Solucionario Zill

CHAPTER 5. INTEGRALS

Given f (x) = √ x 4 + 1 = (x 4 + 1) 1/2 , we have f (x) = 2x 3 (x 4 + 1) −1/2 . Thus,

To check this, take the derivative of the above function, yielding 96x 2 √ 256x 4 + 1 − 16384x 6 (256x 4 + 1) 3 , which should be the same as f (4x). Since f (x) = 6x 2 √ x 4 + 1 − 4x 6 (x 4 + 1) 3 , we have f (4x) = 6(4x) 2

76. First evaluating sec 2 3x dx, we get sec 2 3x dx = 1 3 (sec 2 3x)(3 dx) u = 3x, du = 3 dx = 1 3 sec 2 u du = 1 3 tan u + C = 1 3 tan 3x + C Next, evaluating sec 2 3x dx dx = 1 3 tan 3x + C dx, we get 1 3 tan 3x + C dx = (Cx + C 1 ) + 1 3 tan 3x dx = (Cx + C 1 ) + 1 9 (tan 3x)(3 dx) u = 3x, du = 3 dx = (Cx + C 1 ) + 1 9 tan u du = (Cx + C 1 ) − 1 9 ln | cos u| + C 2 = Cx − 1 9 ln | cos 3x| + C 3 . For A 2 , we have Δx = 3 n , f a + k b − a n = 2 + 3k n , and For A 2 , we have Δx = 2 n , f a + k b − a n = 3 + 2k n , and n(n + 1)(2n + 1) 6 + 27 n 3 n(n + 1) 2 − 9 4n 3 n = lim n→∞ 9 + 9 1 + 1 n − 9 n − 9 2 1 + 1 n 2 + 1 n + 27 2n 1 + 1 n − 9 4n 2 = 9 + 9 − 0 − 9 + 0 − 0 = 9. (c) Comparing the results of (a) and (b), we find that equating them leads to summation formula (ii): 2 n k=1 k + n = n 2 + 2n; n k=1 k = n 2 + 2n − n 2 = n 2 + n 2 = n(n + 1) 2

The Area Problem

THE AREA PROBLEM

305

Using f (k) = (k + 1) 3 similarly to (a), we obtain n k=1

[(k + 1) 3 − k 3 ] = (n + 1) 3 − 1 3 = n 3 + 3n 2 + 3n.

Analogously for (b), we also have n k=1

[(k + 1) 3 − k 3 ] = n k=1 (3k 2 + 3k + 1) = 3 n k=1 k 2 + 3 n k=1 k + n.

Combining these, we obtain 3 n k=1 k 2 + 3 n k=1 k + n = n 3 + 3n 2 + 3n 3 n k=1 k 2 + 3n(n + 1) 2 + n = n 3 + 3n 2 + 3n 3 n k=1 k 2 = n 3 + 3n 2 + 2n − 3n 2 + 3n 2 n k=1 k 2 = 2n 3 + 6n 2 + 4n − 3n 2 − 3n 6 = 2n 3 + 3n 2 + n 6 = n(2n 2 + 3n + 1) 6 = n(n + 1)(2n + 1) 6 .

56. The pattern illustrated in Figure 5.3.9 indicates that the summation of cubes is the square of the summation of the numbers being cubed. That is:

Using Δx = b n and f a + k b − a n = f kb n = k(h 2 − h 1 ) n + h 1 we find

58. Since the total number of cans is 136 and there is one additional can per row, we have 60. We note that A 2 = 1 − A 1 where A 1 is the area under y = x 2 from 0 to 1. Using Δx = 1 n and f a + k b − a n = k 2 n 2 we find Thus, A 2 = 1 − 1 3 = 2 3 . e (k−1)/n · 1 n = lim n→∞ 1 n 1 + e 1/n + e 2/n + · · · + e (n−1)/n = lim n→∞ 1 n 1 + e 1/n + (e 1/n ) 2 + · · · + (e 1/n ) n−1 .

Using a = 1, r = e 1/n , we obtain

THE DEFINITE INTEGRAL

309 64. 1 + 3 + 5 + · · · + 2n − 1 = n k=1 (2k − 1) = 2 n k=1 k − n k=1 1 = n(n + 1) − n = n 2

The total distance moved is thus proportional to 1 + 3 + 5 + · · · + 2n − 1 = n 2 .

The Definite Integral

1. From Δx 1 = 1, Δx 2 = 2/3, Δx 3 = 2/3, and Δx 4 = 2/3 we see that the norm of the partition is P = 1. Using f (x * 1 ) = 5/2, f (x * 2 ) = 5, f (x * 3 ) = 7, and f (x * 4 ) = 9 we compute the Riemann sum (1) + 5 2 3 + 7 2 3 + 9 2 3 = 33 2 .

2. From Δx 1 = 1, Δx 2 = 1/2, Δx 3 = 1, Δx 4 = 5/2, and Δx 5 = 2 we see that the norm of the partition is P = 5/2. Using f (x * 1 ) = −11/2, f (x * 2 ) = −9/2, f (x * 3 ) = −4, f (x * 4 ) = −2, and f (x * 5 ) = 0 we compute the Riemann sum 3. From Δx 1 = 3/4, Δx 2 = 1/2, Δx 3 = 1/2, and Δx 4 = 1/4 we see that the norm of the partition is P = 3/4. Using f (x * 1 ) = 9/16, f (x * 2 ) = 0, f (x * 3 ) = 1/4, and f (x * 4 ) = 49/64 we compute the Riemann sum (1) + 10 1 2 = 331 32 .

5. From Δx 1 = π, Δx 2 = π/2, and Δx 3 = π/2 we see that the norm of the partition is P = π. Using f (x * 1 ) = 1, f (x * 2 ) = −1/2, and f (x * 3 ) = − √ 2/2 we compute the Riemann sum

6. From Δx 1 = π/4, Δx 2 = π/4, Δx 3 = π/3, and Δx 4 = π/6 we see that the norm of the partition is P = π/3. Using f (x * 1 ) = 1/2, f (x * 2 ) = √ 3/2, f (x * 3 ) = √ 2/2, and f (x * 4 ) = 1/2 we compute the Riemann sum 2 n = f a + k b − a n b − a n and f a + k b − a n = f 2k n = 1 + 2k n .

Taking f (x) = x + 1 we have lim 14. Using b − a n = 3 n and f a + k b − a n = 3k n we have 18. Using b − a n = 2 n and f a + k b − a n = 3 − 8k 3 n 3 we have 19. Using f a + k b − a n = a + k(b − a) n we have

21.

f (x) dx = −2.5 + 3.9 = 1.4 48.

-2 2 2 49. From the figure, we see that the area under the graph is a triangle with a base and height of 6. Thus, the area from geometry is A = bh 2 = 6(6) 2 = 18. 50. From the figure, we see that the area under the graph consists of two triangles; one has a base and height of 1 while the other has a base and height of 2. Thus, the area from geometry is 52. From the figure, we see that the area under the graph consists of a semicircle of radius 3 above a rectangle of width 6 and height 2. Thus, the area from geometry is A = πr 2 2 + wh = π(3) 2 2 + 6(2) = 9π 2 + 12. 57. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 1 and a height of 2 subtracted from the area of a triangle with a base of 4 and a height of 8. Thus, the net signed area from geometry is

-2 2 4 -2 2 4 6 8 58. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 4 and a height of 2 subtracted from the area of a triangle with a base of 4 and a height of 2. Thus, the net signed area from geometry is

(2) 2 − 4(2) 2 = 0. 1 − x 2 dx, so the net signed area of the graph below left is the same as the difference between the net signed areas of the graphs below right. This difference, in turn, is the area of a semicircle of radius 1 subtracted from the net signed area of two triangles with bases and heights of 1. From geometry, this is 60. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 1 and a height of 1 subtracted from the area of a triangle with a base of 2 and a height of 1. Thus, the net signed area from geometry is

-2 2 -2 2 2 61. From the figure, we see that the net signed area under the graph is the negative of the area of a triangle with a base of 2 and a height of 2. Thus, the net signed area from geometry is

-3 3 -3 3 62. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 1 and a height of 1 subtracted from the area of a triangle with a base of 3 and a height of 3. Thus, the net signed area from geometry is 63. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 4 and a height of 4 subtracted from the sum of the areas of a triangle with a base of 3 and a height of 3, and a rectangle of width 2 and height 3. Thus, the net signed area from geometry is

] − 4(4) 2 = 5 2 = 2.5.

First, choosing each x * k to be rational, we obtain = lim n→∞ π 2n cos π 4n + cos 3 π 4n + · · · + cos (2n − 1) π 4n = lim n→∞ π 2n ⎡ ⎣ sin 2n · π 4n 2 sin π 4n ⎤ ⎦ = π 4 lim n→∞ sin π 2 n sin π 4n = π 4 lim n→∞ 1 n sin π 4n = π 4 · 4 π = 1.

Fundamental Theorem of Calculus

5

3 √ s 2 + 2 √ s 3 ds = (5s −2/3 + 2s −3/2 ) ds = 15s 1/3 − 4s −1/2 + C 13.

(4x + 1) 2 dx = (16x 2 + 8x + 1) dx = 16 3 2x 3 − x 2 + 2x + 4 1 + x 2 dx = 2x − 1 + 5

43. y = (6x 2 + 9) dx = 2x 3 + 9x + C 44. y = (10x + 3x 1/2 ) dx = 5x 2 + 2x 3/2 + C

THE INDEFINITE INTEGRAL

289 46. y = (2 + x) 2

Solving 3 = f (2) = 4 − 2 + C = 2 + C we obtain C = 1. Thus f (x) = x 2 − x + 1.

