u = π/2 − x)
=
π/2
0
cosn
u du =
π/2
0
cosn
x dx (by replacing u by x)
42. u = 1 − x, −
0
1
(1 − u)un
du =
1
0
(1 − u)un
du =
1
0
(un
− un+1
)du =
1
n + 1
−
1
n + 2
=
1
(n + 1)(n + 2)

43. (a) V 2
rms =
1
1/f − 0
1/f
0
V 2
p sin2
(2πft)dt =
1
2
fV 2
p
1/f
0
[1 − cos(4πft)]dt
=
1
2
fV 2
p t −
1
4πf
sin(4πft)
1/f
0
=
1
2
V 2
p , so Vrms = Vp/
√
2
(b) Vp/
√
2 = 120, Vp = 120
√
2 ≈ 169.7 V
44. Let u = t − x, then du = −dx and
t
0
f(t − x)g(x)dx = −
0
t
f(u)g(t − u)du =
t
0
f(u)g(t − u)du;
the result follows by replacing u by x in the last integral.
45. (a) I = −
0
a
f(a − u)
f(a − u) + f(u)
du =
a
0
f(a − u) + f(u) − f(u)
f(a − u) + f(u)
du
=
a
0
du −
a
0
f(u)
f(a − u) + f(u)
du, I = a − I so 2I = a, I = a/2
(b) 3/2 (c) π/4
46. x =
1
u
, dx = −
1
u2
du, I =
1
−1
1
1 + 1/u2
(−1/u2
)du = −
1
−1
1
u2 + 1
du = −I so I = 0 which is
impossible because
1
1 + x2
is positive on [−1, 1]. The substitution u = 1/x is not valid because u
is not continuous for all x in [−1, 1].
47.
1
0
sin πxdx = 2/π
49. (a) Let u = −x then
a
−a
f(x)dx = −
−a
a
f(−u)du =
a
−a
f(−u)du = −
a
−a
f(u)du
so, replacing u by x in the latter integral,
a
−a
f(x)dx = −
a
−a
f(x)dx, 2
a
−a
f(x)dx = 0,
a
−a
f(x)dx = 0
The graph of f is symmetric about the origin so
0
−a
f(x)dx is the negative of
a
0
f(x)dx
thus
a
−a
f(x)dx =
0
−a
f(x) dx +
a
0
f(x)dx = 0
(b)
a
−a
f(x)dx =
0
−a
f(x)dx +
a
0
f(x)dx, let u = −x in
0
−a
f(x)dx to get
0
−a
f(x)dx = −
0
a
f(−u)du =
a
0
f(−u)du =
a
0
f(u)du =
a
0
f(x)dx
so
a
−a
f(x)dx =
a
0
f(x)dx +
a
0
f(x)dx = 2
a
0
f(x)dx
The graph of f(x) is symmetric about the y-axis so there is as much signed area to the left
of the y-axis as there is to the right.

224 Chapter 5
50. (a) By Exercise 49(a),
1
−1
x cos(x2) dx = 0
(b) u = x − π/2, du = dx, sin(u + π/2) = sin u, cos(u + π/2) = − sin u
π
0
sin8
x cos5
x dx =
π/2
−π/2
sin8
u(− sin5
u) du = −
π/2
−π/2
sin13
u du = 0 by Exercise 49(a).
CHAPTER 5 SUPPLEMENTARY EXERCISES
5. If the acceleration a = const, then v(t) = at + v0, s(t) = 1
2 at2
+ v0t + s0.
6. (a) Divide the base into n equal subintervals. Above each subinterval choose the lowest and
highest points on the curved top. Draw a rectangle above the subinterval going through the
lowest point, and another through the highest point. Add the rectangles that go through the
lowest points to obtain a lower estimate of the area; add the rectangles through the highest
points to obtain an upper estimate of the area.
(b) n = 10: 25.0 cm, 22.4 cm
(c) n = 20: 24.4 cm, 23.1 cm
7. (a)
1
2
+
1
4
=
3
4
(b) −1 −
1
2
= −
3
2
(c) 5 −1 −
3
4
= −
35
4
(d) −2
(e) not enough information (f) not enough information
8. (a)
1
2
+ 2 =
5
2
(b) not enough information
(c) not enough information (d) 4(2) − 3
1
2
=
13
2
9. (a)
1
−1
dx +
1
−1
1 − x2 dx = 2(1) + π(1)2
/2 = 2 + π/2
(b)
1
3
(x2
+ 1)3/2
3
0
− π(3)2
/4 =
1
3
(103/2
− 1) − 9π/4
(c) u = x2
, du = 2xdx;
1
2
1
0
1 − u2du =
1
2
π(1)2
/4 = π/8
10.
1
2
y
x
0.2
0.4
0.6
0.8
1
0.2 0.6 1