50. We have f (x) = x −1/2 dx = 2x 1/2 + C. Solving 1 = f (9) = 2 √ 9 + C = 6 + C we obtain

. We have f (x) = 6 dx = 6x + C. Solving 2 = f (−1) = −6 + C we obtain C = 8. Then f (x) = 6x+8 and f (x) = (6x+8) dx = 3x 2 +8x+C. Solving 0 = f (−1) = 3−8+C = −5+C we obtain C = 5. Thus f (x) = 3x 2 + 8x + 5.

53. We have f (x) = (12x 2 + 2) dx = 4x 3 + 2x + C. Solving 3 = f (1) = 6 + C we obtain C = −3. Then f (x) = 4x 3 + 2x − 3 and f (x) = (4x 3 + 2x − 3) dx = x 4 + x 2 − 3x + C.

Solving 1 = f (1) = −1 + C we obtain C = 2. Thus f (x) = x 4 + x 2 − 3x + 2.

54. f (x) = a n−1 x n−1 + a n−2 x n−2 + · · · + a 1 x + a 0 55. G is an antiderivative of f . In other words, since G (x) = f (x), f is the slope function for G. Observe where G is increasing, and the graph of f is always positive. Also, G appears to have no relative extrema on the interval shown, and correspondingly the graph of f does not cross the x-axis. 56. F is an antiderivative of f . In other words, since F (x) = f (x), f is the slope function for F . Observe where the tangent lines to the graph of F have positive (negative) slope, the graph of f is positive (negative). Also, the graph of F has two relative extrema and the graph of f correspondingly crosses the x-axis.

57. y = ω 2 g x dx = ω 2 2g

x 2 + C. From Figure 5.1.5 we see that y(0) = 0. Thus, 0 = y(0) = C, and y = ω 2 x 2 2g .

58. We have f (x) = qL 2EI

x − q 2EI

x 2 dx = qL 4EI

x 2 − q 6EI

x 3 + C.

x 3 − qL 3 24EI and

x 3 + C. Since y = 4x + 7 is a tangent line to the graph of f , then 4x + 7 = 1 3

x 3 + C at some point on f . In addition, the slope at this point

Thus, 4(±2) + 7 = 1 3 (±2) 3 + C, so C = 37/3 or 5/3. Thus,

62. e 4 dx/x = e 4 ln |x|+C = e ln x 4 e C = C 1 e ln

Thus, both results are correct.

64. Since d dx sin πx = π cos πx, the antiderivative F of cos πx would be of the form 1 π sin πx + C.

Integration by the u-Substitution

INTEGRATION BY THE u-SUBSTITUTION

295 55.

(3 − 2 sin x) 2 dx = (9 − 12 sin x + 4 sin 2 x) dx = 9x + 12 cos x + 4 1 2 (1 − cos 2x) dx = 9x + 12 cos x + 2 x − 1 2 sin 2x + C = 11x + 12 cos x − sin 2x + C 56.

(1 + cos 2x) 2 dx = (1 + 2 cos 2x + cos 2 2x) dx = x + sin 2x + 1 2 (1 + cos 4x) dx

60. We have f (x) = (1 + 2x) 5 dx = 1 12

(1 + 2x) 6 + C. Solving 0 = f (0) = 1 12 + C we obtain

Solving 0 = f (0) = 1 12

To check this, take the derivative of the above function, yielding

76. First evaluating sec 2 3x dx, we get

FUNDAMENTAL THEOREM OF CALCULUS

65. Using the fact that f (x) = | sin x| is an even function on [−π, π] and sin

, and so Si (x) = 0 for x = nπ, n = 1, 2, . . . . The first four positive critical numbers are then π, 2π, 3π, and 4π. Now,

shows that there are relative maxima at x = π and x = 3π and relative minima at x = 2π and x = 4π.

327 76. Letting Δx k = 2/n we have 81. The reasoning is flawed at the point that sin t is substituted with √ 1 − cos 2 t. The use of the square root loses sin t's sign changes.

321

29.

40. Using the fact that f (x) = cos 2 x is even, we have

41.

59. Using the fact that f (x) is an even function on [−2, 2], we have

60.

= (2 − 1) + (6 − 4) + (12 − 9) = 6

(a) Since f is odd and continuous at

328

(a)

d dx x

sin 3 x dx = 0 84. As this project's exact results may vary for every "run" of the exercise, no exact solution is given. In general, the student should see the empirical probability n/N approach the area of the region as the number of random points increases.

85. (a) At time n the radius of the circle is r 0 + cu and the area is A(u) = π(r 0 + cu) 2 . Then

(b) Substituting RT /P v = 1.9 × 10 6 , K = 0.01 × 10 −3 , c = 0.01, r 0 = 100, and V 0 = 10, 000, we find t ≈ 2, 617, 695 seconds, or t ≈ 30 days and 7 hours. sec x dx can be found in the derivation of the projection, whose key properties are that it is conformal (i.e., angle-preserving) and that it represent lines of constant course as straight segments. 27

24.

27. Since f (x) = 1 1 + 3x 2 is an even function,

28. 32. (a) The outer radius of the kth disk (from the top) is r k = 1.5(k + 1) 2 cm, its inner radius is 1.5 2 cm, and its thickness is 1.5 cm. Then its volume is π(1.5) r 2 k − 1.5 2 2 = π(1.5) 3 k 2 + 2k 4 . Thus, the total volume is

(1.5) 3 n(n + 1)(2n + 1) 6 + 2 n(n + 1) 2 = π 4 (1.5) 3 n(n + 1)(2n + 7) 6 , and therefore the value of the gold is

(1.5) 3 n(n + 1)(2n + 7) 6 ≈ 38πn(n + 1)(2n + 7).

(b) For n = 64, the value of the gold is approximately 38π(64)(64 + 1)[2(64) + 7] = 38π(64 · 65 · 135) = 21, 340, 800π ≈ $67, 044, 100.50.

33. From the figure we note that

Since Techniques of Integration 7.1 Integration -Three Resources

13. cot 10x dx = 1 10 cot 10x(10 dx) u = 10x, du = 10 dx = 1 10 cot u du = 1 10 (ln | sin u|) + C = 1 10 ln | sin 10x| + C

The area of each of the four triangles between the (n − 1)st and n-th squares is

Thus, the area of the n-th square is A n−1 − 4 1 8

. . , and so on. In general, A n = 1 2 n−1 .

(b) S 1 = 1, S 2 = 1 + In the bottom three rows of the table, each number (after the second one) is the sum of the two preceding numbers in that row.

INTEGRATION -THREE RESOURCES

403

14.

x csc 2 (1 + tan x) 2 sec 2 x dx u = 1 + tan x, du = sec 2 x dx

Integration by Substitution

1.

x 2(x + 1/2) − 4/3 (x + 1/2) 2 + 9/4 dx = 3 4

2(x + 1/2) (x + 1/2) 2 + 9/4 dx − 1 (x + 1/2) 2 + 9/4 dx

31. 32. 33.

37. 38. 41.

x 2

u 9/2 − 6 7 u 7/2 + 6 5 u 5/2 − 2 3 u 3/2 1 0 + 2 9 u 9/2 − 6 7 u 7/2 + 6 5

Assuming that A − Bw

Integration by Parts

x sin 3x − 2 27 cos 3x + C 24.

Solving for the integral, we have 17 e x sin 4x dx = e x sin 4x − 4e x cos 4x + C or

Solving for the integral, we have e −2θ cos θ dθ = sin θ − 2 cos θ 5e 2θ + C.

28.

Solving for the integral, we have e αx sin βx dx = e αx (α sin βx − β cos βx) α 2 + β 2 + C.

29.

θ sec θ tan θ dθ u = θ, du = dθ; dv = sec θ tan θ dθ, v = sec θ = θ sec θ − sec θ dθ = θ sec θ − ln | sec θ + tan θ| + C Solving for the integral, we have sin x cos 2x dx = 1 3 cos x cos 2x + 2 3 sin x sin 2x + C.

32

.

Solving for the integral, we have cosh x cosh 2x dx = 2 3 sinh 2x cosh x − 1 3 sinh x cosh 2x + C.

33.

x

Solving for the integral, we have sin(ln

Solving for the integral, we have

38. ln(x 2 + 1) dx u = ln(x 2 + 1), du = 2x

Solving for the integral, we have π −π e x cos x dx = 1 2 (e −π − e π ).

45. 50. V = 2π x sin x dx

.

Solving for the integral, we have

INTEGRATION BY PARTS

73.

Solving for the integral, we have

Then

80. We first compute e −x cos 2x dx:

Solving for the integral, we have

Solving for the integral, we have

with n = 4 and n = 2.

(a) Graph shown at right.

(b) For x > 0, the first and second points of intersection of y = x sin x and y = x cos x are x 1 = π/4 and x 2 = 5π/4.

427

Solving for the integral sin n x dx, we have

Solving for the integral cos n x dx, we have

Solving for sec n x dx, we have sec n x dx = sec n−2 x tan x n − 1 + n − 2 n − 1 sec n−2 x dx, n = 1.

67. Using sin n x dx = − sin n−1 x cos x n + n − 1 n sin n−2 x dx with n = 3,

68. Using sec n x dx = sec n−2 x tan x n − 1 + n − 2 n − 1 sec n−2 x dx with n = 4,

428

CHAPTER 7. TECHNIQUES OF INTEGRATION

69. Using cos n x dx = cos n−1 x sin x n + n − 1 n cos n−2 x dx with n = 3, cos 3 10x dx u = 10x, du = 10 dx = 1 10 cos 3 u du = cos 2 u sin u 10 · 3 + 2 10 · 3 cos u du = cos 2 u sin u 30 + 1 15 sin u + C = cos 2 10x sin 10x 30 + 1 15 sin 10x + C.

70. Using cos n x dx = cos n−1 x sin x n + n − 1 n cos n−2 x dx with n = 4,

where k = 0 when n is even and n ≥ 2, and k = 1 when n is odd and n ≥ 3. Thus, we get, respectively:

(1 − cos θ − cos 8θ + cos 8θ cos θ) dθ cos 7 x dx all appear to be 0. We note that, for every t

thus lending credence to this conjecture.

60. Based on Problem 59, we conjecture that the value of π 0 cos n x dx, where n is a positive odd integer, is 0. To prove this, we evaluate π 0 cos n x dx:

Since n is a positive odd integer, n − 1 is guaranteed to be even, and thus k = n − 1 2 is an integer. Further, we substitute cos 2 x = 1 − sin 2 x:

By the binomial theorem, (1 − sin 2 x) k expands into an expression of the form 1 + c 1 sin 2 x + c 2 (sin 2 x) 2 + c 3 (sin 2 x) 3 + · · · + c k (sin 2 x) k . For this particular proof, it is not necessary to specify the precise values of the binary coefficients c i . Using the u substitution u = sin x, du = cos x dx, integration is accomplished as follows:

Evaluating this, we note that sin π = sin 0 = 0, and therefore π 0 cos n x dx = 0.

To find π/4 0 sec 5 θ dθ we use integration by parts.

. Now y(10) = 0 and a = 10, so 10 ln 10 − √ 100 − 100 10 + √ 100 − 100 = 0 + C 1 and C 1 = 0. Thus

Note: If the substitution y = a cos θ is used, we obtain the equivalent solution

58. Using symmetry with respect to the x-axis, we have

62. The circle of radius a, which is centered on the origin, is defined by x 2 +y 2 = a 2 . Let c be the distance between the centers of the two circles. The circle of radius b is thus defined by

By using the substitution x = r sin θ, dx = r cos θ dθ , we find

Substituting b and a respectively, we get

From the figure, it can be seen that a 2 = b 2 + c 2 or c = √ a 2 − b 2 . We substitute to simplify further:

The special-case lune of Hippocrates specifies a lune where the triangle formed by the origin and the intersections of the two circles is a right isosceles triangle. For this lune, b = well-known result that the area of the lune of Hippocrates is the same as the area of the right isosceles triangle that defines it, or 1 2 a 2 .

17. Write

Setting x = 0, x = 1, and x = −1 gives A = 6, B = −3/2, and C = −7/2. Thus

Setting x = 0, x = 2, and x = −2 gives A = −1/4, B = 19/8, and C = 23/8. Thus

Setting x = 1/2, x = −1/2, and x = −7 gives A = 1/15, B = −1/13, and C = 1/195. Thus

Solving

gives A = −2, B = 1, and C = 6. Thus

Solving

Solving

gives A = −1/32, B = 1/16, C = 1/32, and D = 1/16. Thus

gives A = −1/36, B = −79/1125, C = 14/75, D = 49/500, and E = 11/50. Thus

.

Solving

gives A = 0, B = −1/2, C = 0, and D = 1/2. Thus

.

.

Then 1 = (Ax + B)(x 2 + 4) + (Cx + D)(x 2 + 1)

Solving

gives A = 0, B = 1/3, C = 0, and D = −1/3. Thus 1

Solving

gives A = 0, B = −1/5, C = 0, and D = 1/5. Thus

.

38. Write 81

.

Then

Solving

gives A = 0, B = 1, C = −1, and D = 0. Thus

43. For this and possibly later problems, we will encounter cos 2 θ dθ. Using Example 12 of Section 5.2 in the text, we have

Then

gives A = 1, B = −2, C = −3, and D = 5. Thus

Then

Solving

gives A = 0, B = 1, C = 0, and D = −3. Thus

Then

.

, and x = −3 gives A = −9/16, B = −9/16, C = 9/16, and D = 9/16. Thus

Then 1 = A(x − 5) + B(x − 1). Setting x = 1 and x = 5 gives A = −1/4 and B = 1/4. Thus

Then 1 = A(x + 2) + B(x − 2). Setting x = 2 and x = −2 gives A = 1/4 and B = −1/4.

gives A = 6, B = −6, and C = −4. Thus

dx. Partial fraction decomposition is unnecessary because the substitution u = x 2 + 4, du = 2x dx will suffice for the first term, while the second term corresponds to formula 24 in Table 7.1.1.

(c)

x (x 2 + 5) 2 dx = 1 2 1 (x 2 + 5) 2 (2x dx). Partial fraction decomposition is unnecessary because the substitution u = x 2 + 5, du = 2x dx will suffice.

2x 3 + 5x x 4 + 5x 2 + 6 dx = 1 2 1 x 4 + 5x 2 + 6 (4x 3 + 10x dx). Partial fraction decomposition is unnecessary because the substitution u = x 4 + 5x 2 + 6, du = 4x 3 + 10x dx will suffice.

73. Rewrite the integral as

The integrand in Problem 53 is an odd function and its definite integral is symmetric about the y-axis. Thus, the definite integral's value is known to be 0.

gives A = −1/3, B = 1/2, C = −1/6, and D = −1/2. Thus

23.

tan 10 x sec 4 x dx = tan 10 x(tan 2 x + 1) sec 2 x dx u = tan x, du = sec 2 x dx = u 10 (u 2 + 1) du = (u 12 + u 10 ) du = 1 13 u 13 + 1 11

Solving for the integral,

44.

49.

x 2 + 2x + 5 dx = (x + 1) 2 + 4 dx x + 1 = 2 tan θ, dx = 2 sec 2 θ dθ 2

x + 1 θ x 2 + 2x + 5 = 4 tan 2 θ + 4(2 sec 2 θ dθ) = 4 sec 3 θ dθ See Section 7.3, Example 5 = 2 sec θ tan θ + 2 ln | sec θ + tan θ| + C = 1 2 (x + 1) x 2 + 2x + 5 + 2 ln

= 3 cos θ (9 − 9 sin 2 θ) 3/2 dθ = 1 9 cos θ cos 3 θ dθ = 1 9 sec 2 θ dθ

(1 + 2 cos x + cos 2 x) dx = 1 4 1 + 2 cos x + 1 + cos 2x 2 dx = 1 4 3 2 + 2 cos x + 1 2 cos 2x dx

x 2 dx x = 3 tan θ, dx = 3 sec 2 θ dθ = √ 9 tan 2 θ + 9 9 tan 2 θ (3 sec 2 θ dθ) = sec 3 θ tan 2 θ dθ = sec 3 θ sec 2 θ − 1 dθ = sec θ + sec θ sec 2 θ − 1 dθ = ln | sec θ + tan θ| + sec θ tan 2 θ dθ = ln | sec θ + tan θ| + cos θ sin 2 θ dθ u = sin θ, du = cos θ dθ

Solving for the integral, x(1 + ln x) 2 dx u = (1 + ln x) 2 , du = 2(1 + ln x) The height of the building is s(0) = 256 ft.

22. Let the depth of the well be h.

If t r is the time for the rock to hit the water, then 0 = s(t r ) = −16t 2 r + h, and h = 16t 2 r . Since the speed of sound is 1080 ft/s and the sound is heard after 2 seconds, h = 1080(2 − t r ). Then 16t 2 r = 1080(2 − t r ) or 2t 2 r + 135t r − 270 = 0. Using the quadratic formula to find the positive root, we obtain Solving v(t) = −32t + 32 = 0 we see that the maximum height is attained when t = 1 second. The maximum height is s(1) = 400 ft. Setting s(t) = −16t 2 + 32t + 384 = 0, we have t 2 − 2t − 24 = (t − 6)(t + 4) = 0. Thus, the ball hits the ground at 6 seconds.

Solving s(t) = 1 2 at 2 + v 0 t = 0, we obtain t = 0 and t = − 2v 0 a . Then v(−2v 0 /a) = −v 0 , and the speed at impact with the ground is the initial velocity v 0 . 33. Let a be the acceleration of gravity on the earth, v(0) = v 0 , and s(0

To find the maximum height reached on earth, we solve v(t) = at + v 0 = 0. The maximum height is reached when t = −v 0 /a and is s( for ϑ 0 we see that the initial velocity on earth must be √ 2v 0 . (2 + y − y 2 ) dy = 2y + 1 2

Powers of Trigonometric Functions

POWERS OF TRIGONOMETRIC FUNCTIONS

435 15. tan 3 2t sec 4 2t dt = tan 3 2t sec 2 2t sec 2 2t dt = tan 3 2t(1 + tan 2 2t) sec 2 2t dt = 1 2 tan 3 2t(2 sec 2 2t dt) + 1 2 tan 5 2t(2 sec 2 2t dt)

Solving for the integral, we have

18. tan 2 3x sec 2 3x dx = 1 3 tan 2 3x sec 2 3x(3 dx) u = tan 3x, du = sec 2 3x(3 dx)

Thus,

The last integral is evaluated in Example 8 of Section 7.4. Thus,

Solving for the integral, we have

439

From Exercise 33,

53. If m = n, then using the fact that sin mx sin nx is an even function we have

54. Since sin mx cos nx is an odd function, π −π sin mx sin nx dx = 0.

442

Trigonometric Substitutions

Alternatively, the substitution x = 6 sec θ could have been used.