Chapter 5 Supplementary Exercises 225
11. The rectangle with vertices (0, 0), (π, 0), (π, 1) and (0, 1) has area π and is much too large; so
is the triangle with vertices (0, 0), (π, 0) and (π, 1) which has area π/2; 1 − π is negative; so the
answer is 35π/128.
12. (a)
1
n
n
k=1
k
n
=
n
k=1
f(x∗
k)∆x where f(x) =
√
x, x∗
k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
lim
n→+∞
1
n
n
k=1
k
n
=
1
0
x1/2
dx =
2
3
(b)
1
n
n
k=1
k
n
4
=
n
k=1
f(x∗
k)∆x where f(x) = x4
, x∗
k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
lim
n→+∞
1
n
n
k=1
k
n
4
=
1
0
x4
dx =
1
5
13. left endpoints: x∗
k = 1, 2, 3, 4;
4
k=1
f(x∗
k)∆x = (2 + 3 + 2 + 1)(1) = 8
right endpoints: x∗
k = 2, 3, 4, 5;
4
k=1
f(x∗
k)∆x = (3 + 2 + 1 + 2)(1) = 8
14. The direction ﬁeld is clearly an even function, which means that the solution is even, its derivative
is odd. Since sin x is periodic and the direction ﬁeld is not, that eliminates all but x, the solution
of which is the family y = x2
/2 + C.
15. (a) 1 · 2 + 2 · 3 + · · · + n(n + 1) =
n
k=1
k(k + 1) =
n
k=1
k2
+
n
k=1
k
=
1
6
n(n + 1)(2n + 1) +
1
2
n(n + 1) =
1
3
n(n + 1)(n + 2)
(b)
n−1
k=1
9
n
−
k
n2
=
9
n
n−1
k=1
1 −
1
n2
n−1
k=1
k =
9
n
(n − 1) −
1
n2
·
1
2
(n − 1)(n) =
17
2
n − 1
n
;
lim
n→+∞
17
2
n − 1
n
=
17
2
(c)
3
i=1


2
j=1
i +
2
j=1
j

 =
3
i=1
2i +
1
2
(2)(3) = 2
3
i=1
i +
3
i=1
3 = 2 ·
1
2
(3)(4) + (3)(3) = 21
16. (a)
14
k=0
(k + 4)(k + 1) (b)
19
k=5
(k − 1)(k − 4)
17. For 1 ≤ k ≤ n the k-th L-shaped strip consists of the corner square, a strip above and a strip to
the right for a combined area of 1+(k −1)+(k −1) = 2k −1, so the total area is
n
k=1
(2k −1) = n2
.

226 Chapter 5
18. 1 + 3 + 5 + · · · + (2n − 1) =
n
k=1
(2k − 1) = 2
n
k=1
k −
n
k=1
1 = 2 ·
1
2
n(n + 1) − n = n2
19. (35
− 34
) + (36
− 35
) + · · · + (317
− 316
) = 317
− 34
20. 1 −
1
2
+
1
2
−
1
3
+ · · · +
1
50
−
1
51
=
50
51
21.
1
22
−
1
12
+
1
32
−
1
22
+ · · · +
1
202
−
1
192
=
1
202
− 1 = −
399
400
22. (22
− 2) + (23
− 22
) + · · · + (2101
− 2100
) = 2101
− 2
23. (a)
n
k=1
1
(2k − 1)(2k + 1)
=
1
2
n
k=1
1
2k − 1
−
1
2k + 1
=
1
2
1 −
1
3
+
1
3
−
1
5
+
1
5
−
1
7
+ · · · +
1
2n − 1
−
1
2n + 1
=
1
2
1 −
1
2n + 1
=
n
2n + 1
(b) lim
n→+∞
n
2n + 1
=
1
2
24. (a)
n
k=1
1
k(k + 1)
=
n
k=1
1
k
−
1
k + 1
= 1 −
1
2
+
1
2
−
1
3
+
1
3
−
1
4
+ · · · +
1
n
−
1
n + 1
= 1 −
1
n + 1
=
n
n + 1
(b) lim
n→+∞
n
n + 1
= 1
25.
n
i=1
(xi − ¯x) =
n
i=1
xi −
n
i=1
¯x =
n
i=1
xi − n¯x but ¯x =
1
n
n
i=1
xi thus
n
i=1
xi = n¯x so
n
i=1
(xi − ¯x) = n¯x − n¯x = 0
26. S − rS =
n
k=0
ark
−
n
k=0
ark+1
= (a + ar + ar2
+ · · · + arn
) − (ar + ar2
+ ar3
+ · · · + arn+1
)
= a − arn+1
= a(1 − rn+1
)
so (1 − r)S = a(1 − rn+1
), hence S = a(1 − rn+1
)/(1 − r)