Solving for the integral csc 3 θ dθ, we have

3 sin θ − 5 (9 − 9 sin 2 θ) 3/2 (3 cos θ dθ) = 9 sin θ cos θ − 15 cos θ 27 cos 3 θ dθ

x − 3 = 3 sin θ, dx = 3 cos θ dθ = 9 − 9 sin 2 θ 3 cos θ dθ = 9 cos 2 θ dθ

(2 + 2 cos 2θ) dθ = (2θ + sin 2θ)

(1 + cot 2 θ) csc 2 θ dθ

. We find the area in the first quadrant and use symmetry.

. We find the area in the first quadrant and use symmetry.

TRIGONOMETRIC SUBSTITUTIONS

53. Using the shell method,

From Section 7.3, Example 5 we obtain

2.3,

Partial Fractions

This expression does not fall under any of the four partial fraction decomposition cases covered in Section 7.6.

Setting x = 0 and x = 2 gives A = −1/2 and B = 1/2. Thus

Setting x = 0 and x = −3/2 gives A = 1/3 and B = −2/3. Thus

Setting x = 0 and x = 1/2 gives A = −2 and B = 5. Thus

12. Write 3x + 10

. Then 3x + 10 = A(x + 2) + Bx.

Setting x = 0 and x = −2 gives A = 5 and B = −2. Thus 3x + 10

Setting x = −4 and x = 4 gives A = 3/8 and B = 5/8. Thus

14. Write 1

Setting x = −5/2 and x = 5/2 gives A = −1/10 and B = 1/10. Thus

Setting x = −1/2 and x = −2 gives A = −1/3 and B = 2/3. Thus

.

Setting x = −4, x = 1, and x = −1 gives A = 1/15, B = 3/5, and C = −2/3. Thus

462

PARTIAL FRACTIONS

Solving

gives A = −8, B = 4, C = 8, and D = 4. Thus

Then 1 = A(x + 4) + B(x + 1). Setting x = −1 and x = −4 gives A = 1/3 and B = −1/3. Thus

Then 8x = (Ax + B)(x 2 + 4) + (Cx + D)(x 2 + 1)

Solving

gives A = 8/3, B = 0, C = −8/3, and D = 0. Thus

Then 1 = A(u + 2) + B(u + 1). Setting u = −1 and u = −2 gives A = 1 and B = −1. Thus cos x sin 2 x + 3 sin x + 2 dx = 1 u + 1 du − 1 u + 2 du = ln |u + 1| − ln |u + 2| + C = ln u + 1 u + 2 + C = ln sin x + 1 sin x + 2 + C.

465

Solving

gives A = 1/36, B = −1/25, C = 11/900, and D = 1/30. (Note that A and B can be easily obtained by substituting x = 4 and x = 3, respectively, in the initial equation.) Thus

gives A = −19/16, B = 19/8, C = −11/4, D = 9/2, and E = 35/16. Thus

52.

1 0

x 2 55.

Solving

gives A = 1, B = 1, C = −1, and D = 1. Thus

. Setting x = −3 and x = 1 gives A = −1/4 and B = 1/4. Thus

Then

Solving

Setting x = −2 and x = −3 gives A = −2 and B = 3. Thus = 7 ln 2 − 8 ln 3 + 3 ln 4 = ln 8192 6561 ≈ 0.2220.

.

68

.

Solving

Solving

gives A = −1/9, B = −1/3, and C = 1/9. Thus

Then 1 = A(u + 1) + B(u − 1). Setting u = 1 and u = −1 gives A = 1/2 and B = −1/2.

. Partial fraction decomposition is unnecessary because the substitution u = x 4 − 1, du = 4x 3 dx will suffice.

480

Improper Integrals

In this exercise set, the symbol " h =" is used to denote the fact that L'Hôpital's Rule was applied to obtain the equality.

1.

The integral diverges.

3.

With b = 17 and c = 299.13 we find the percentage of total deaths within the first 34 weeks is 2

(1 + 2 4/3 + 2 5/3 + 2 √ 2 + 2 7/3 + 2 8/3

3/2 5/2 7/2 9/2 11/2 f (x k ) 2/3 2/5 2/7 2/9 2/11

Since the graph of f (x) = c 1 x + c 0 is a straight line and the Trapezoidal Rule uses straight line approximations to the curve, it will give the exact value.

36. From the derivation of Simpson's Rule in the text, it suffices to show that it will give the

In this case, the exact value is

and Simpson's Rule gives

37.

Setting x = 0, x = 2, and x = −2 gives A = 1/4, B = 5/8, and C = −7/8. Thus

. Then 1 = A(u 2 + 3u + 9) + (Bu + C)(u − 3) = (A + B)u 2 + (3A − 3B + C)u + (9A − 3C).

gives A = 1/27, B = −1/27, and C = −2/9. Thus

.

gives A = −1/27, B = −1/9, C = −1/3, and D = 1/27. Thus

17.

ln(x 2 + 4) dx u = ln(x 2 + 4), du = 2x

Solving

gives A = −2/125, B = 1/25, C = 2/125, and D = 1/25. Thus

Solving

gives A = −1/12, B = 1/2, and C = 1/12. Thus

27.

y cos y dy u = y, du = dy; dv = cos y dy, v = sin y = y sin y − sin y dy = y sin y + cos y + C

28. (1 + sin 2 t) cos 3 t dt = (1 + sin 2 t)(1 − sin 2 t) cos t dt = (1 − sin 4 t) cos t dt u = sin t, du = cos t dt

. Then 1 = A(u + 1) + B(u − 1). Setting u = 1 and u = −1

gives A = 1/2 and B = −1/2. Thus

(3 − sec x) 2 dx = (9 − 6 sec x + sec 2 x) dx = 9 − 6 ln | sec x + tan x| + tan x + C 35.

Alternatively, the substitution u = tan x 2 leads to the equivalent solution sin x 1 + sin x dx = x + 2 1 + tan x/2 + C.

38

. cos x 1 + sin x dx u = 1 + sin x, du = cos x dx

Then 1 = A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2).

Setting x = −1, x = −2, and x = −3 gives A = 1/2, B = −1, and C = 1/2. Thus

Solving for the integral, e x cos 3x dx = 3 10 e x sin 3x + 1 10 e x cos 3x + C.

42.

x(x − 5) 9 dx u = x − 5, du = dx = (u + 5)u 9 du = (u 10 + 5u 9 ) du = 1 11 u 11 + 1 2 u 10 + C

Solving for the integral, cos(ln t) dt = 1 2 t cos(ln t) + 1 2 t sin(ln t) + C.

518

Area Revisited

3 cos x dx

55.

59. The area of the ellipse is four times the area in the first quadrant portion of the ellipse. Thus, (1, 1) (2 − e x ) dx.

The area with respect to y is A y = 2 1 ln y − ln y + 1 2 dy.

If integration with respect to x is chosen, we get

ln 2 ln 3/2 = 3 2 − ln 3 2 − 1 + 2 ln 2 − 2 − 2 ln 3 2 + 3 2 = −3 ln 3 + 5 ln 2 ≈ 0.1699.

If integration with respect to y is chosen, we get

ln y − ln y + 1 2 dy = y ln y − y − (y + 1) ln y + 1 2 + (y + 1) The area is then

65. At P (x 0 , 1/x 0 ) the slope of the line segment is −1/x 2 0 . The equation of the line through Q and R is then y = −x/x 2 0 + 2/x 0 . Setting y = 0 we see that the x-intercept is 2x 0 . The area is

which does not depend on x 0 . 70. This project involves a research report, and thus a preset solution is not applicable. It is noted, however, that Cavalieri's Principle relates directly to the situations presented in Problems 68 and 69. Using similar triangles, we have

66.

VOLUMES OF SOLIDS: SLICING METHOD

( 1 − y) 2 dy = π 1 -x

41. The volume of the right circular cylinder is πr 2 h. Placing the center of the red circular cylinder's base in Figure 6.3.19 on the origin, we see that A = πr 2 for every slice from y = 0 to h. Thus, the volume V of the cylinder is

3.3.19

358

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

43. (a) Using Mathematica, we obtain with the disk method

(5a 2 + 9b 2 + 21c 2 + 105d 2 + 18ac + 42bd).

(b) The weight of the ball is 4 3 πr 3 ρ ball and the weight of water displaced is π 3 (3rh 2 −h 3 )ρ water . 45. (a) Each eighth of the bicylinder can be sliced into squares whose sides follow the perimeter of a quadrant of the cylinders' base; that is, x 2 + y 2 = r 2 , one side of the square is y = √ r 2 − x 2 , and its area is y 2 = r 2 − x 2 . Using symmetry, the volume common to the cylinders is thus

(b) This item involves a research report, and thus a preset solution is not applicable. x( 18. y 1 = x 1/3 + 1;

VOLUMES OF SOLIDS: SHELL METHOD

. We use the shell method.

32. The equation of the line through (r 1 , h) and (r 2 , 0) is x = 1 h (r 1 − r 2 )y + r 2 . We use the disk method.

33. We use the disk method.

34. The equation of the line is y = 1 a √ r 2 − a 2 x and the equation of the circle is y = √ r 2 − x 2 . We use the disk method.

35. The equation of the ellipse is y = b 1 − x 2 a 2 . We use the disk method.

Since the solid is symmetric with respect to the x-axis, we will find the volume of the upper hemispheroid and multiply by 2. We use the shell method. 37. y 1 = ω 2 x 2 2g . The depth of the liquid below the x-axis is y 2 = h − ω 2 r 2 2g .

So the volume is

359

Volumes of Solids: Shell Method

The liquid will touch the bottom of the bucket when y

volume of the liquid is then

The graph is symmetric with respect to both coordinate axes, so

Letting r = 1 we have 2π = 4 1 + x 6 dx = s(2 + 0.1) ≈ s(2) + s (2) dx = 0 + 1 + 2 6 (0.1) = 0.1 √ 65 ≈ 0.8062. x 1 + 4x 2 dx = π 6 (1 + 4x 2 ) 3/2 3 0 = π 6 (37 3/2 − 1) ≈ 117.3187

Area of a Surface of Revolution

(1 + 4x 2 ) 3/2 2 0 = π 6 (17 3/2 − 1) ≈ 36.1769

PROBLEMAS 3.6

AREA OF A SURFACE OF REVOLUTION

13. For x < −2, y = −x − 2 and y = −1. For x > 2, y = x + 2 and y = 1.

14. Since the graph is symmetric with respect to the y-axis, we will find the area on [0, a] and multiply by 2.

15. Let θ be the angle formed when the cone is cut and flattened out. The length of the arc of the sector is 2πr, the circumference of the base of the cone. Then θ/2πr = 2π/2πL (the angle subtended by the sector is to the length of the sector as 2π radians is to the circumference of the circle of radius L), and θ = 2πr/L.

Using the hint in the text, the lateral surface area is 1 2

16. If L is the slant height, then L 2 = r 2 + h 2 and the surface area is πrL = πr √ r 2 + h 2 . The surface can also be obtained by revolving the line y = r h x about the x-axis:

17. By similar triangles, r 1 L 1 = r 2 L 2 , r 1 L 2 = r 2 L 1 , and r 1 L 2 − r 2 L 1 = 0.

From Problem 15, the lateral surface area of the frustum is

18. By the Pythagorean Theorem, L 2 = h 2 + (r 2 − r 1 ) 2 or L = h 2 + (r 2 − r 1 ) 2 . From (1) in Section 6.6,

19. We need to extend (3) in Section 6.6 to include functions which are not necessarily nonnegative. In this case we have S = 2π b a |f (x)| 1 + [f (x)] 2 dx. Next, we require the fact that the surface area obtained by revolving f around y = L is the same as that obtained by revolving

20. y = 2 3

x −1/3 ; S = 2π gives y B = R 2 R + h . Now, revolving x = R 2 − y 2 for y B ≤ y ≤ R around the y-axis, we obtain the surface area

Average Value of a Function

19. f ave = 1 π/2 − π/6 π/2 π/6

Setting f (c) = c 2 + c = 1 3 , we obtain 3c 2 + 6c − 1 = 0. Then c = −6 ± √ 36 + 12

The only solution on [−1, 1] is −1 + 2 3 (

AVERAGE VALUE OF A FUNCTION

377 32. Intuitively, the average value of the linear function f (x) = ax + b on [x 1 , x 2 ] should be the value of f at the midpoint of that interval, or X = x 1 + x 2 2 . This can be proven as follows:

= a n + a n−1 + · · · + a + 1

and therefore The average value of f on the interval [0, n] appears to be 1 2 (n − 1), which can be proven as follows:

37. There is no unique answer to this question; among several possible approaches, here is probably the simplest. Suppose the circle is centered at the origin and that one of the points on the circle is (−1, 0). If (x, y) is any other point on the circle, then the length of the chords between (−1, 0) and (x, y) is the distance between the points:

The average chord length L ave is then any x in [a, b] the circumference of a circular cross-section is 2πf (x). The average

(a) The surface area is

The first integral represents the area of the semicircle and is thus 1 2 π(3) 2 . The second integral has an odd integrand and is thus 0. Therefore W = 1497.6 5 · 1 2 π(3) 2 = 33, 696π ft-lb. (2, 700, 000 − 100x) dx = 2, 700, 000x − 50x 2 1000 0 = 2, 650, 000, 000 ft-lb

where p 1 and p 2 are the pressures corresponding to volumes v 1 and v 2 , respectively.

24. Using Newton's second law F = ma = mg, we have

25. Since the distance moved is 0, no work is done.

The force is

. Using symmetry,

x dx = 124.8 − √ 3 9

x 3 + 5 √ 3 12

x 8. Solving x = y 2 and y = −x + 2 simultaneously, we find that the graphs intersect at (4, −2) and (1, 1). To compute the force we divide the plate into two parts with the line x = 1. For the upper part we use symmetry. The first integral has an odd integrand and is thus 0. The second integral represents the area of the circle and is thus π(4) 2 . Therefore F = 624(16π) = 9984π lb. x 9 − 9 4

x 2 dx + 10ρ 2 −2 2 9 − 9 4

x 2 dx.

The first integral has an odd integrand and is thus 0. The second integral represents the area of the ellipse and is thus π(2)(3). Therefor F = 10ρ(6π) = 60πρ lb.

CENTERS OF MASS AND CENTROIDS

x = 9/4 9/2 = 1 2 ; y = 36/5 9/2 = 8 5

x = 3/20 1/3 = 9 20 ; y = 3/20 1/3 = 9 20

393

x = −27/10 9/2 = − 3 5 ; y = 9/4 9/2 = 1 2 2 -/ 2 / 2 39. By symmetry, x = 0. . To verify this result we compute the volume using the washer method.

43. We identify A = πa 2 . The centroid of the region R is b units from L, so V = 2πb(πa 2 ) = 2π 2 a 2 b.

-3 3 6 9 5 10 44. Since the graph is symmetric around x = 3, we would expect the center of mass to occur at the center of the rod.

45. Thinking geometrically, the centroid of a triangle would appear to be the intersection of its three medians (a median is a line segment from one of the triangle's vertices to the midpoint of the opposing side). Doing some research on the centroid of a triangle shows this to be true, and in fact this intersection is the mean of the coordinates of the triangle's vertices.

46. Decomposing the region R into three 1 × 1 squares whose centers are (1/2, 3/2), (3/2, 3/2), and (3/2, 1/2), we get the centroid of R = ([1/2 + 3/2 + 3/2]/3, [3/2 + 3/2 + 1/2]/3) = ([7/2]/3, [7/2]/3) = (7/6, 7/6).

Chapter 6 in Review

True

PROBLEMAS DE REPASO DE LA UNIDAD 3

17. Solving sin x = sin 2x, we get sin x = 2 sin x cos x, cos x = 1 2 , and x = π 3 on (0, π). Thus, . At a rate of loss of 1/4 pound per second, it will take 120 seconds to lose the entire 30 pounds.

In 120 seconds, the bucket will be raised 120 feet. 13. Let > 0. Then n n + 1 − 1 = − 1 n + 1 < implies 1 n + 1 < or n > 1 − 1. Take N to be the smallest integer greater than 1 − 1.

14. Let > 0. Then e n + 1 e n − 1 = 1 e n < implies 1 e n < or n > ln 1 . Take N to be the smallest integer greater than ln 1 . 29. Since the terms alternate between −1 and 1, the sequence diverges.

lim

30. Since sin nπ = 0 for all integers n, the sequence converges to 0.

Since the terms alternate between values that are increasingly greater than and less than 0, the sequence diverges.

n→∞ a n+1 a n = lim n→∞ 5 n+1 x n+1 /(n + 1)! 5 n x n /n! = lim n→∞ 5 n + 1 = 0

The series is absolutely convergent on (−∞, ∞).

n→∞ a n+1 a n = lim n→∞ (x − 3) n+1 /(n + 1) 3 (x − 3) n /n 3 = lim n→∞ n n + 1

The series is absolutely convergent for |x − 3| < 1 or on (2, 4). At x = 2, the series . Thus, the series converges on [3, 5)..

n→∞ a n+1 a n = lim n→∞ (n + 1)!2 n+1 x n+1 n!2 n x n = lim n→∞ 2(n + 1)|x| = ∞, x = 0

The series converges only at x = 0.

10. lim n→∞ a n+1 a n = lim n→∞ nx n+1 /(n + 1) 2(n+1) (n − 1)x n /n 2n = lim n→∞ nn 2n (n − 1)(n + 1)2 2n+2 |x| = lim n→∞ n (n + 1)(n + 1) 2 n n + 1 2n |x| = lim n→∞ n (n − 1)(n + 1) 2 · 1 n + 1 n n 2 |x| = lim n→∞ n (n − 1)(n + 1) 2 · 1 [(1 + 1/n) n ] 2 |x| = 0 · 1 e 2 |x| = 0 The series is convergent on (−∞, ∞).

n→∞ a n+1 a n = lim n→∞ (3x − 1) n+1 /[(n + 1) 2 + (n + 1)] (3x − 1) n /(n 2 + n) = lim n→∞ n 2 + n n 2 + 3n + 2 |3x − 1| = |3x − 1|

The series is absolutely convergent for |3x − 1| < 1 or on (0, 2/3). At x = 0, the series with the p-series ∞ k=0 1 k 2 . At x = −13/3, the series ∞ k=1 3 k (−2) k k(k + 1) 2 3 k = ∞ k=1 (−1) k k(k + 1) converges by the alternating series test. Thus, the given series converges on [−17/3, −13/3].

n→∞ a n+1 a n = lim n→∞ (x − 2) n+1 /(n + 1)!(n + 1)!3 n+1 (x − 2) n /n!n!3 n = lim n→∞ 1 3(n + 1) 2 |x − 2| = 0 The series converges on (−∞, ∞). 23. lim n→∞ a n+1 a n = lim n→∞ x 2(n+1)+1 /9 n+1 x 2n+1 /9 n = lim n→∞ 1 9

x 2 = 1 9

x 2

The series is absolutely convergent for 1 9

x 2 < 1 or on (−3, 3). At x = −3 the series (−1) k 3 diverges by the n-th term test.