Chapter 5 Supplementary Exercises 227
27. (a)
19
k=0
3k+1
=
19
k=0
3(3k
) =
3(1 − 320
)
1 − 3
=
3
2
(320
− 1)
(b)
25
k=0
2k+5
=
25
k=0
25
2k
=
25
(1 − 226
)
1 − 2
= 231
− 25
(c)
100
k=0
(−1)
−1
2
k
=
(−1)(1 − (−1/2)101
)
1 − (−1/2)
= −
2
3
(1 + 1/2101
)
28. (a) 1.999023438, 1.999999046, 2.000000000; 2
(b) 2.831059456, 2.990486364, 2.999998301; 3
29. (a) If u = sec x, du = sec x tan xdx, sec2
x tan xdx = udu = u2
/2 + C1 = (sec2
x)/2 + C1;
if u = tan x, du = sec2
xdx, sec2
x tan xdx = udu = u2
/2 + C2 = (tan2
x)/2 + C2.
(b) They are equal only if sec2
x and tan2
x diﬀer by a constant, which is true.
30.
1
2
sec2
x
π/4
0
=
1
2
(2 − 1) = 1/2 and 1
2 tan2
x
π/4
0
=
1
2
(1 − 0) = 1/2
31. 1 + x−2/3dx = x−1/3
x2/3 + 1dx; u = x2/3
+ 1, du =
2
3
x−1/3
dx
3
2
u1/2
du = u3/2
+ C = (x2/3
+ 1)3/2
+ C
32. (a)
b
a
n
k=1
fk(x)dx =
n
k=1
b
a
fk(x)dx
(b) yes; substitute ckfk(x) for fk(x) in part (a), and then use
b
a
ckfk(x)dx = ck
b
a
fk(x)dx
from Theorem 5.5.4
33. (a)
x
1
1
1 + t2
dt (b)
x
tan(π/4−2)
1
1 + t2
dt
34. (a) F (x) =
x − 3
x2 + 7
; increasing on [3, +∞), decreasing on (−∞, 3]
(b) F (x) =
7 + 6x − x2
(x2 + 7)2
=
(7 − x)(1 + x)
(x2 + 7)2
; concave up on (−1, 7), concave down on (−∞, −1)
and (7, +∞)

228 Chapter 5
(c) F (x) =
x − 3
x2 + 7
= 0 when x = 3, which is a relative minimum, and hence the absolute
minimum, by the ﬁrst derivative test.
(d) F(x)
x
2
3
-10 20
35. F (x) =
1
1 + x2
+
1
1 + (1/x)2
(−1/x2
) = 0 so F is constant on (0, +∞).
36. (−3, 3) because f is continuous there and 1 is in (−3, 3)
37. (a) The domain is (−∞, +∞); F(x) is 0 if x = 1, positive if x > 1, and negative if x < 1, because
the integrand is positive, so the sign of the integral depends on the orientation (forwards or
backwards).
(b) The domain is [−2, 2]; F(x) is 0 if x = −1, positive if −1 < x ≤ 2, and negative if
−2 ≤ x < −1; same reasons as in Part (a).
38. The left endpoint of the top boundary is ((b − a)/2, h) and the right endpoint of the top boundary
is ((b + a)/2, h) so
f(x) =