Thus, the given series converges on (−3, 3).

n→∞ a n+1 a n = lim n→∞ 5 n+1 x 2(n+1) /(2n + 1)! 5 n x 2n /(2n)! = lim n→∞ 5x 2 (2n + 2)(2n + 1) = 0

The series is absolutely convergent on (−∞, ∞). The series is absolutely convergent on (−∞, ∞). Thus, ln 1+1/k) ≤ 1/k and since e x is an increasing function, (1+1/k) ≤ e 1/k and (1+1/k) k ≤ e. Hence, e (1 + 1/k) k ≥ 1 and the series 29. lim n→∞ a n+1 a n = lim n→∞ (n + 1)!(x/2) n+1 1 · 3 · 5 · · · (2n − 1)(2n + 1) n!(x/2) n 1 · 3 · 5 · · · (2n − 1) = lim n→∞ n + 1 2n + 1

The series has radius of convergence 4.

n→∞ a n+1 a n = lim n→∞ 1 · 3 · 5 · · · (2n − 1)(2n + 1)(x − 1) n+1 3 n+1 (x + 1)! 1 · 3 · 5 · · · (2n − 3)(2n − 1)(x − 1) n 3 n n! = lim n→∞ 2n − 1 3(n + 1)

The series has radius of convergence 3/2.

{(−1) n 2}

Since the terms alternate between −2 and 2, the sequence diverges. 58. a n+1 = 1 2 a n + 5 a n =⇒ lim n→∞ a n+1 = 1 2 lim n→∞ a n + 5 lim n→∞ a n

CHAPTER 9. SEQUENCES AND SERIES

59. For a n = 5 n n! , we have a n+1 = 5 n+1 (n + 1)! . Expanding, we get a n+1 = 5 · 5 n (n + 1) · n! = 5 n + 1 5 n n! .

The last factor of the last term is a n , so a n+1 = 5 n + 1 a n .

60. Starting with a 1 = √ 3, we are given a 2 = √ 3 + a 1 , a 3 = √ 3 + a 2 , and so on. Thus, the recursion formula is a n+1 = √ 3 + a n .

61. Let a n = 0, b n = sin 2 n 4 n , and c n = 1 4 n . Then lim n→∞ a n = lim n→∞ c n = 0, so lim n→∞ sin 2 n 4 n = 0.

62. Let a n = 4, b n = 16 + 1 n 2 , and c n = 4 + 1 n .

[To see that b n ≤ c n , note that for x, y ≥ 0,

x 2 + y 2 ≤ x 2 + 2xy + y 2 = (x + y) 2 . Thus, x 2 + y 2 ≤ x + y.] Then lim n→∞ a n = lim n→∞ c n = 4, so lim n→∞ 16 + 1 n 2 = 4.

63. Let a n = 0, b n = ln n n(n + 2)

, and c n = n n(n + 2) = 1 n + 2 . Then, for n ≥ 1, a n ≤ b n ≤ c n .

Since lim n→∞ a n = lim n→∞ c n = 0, we have lim n→∞ ln n n(n + 2) = 0.

64. Let a n = 0, b n = n! n n = 1 n 2 n · 3 n · 4 n · · · n n , and c n = 1 n .

[To see that b n ≤ c n , note that 75. Since {a n } converges, then lim n→∞ a n is some value L. Since the limit of a product is the same as the product of the factors' limits, we have lim n→∞ a 2 n = lim n→∞ (a n ·a n ) = ( lim n→∞ a n )( lim n→∞ a n ) = L 2 .

Thus, a 2 n actually does converge.

15. Since a n+1 a n = 3 n+1 /(1 + 3 n+1 ) 3 n /(1 + 3 n ) = 3 + 3 n+1 1 + 3 n+1 = 1 + 2 1 + 3 n+1 > 1, the sequence is monotonic.

Using 3 n 1 + 3 n > 0 and 3 n 1 + 3 n = 1 − 1 1 + 3 n < 1, we see that the sequence is bounded. Thus, the sequence converges.

16. Since a n+1 a n = (n + 1)/5 n+1 n/5 n = n + 1 5n = 1 5 1 + 1 n < 1, the sequence is monotonic. Using n 5 n > 0 and n 5 n ≤ 1 5 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

17. Let f (x) = e 1/x . Then f (x) = −e 1/x x 2 < 0 and the sequence is monotonic. Using e 1/n > 0 and e 1/n ≤ e (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

Thus, 0 < S n < 2 − 1 n . Since n > 0, then 0 < S n < 2 for all n and so {S n } is bounded.

Because S n is a partial sum whose addends are all positive, {S n } is also monotonic.

58. Integrating by parts, we obtain

Then, using L'Hôpital's Rule,

68. AP 1 + P 1 P 2 + P 2 P 3 + P 3 P 4 + P 4 P 5 + P 5 P 6 + · · · = √ 2 + 1 + √ 2 2 + 1 2 + √ 2 4 + 1 4 + · · · = √ 2 1 + 1 2 + 1 2 2 + · · · + 1 + 1 2 + 1 2 2 + · · · = (1 + √ 2) 1 + 1 2 + 1 2 2 + · · · = (1 + √ 2) 1 1 − 1/2 = 2 + 2 √ 2 69. AP 1 + P 1 P 2 + P 2 P 3 + P 3 P 4 + P 4 P 5 + P 5 P 6 + · · · = sin 30 • + (cos 30 • ) sin 30 • + (cos 30 • ) 2 sin 30 • + (cos 30 • ) 3 sin 30 • + · · · = 1 2 +

70. The function f is nonzero on the intervals

(1 − 1/2, 1 + 1/2), (2 − 1/4, 2 + 1/4), (3 − 1/8, 3 + 1/8), . . . , (n − 1/2 n , n + 1/2 n ), . . .

where A(n) is the area of the isosceles triangles whose base is centered at n. Thus, denotes the x-coordinate of the center of mass of the kth book, then the centers of mass are defined by x n = mx 1 + mx 2 + · · · + mx n nm = 1 n n k=1

x k .

Therefore,

In other words, the center of mass for each stack of books is at the edge of the (c) The overhang of n books from the edge of the table is

where H n = n k=1 1 k are the harmonic numbers. The x-coordinate of the center of mass is

Since the overhang of the first (or bottom) book in the stack from the edge of the Alternatively, note that the fly flies at a constant rate of 20 mph for 1 hour. Thus, the distance covered is 20(1) = 20 miles.

19. The function f (x) = 1 1 + √ x is continuous and decreasing on [1, ∞). Since

the integral diverges and

20. The function f (x) = 1 √ 1 + x 2 is continuous and decreasing on [1, ∞). Since

the integral diverges and

21. The function f (x) = x (x 2 + 1) 3 is continuous and decreasing on [1, ∞). Since 22. The function f (x) = 1 (4x + 1) 3/2 is continuous and decreasing on [1, ∞). Since

the integral converges and ∞ k=1 1 (kx + 1) 3/2 converges.

29. The function f (x) = 2 e x + e −x is continuous and decreasing on [1, ∞). Since

34. Since k √ k − 1 3 √ k 2 − 2 > 1 k 1/6 for k ≥ 2, the series diverges by comparison with the p-series ∞ k=2 1 k 1/6 . 35. Since 1 9 + sin 2 k > 1 10 for k ≥ 1, the series diverges by comparison with the series ∞ k=1 1 10 .

36. Using the limit comparison test with a n = 3 n 3 2n − 1 and b n = 1 3 n , we have lim n→∞ a n b n = lim 40. Since (0.9) k k ≤ (0.9) k , the series converges by comparison with the geometric series ∞ k=1 (0.9) k . 41. Since a k converges, then lim k→∞ a k = 0. Therefore, for a sufficiently large n and k ≥ n, we can say that 0 < a k < 1 and so a 2 k < a k . Thus, a 2 k converges by the direct comparison test.

42. Assuming q(k) = 0 for k ≥ 1, the series will converge if m ≥ n + 2.

43. The statement is false. The condition of a positive-term series is missing: as a counterexample, consider the convergent series b k = 0 = 0 and the divergent series a k = − 1 k .

44. For a k > 0 for all k, the limit comparison test lim n→∞ ln(1 + a n ) a n = lim 48. This exercise involves a research report, and thus a preset solution is not given. 11. Since lim n→∞ a n+1 a n = lim n→∞ 99 n+1 ((n + 1) 3 + 1) 99 n (n 3 + 1) · n 2 100 n (n + 1) 2 100 n+1 = lim n→∞ 99 100 · (n + 1) 3 + 1 n 3 + 1 · n 2 (n + 1) 2 = 99 100 lim n→∞ (n + 1) 3 + 1 n 3 + 1 lim n→∞ n 2 (n + 1) 2 = 99 100 · 1 · 1 = 99 100 < 1, the series converges by the ratio test.

Then a k+1 = e (1 + 1/k) k a k . We will now show that (1 + 1/k) k < e. Using the definition of ln x, we have ln(1 + 1/k) = Thus, ln(1+1/k) ≤ 1/k and since e x is an increasing function, (1+1/k) ≤ e 1/k and (1+1/k) k ≤ e. Hence, e (1 + 1/k) k ≥ 1 and a k+1 = e (1 + 1/k) k a k ≥ a k . Therefore, the two series diverge because their terms do not approach 0. The given series converges on (∞, −1/e) (1/e, ∞).

37. This is a geometric series with common ratio r = x x and will converge for |e x | = e x < 1 or

x < 0. Thus, the series converges on (−∞, 0).

2, 3, 5, 8, 13, · · ·

The recursion formula matches the pattern in Problem 78. Not surprisingly, the resulting sequence is called the Fibonacci sequence.

80. (a) a n = 1 + 1 a n−1 =⇒ lim n→∞ a n = 1 + 1 lim n→∞ a n−1

By the quadratic formula, the solutions are −(−1) ± (−1) 2 − 4(1)(−1) 2(1) = 1 ± √ 5 2 .

The limit must be positive, so φ = 1 + √ 5 2 .

(b) Since this portion of the exercise involves a research report, no solution is given. The relationship between φ and the shape of the multi-chambered nautilus shell lies in how the shell's successive spirals grow at a rate that approximates φ for every quarter turn that they make.

Monotonic Sequences

1. a n+1 − a n = n + 1 3n + 4 − n 3n + 1 = 1 (3n + 4)(3n + 1) > 0. The sequence is increasing.

2. a n+1 − a n = 11 + n n + 1 − 10 + n n = −10 n(n + 1) < 0. The sequence is decreasing.

3. a 1 = −1, a 2 = √ 2, a 3 = − √ 3. The sequence is not monotonic.

PROBLEMAS 4.2

MONOTONIC SEQUENCES

The sequence is nondecreasing.

The sequence is increasing.

. Since f (x) < 0 for x < 5 and f (x) > 0 for x > 5, the sequence is not monotonic.

7. a 1 = 2, a 2 = 2, a n+1 a n = 2 n+1 /(n + 1)! 2 n /n! = 2 n + 1 < 1 for n > 1.

The sequence is nonincreasing.

8. a n+1 a n = 2 2n+2 [(n + 1)!] 2 /(2n + 2)! 2 2n (n!) 2 /(2n)! = 4(n + 1) 2 (2n + 2)(2n + 1) = 4n 2 + 8n + 4

The sequence is increasing.

The sequence is increasing.

10. a 1 = 0, a 2 = 6, a 3 = 6 a n+1 − a n = (n + 1) 2 + (−1) n+1 (n + 1) − n 2 − (−1) n n

The sequence is nondecreasing.

11. Since n < π for n = 1, 2, 3, we have sin n > 0 for n = 1, 2, 3. Since π < n < 2π for n = 4, 5, we have sin n < 0 for n = 4, 5. Thus, a 3 > 0, a 4 < 0, and a 5 > 0. The sequence is not monotonic.

12. a n+1 − a n = ln n + 3 n + 2 − ln n + 2 n + 1 = ln (n + 3)(n + 1) (n + 2) 2 = ln n 2 + 4n + 3 n 2 + 4n + 4

Since n 2 + 4n + 3 n 2 + 4n + 4 < 1, ln n 2 + 4n + 3 n 2 + 4n + 4 < 0 and the sequence is decreasing.

13. Since a n+1 − a n = 4n + 3 5n + 7 − 4n − 1 5n + 2 = 13 (5n + 7)(5n + 2) > 0, the sequence is monotonic.

Using 4n − 1 5n + 2 > 0 and 4n − 1 5n + 2 < 4n 5n + 2 < 4n 5n = 4 5 , we see that the sequence is bounded.

Thus, the sequence converges.

14. Since a n+1 − a n = 6 − 4(n + 1) 2 1 + (n + 1) 2 − 6 − 4n 2 1 + n 2 = 10[n 2 − (n + 1) 2 ] [1 + (n + 1) 2 ][1 + n 2 ] < 0, the sequence is monotonic. Using 6 − 4n 2 1 + n 2 < 6 and 6 − 4n 2 1 + n 2 > −4 − 4n 2 1 + n 2 = −4(1 + n 2 ) 1 + n 2 = −4, we see that the sequence is bounded. Thus, the sequence converges.

578

Since

a n+1 a n = (n + 1)!/(n + 1) n+1 n!/n n = (n + 1)n n (n + 1) n+1 = n n + 1 n < 1, the sequence is monotonic. Using n! n n > 0 and n! n n ≤ 1 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

a n+1 a n = (n + 1)!/[1 · 3 · 5 · · · (2n + 1)] n!/[1 · 3 · 5 · · · (2n − 1)] = n + 1 2n + 1 < 1, the sequence is monotonic. Using n! 1 · 3 · 5 · · · (2n − 1) > 0 and n! 1 · 3 · 5 · · · (2n − 1) ≤ 1 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

20. Since a n+1 a n = [2 · 4 · 6 · · · (2n + 2)]/[1 · 3 · 5 · · · (2n + 3)] [2 · 4 · 6 · · · 2n]/[1 · 3 · 5 · · · (2n + 1)] = 2n + 2 2n + 3 < 1, the sequence is monotonic. Using 2 · 4 · 6 · · · 2n 1 · 3 · 5 · · · (2n + 1) > 0 and 2 · 4 · 6 · · · 2n 1 · 3 · 5 · · · (2n + 1) ≤ 2 3 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

21. Let f (x) = tan −1 x. Then f (x) = 1 1 + x 2 > 0 and the sequence is monotonic. Since | tan −1 n| < π 2 , we see that the sequence is bounded. Thus, the sequence converges.

22. Let f (x) = ln(x + 3)

x + 3 . Then f (x) = 1 − ln(x + 3) (x + 3) 2 < 0 for x ≥ 0 and the sequence is monotonic. Using ln(n + 3) n + 3 > 0 and ln(n + 3) n + 3 < ln 4 4 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

23. The sequence is {(0.8) n }. Since a n+1 a n = (0.8) n+1 (0.8) n = 0.8 < 1, the sequence is monotonic. Using (0.8) n > 0 and (0.8) n ≤ 0.8 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

b n is a p-series with p = 1 2 < 1, it diverges and ∞ k=1 1 10 + √ k diverges.

17. Using the limit comparison test with a n = 1 n √ n 2 − 1 and b n = 1 n 2 , we have lim n→∞ a n b n = lim n→∞ 1/n √ n 2 − 1 1/n 2 = lim n→∞ n 2 n √ n 2 − 1 = lim n→∞ 1 1 − 1/n 2 = 1.

Since ∞ n=1 b n is a p-series with p = 2 > 1, it converges and ∞ n=2 1 n √ n 2 − 1 converges.

18. Using the limit comparison test with a n = 1 (n + 1)(n + 2) and b n = 1 n , we have lim n→∞ a n b n = lim n→∞ 1/ (n + 1)(n + 2) 1/n = lim n→∞ n √ n 2 + 3n + 2 = lim n→∞ 1 1 + 3/n + 2/n 2 = 1.

Since ∞ n=1 1 n diverges, ∞ n=1 1 (n + 1)(n + 2) diverges.

19. Using the limit comparison test with a n = n 2 − n + 2 3n 5 + n 2 and b n = 1 n 3 , we have lim n→∞ a n b n = lim n→∞ (n 2 − n + 2)/(3n 5 + n 2 ) 1/n 3 = lim n→∞ n 5 − n 4 + 2n 3 3n 5 + n 2 = lim n→∞ 1 − 1/n + 2/n 2 3 + 1/n 3 = 1 3 .

Since ∞ n=1 b n is a p-series with p = 3 > 1, it converges and ∞ n=1 n 2 − n + 2 3n 5 + n 2 converges.

20. Using the limit comparison test with a n = n (4n + 1) 3/2 and b n = 1 n 1/2 , we have lim n→∞ a n b n = lim n→∞ n/(4n + 1) 3/2 1/n 1/2 = lim n→∞ n 3/2 (4n + 1) 3/2 = lim

b n is a p-series with p = 1 2 < 1, it diverges and 23. Using the limit comparison test with a n = n + ln n n 3 + 2n − 1 and b n = 1 n 2 , we have lim n→∞ a n b n = lim n→∞ (n + ln n)/(n 3 + 2n − 1) 1/n 2 = lim n→∞ n 3 + n 2 ln n n 3 + 2n − 1 = lim n→∞ 1 + (ln n)/n 1 + 2/n − 1/n = 1.

the series can be written as

b k is a p-series with p = 1 2 < 1, it diverges and the given series is not absolutely

and a k+1 < a k . Since lim k→∞ a k = lim 42. The error will be less than a 7 = 1 7(2 7 ) ≈ 0.00112.

43. This is not an alternating series since, for k = 1 to k = 5, the terms are positive, while for k = 7 to k = 11, the terms are negative. Since |a k+1 | ≤ 1 k 2 , the series is absolutely convergent by comparison with the p-series ∞ k=1 1 k 2 . Hence, the series is convergent.

44. This is not an alternating series since for k ≤ 6, the terms are positive. For k ≥ 7, the terms alternate but do not satisfy |a k+1 | ≤ |a k | since a 7 = 100 − 2 7 3 7 ≈ −0.013 and a 8 = 100 + 2 8 3 8 ≈ 0.054. Write 45. This is not an alternating series. Since |a k | = 1 2 k−1 , ∞ k=1 |a k | is a geometric series with r = 1 2 < 1, and the original series is absolutely convergent.

46. This is not an alternating series. Since |a k | = 1 k 2 , ∞ k=1 |a k | is a p-series with p = 2 > 1, and the original series is absolutely convergent. 48. This is not an alternating series. The sequence of partial sums S 2 , S 5 , S 9 , S 14 , S 20 , . . . is 1, 0, 1, 0, 1, . . . . Since the sequence diverges, so must {S n }. Thus, the original series diverges.