2hx/(b − a), x < (b − a)/2
h, (b − a)/2 < x < (b + a)/2
2h(x − b)/(a − b), x > (a + b)/2
The area of the trapezoid is given by
(b−a)/2
0
2hx
b − a
dx+
(b+a)/2
(b−a)/2
hdx+
b
(b+a)/2
2h(x − b)
a − b
dx = (b−a)h/4+ah+(b−a)h/4 = h(a+b)/2.
39. (a) no, since the velocity curve is not a straight line
(b) 25 < t < 40 (c) 3.54 ft/s (d) 141.5 ft
(e) no since the velocity is positive and the acceleration is never negative
(f) need the position at any one given time (e.g. s0)
40. w(t) =
t
0
τ/7 dτ = t2
/14, assuming w0 = w(0) = 0; wave =
1
26
52
26
t2
/7 dt =
1
26
t3
21
52
26
= 676/3
Set 676/3 = t2
/14, t = ±
26
3
√
21, so t ≈ 39.716, so during the 40th week.
41. u = 5 + 2 sin 3x, du = 6 cos 3xdx;
1
6
√
u
du =
1
3
u1/2
+ C =
1
3
√
5 + 2 sin 3x + C
42. u = 3 +
√
x, du =
1
2
√
x
dx; 2
√
udu =
4
3
u3/2
+ C =
4
3
(3 +
√
x)3/2
+ C

Chapter 5 Horizon Module 229
43. u = ax3
+ b, du = 3ax2
dx;
1
3au2
du = −
1
3au
+ C = −
1
3a2x3 + 3ab
+ C
44. u = ax2
, du = 2axdx;
1
2a
sec2
udu =
1
2a
tan u + C =
1
2a
tan(ax2
) + C
45. −
1
3u3
−
3
u
+
1
4u4
−1
−2
= 389/192 46.
1
3π
sin3
πx
1
0
= 0
47. With b = 1.618034, area =
b
0
(x + x2
− x3
)dx = 1.007514.
48. (a) f(x) =
1
3
x2
sin 3x −
2
27
sin 3x +
2
9
x cos 3x − 0.251607
(b) f(x) = 4 + x2 +
4
√
4 + x2
− 6
49. (a) Solve
1
4
k4
− k − k2
+
7
4
= 0 to get k = 2.073948.
(b) Solve −
1
2
cos 2k +
1
3
k3
+
1
2
= 3 to get k = 1.837992.
50. F(x) =
x
−1
t
√
2 + t3
dt, F (x) =
x
√
2 + x3
, so F is increasing on [1, 3]; Fmax = F(3) ≈ 1.152082854
and Fmin = F(1) ≈ −0.07649493141
51. (a)
1 2 4 6 7 8
-0.5
0.5
1
y = J0(x)
x
y (b) 0.7651976866 (c) J0(x) = 0 if x = 2.404826
52. lim
n→+∞
n
k=1
25(k − 1)
n
−
25(k − 1)2
n2
5
n
=
125
6
CHAPTER 5 HORIZON MODULE
1. vx(0) = 35 cos α, so from Equation (1), x(t) = (35 cos α)t; vy(0) = 35 sin α, so from Equation (2),
y(t) = (35 sin α)t − 4.9t2
.
2. (a) vx(t) =
dx(t)
dt
= 35 cos α, vy(t) =
dy(t)
dt
= 35 sin α − 9.8t
(b) vy(t) = 35 sin α − 9.8t, vy(t) = 0 when t = 35 sin α/9.8;
y = vy(0)t − 4.9t2
= (35 sin α)(35 sin α)/9.8 − 4.9((35 sin α)/9.8)2
= 62.5 sin2
α, so
ymax = 62.5 sin2
α.

230 Chapter 5
3. t = x/(35 cos α) so y = (35 sin α)(x/(35 cos α)) − 4.9(x/(35 cos α))2
= (tan α)x −
0.004
cos2 α
x2
;
the trajectory is a parabola because y is a quadratic function of x.
4.
15◦
25◦
35◦
45◦
55◦
65◦
75◦
85◦
no yes no no no yes no no
65
0
0 120
5. y(t) = (35 sin α s)t − 4.9t2
= 0 when t = 35 sin α/4.9, at which time
x = (35 cos α)(35 sin α/4.9) = 125 sin 2α; this is the maximum value of x, so R = 125 sin 2α m.
6. (a) R = 95 when sin 2α = 95/125 = 0.76, α = 0.4316565575, 1.139139769 rad ≈ 24.73◦
, 65.27◦
.
(b) y(t) < 50 is required; but y(1.139) ≈ 51.56 m, so his height would be 56.56 m.
7. 0.4019 < α < 0.4636 (radians), or 23.03◦
< α < 26.57◦

Solution Manual : Chapter - 05 Integration

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