49. The terms do not approach 0, so the series diverges.

50. The terms of the series are all 0, so the series converges.

51. All terms of the series after the first are 0, so the series converges.

52. All odd terms of the series are 1 or −1, so the terms do not approach 0 and the series diverges.

53. The statement is true because a positive-term series a k is the same as the series of its terms' absolute values |a k |. If this series is convergent, then it is also absolutely convergent, and so, as stated in the discussion, its terms can be rearranged in any manner and the resulting series will converge to the same number as the original series. Thus, the series is absolutely convergent by the root test. x 2 , is a p-series with p = 2 > 1, which also converges. 61. e −x sin x + e −2x sin 2x + e −3x sin 3x + · · · can be written as 3. lim n→∞ a n+1 a n = lim n→∞ 2 n+1 x n+1 /(n + 1) 2 n x n /n = lim n→∞ 2n n + 1 |x| = 2|x|

579

24. The sequence is 3 1/2 n . When x > 1, √ x > 1, so a n+1 a n = 3 1/2 n+1 3 1/2 n = 1 3 1/2 n+1 < 1 and the sequence is monotonic. Using 3 1/2 n > 0 and 3 1/2 n ≤ √ 3 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges.

25. a n+1 = 1 2 a n + 5, a 1 = 1. We will show that a n < 10 for all n.

For n = 1, we have a 2 = 11 2 < 10. Assume that a k < 10. Then a k+1 = 1 2 a k + 5 < 1 2 (10) + 5 = 10; that is, a k+1 < 10 whenever a k < 10. The sequence is bounded because 0 < a n < 10.

Next, we will show that the sequence {a n } is monotonic. Because a n < 10, necessarily 1 2 a n < 1 2 · 10 = 5. Therefore, from the recursion formula, a n+1 = 1 2 a n + 5 > 1 2 a n + 1 2 a n = a n .

This shows that a n+1 > a n for all n, and so the sequence is increasing.

Since {a n } is bounded and monotonic, it follows from Theorem 9.2.1 that the sequence converges. Because we must have lim n→∞ a n = L and lim n→∞ a n+1 = L, the limit of the sequence can be determined from the recursion formula:

26. a n+1 = √ 2 + a n , a 1 = 0. We will show that a n < 2 for all n.

For n = 1, we have a 2 = √ 2 < 2. Assume that a k < 2. Then a k+1 = √ 2 + a k < √ 2 + 2 = 2; that is, a k+1 < 2 whenever a k < 2. The sequence is bounded because 0 < a n < 2.

Also, the sequence is increasing: a n+1 = √ 2 + a n > √ a n + a n = √ 2a n > √ a n a n = a n .

That is, a n+1 > a n .

By Theorem 9.2.1, the sequence is bounded and monotonic, and so it is convergent. From

27. a n+1 = √ 7a n , a 1 = √ 7. Now 0 < a n < 7 =⇒ √ a n < √ 7, and so a n+1 = √ 7a n = √ 7 √ a n > √ a n √ a n = a n .

Thus, a n+1 > a n for all n. The sequence is therefore monotonic (increasing) and bounded. By Theorem 9.2.1, the sequence converges. From 28. Since a n+1 a n = 1 − 1 n 2 < 1 for n ≥ 2, the sequence is monotonic. Using a n > 0 and a n ≤ 2 (since the sequence is decreasing), we see that the sequence is bounded. Thus, the sequence converges. Taking the limit of the recursion formula shows nothing, since lim n→∞ a n+1 = lim n→∞ 1 − 1 n 2 a n = lim n→∞ a n =⇒ L = L.

(c) If a > b then b a < 1 and by part (b), p n+1 < p n . The sequence is thus monotonically

Since 0 < a a + p n < 1, the distance from p n+1 to b − a is less than the distance from p n to b − a. This means that p n+1 is between b − a and p n . Thus, if 0 < p 0 < b − a, the sequence {p n } is increasing and bounded above by b − a. If 0 < b − a < p 0 , the sequence {p n } is decreasing and bounded below by b − a. In either case, it follows from part (a) that the sequence converges to b − a.

30. {(−1) n } is bounded but not convergent.

31. Since {a n } is convergent, it follows from Definition 9.1.2 that there exists an N such that |a n − L| < 1 whenever n > N. Adding |L| to both sides, we have |a n − L| + |L| < 1 + |L|. By the triangle inequality, |(a n −L)+L| = |a n | ≤ |a n −L|+|L|, and so |a n | ≤ |a n −L|+|L| < 1+|L| for all n > N. For n ≤ N , we have a finite set of numbers that therefore has a maximum value M and minimum value m; thus, {a n } is bounded.

32. Since a n+1 − a n = Partitioning [1, n] at 1, 2, 3, . . . , n, the upper sum is U = 1 + 1 2 + · · · + 1 n − 1 and the lower sum is L = 1 2 + 1 3 + · · · + 1 n . Since L < A < U, we have, for n ≥ 2, 1 2 + 1 3 + · · · + 1 n < ln n < 1 + 1 2 + · · · + 1 n − 1 . Now, a n = 1 + 1 2 + · · · + 1 n − ln n, so 1 2 + 1 3 + · · · + 1 n = a n − 1 + ln n and 1 + 1 2 + · · · + 1 n − 1 = a n − 1 n + ln n. Thus, for n ≥ 2, a n − 1 + ln n < ln n < a n − 1 n + ln n and a n − 1 < 0 < a n − 1 n , or a n < 1 and a n > 1 n > 0. Since a 1 = 1, the sequence is bounded below by 0 and above by 1. To see that the sequence is monotonic, note that a n − 1 + ln n < ln n implies a n+1 − 1 + ln(n + 1) < ln(n + 1). Subtracting, we have [a n+1 − 1 + ln(n + 1)] − (a n − 1 + ln n) < ln(n + 1) − ln n or a n+1 − a n < 0. Since the sequence is bounded and monotonic, it is convergent. 10. 5 + 0 − 7 + 0 − · · · 11. Write a k = 1

Series

13. Write a k = 1/2 2k − 1 − 1/2 2k + 1 . Then

14. Write a k = 1 (k + 3)(k + 4) = 1 k + 3 − 1 k + 4 . Then 17. Identify r = − 1 2 and a = 1. The series converges to 1 1 + 1/2 = 2 3 .

18.

∞ k=1 π k 1 3

. Identify a = π and r = π 3 > 1. The series diverges. 42. Let f (x) = x sin 1 x . Then, using L'Hôpital's Rule,

so the series diverges.

43. This is a geometric series with r = x 2 and will converge for x 2 < 1 or |x| < 2.

44. This is a geometric series with r = 1 x and will converge for 1 x < 1 or |x| > 1.

45. This is a geometric series with r = x + 1 and will converge for |x + 1| < 1 or −2 < x < 0.

46. This is a geometric series with r = 2x 2 and will converge for |2x 2 | < 1 or |x| <

, we obtain N 0 = 1000.

A 0 (e −k ) n . This is a geometric series with a = A 0 and

51. The total amount of the drug immediately after the n-th dose is A n = 15+15(0.2)+15(0.2) 2 + · · · + 15(0.2) n−1 . As n → ∞, the total accumulation of the drug will be lim n→∞ A n = lim , · · · , a n = 1 1 + 1 10 + · · · + 1 10 n−1 + 1 10 n = 1 n k=0 1 10 k

As n → ∞, a n → 1 56. For S n = 1 1 · 1 + 1 2 · 2 + 1 3 · 3 + · · · + 1 n · n , the inequality 0 < S n < 1 + 1 1 · 2 + 1 2 · 3 + · · · + 1 (n − 1) · n is true because, for all n ≥ 1, n > n − 1, n · n > (n − 1) · n, and therefore 1 n · n < 1 (n − 1) · n . The second inequality is true because each individual term

Simplifying this, we get 1 + 1 1 · 2 + 1 2 · 3 + · · · + 1 (n − 1) · n = 1 + 1

4.3.3

586

59.

61. The general term of the series is a n = 62. Since S n = 1 + 1 x dx = ln(n + 1) with their corresponding partial sums. This shows that ln(n + 1) < S n = 1 + 1 2 + 1 3 + 1 4 + · · · + 1 n : This yields the overall inequality ln(n + 1) < 1 + 1 2 + 1 3 + 1 4 + · · · + 1 n < 1 + ln n.

(b) Using the inequality in part (a), we know that S n ≥ 10 when ln(n + 1) ≥ 10. Solving ln(n+1) = 10, we get n+1 = e 10 , and n = e 10 −1. Thus, S n ≥ 10 for n > e 10 −1 ≈ 22026.

Using a calculator yields

This is a geometric series with first term d and common ratio (1 − p). Thus

Since 0 < p < 1, lim 66. Suppose the tortoise starts 10 feet (120 inches) in front of Achilles, the tortoise travels at 1 inch per second, and Achilles runs at 5 feet (60 inches) per second. Assuming constant speeds, conventional reasoning states that Achilles and the tortoise will have reached the same point at some time t such that (1)t + 120 = 60t, 59t = 120, and t = 120/59 ≈ 2.1864 seconds. Achilles then passes the tortoise after approximately 2.1864 seconds.

As phrased by Zeno, it will take 2 seconds for Achilles to reach the tortoise's starting point of 120 inches. By this time, the tortoise will be 2 inches in front of Achilles. It will then take Achilles 1/30 second to reach that point, by which time the tortoise will be 1/30 inches in front of Achilles. The distance travelled by Achilles can thus be written as a geometric series: 120 + 2 + 1/30 + · · · = 120 + 120(60 −1 ) + 120(60 −2 ) + · · · + 120(60 1−n ) + · · ·

In this example, we have a = 120 and r = 1/60; in general, a is the tortoise's head start, and r is the reciprocal of Achilles's speed. |r| < 1, so the series converges. At a constant speed, Achilles will reach this finite distance in a finite amount of time.

The issue with Zeno's statement of the problem is that Achilles reaches the tortoise's previous position in less and less time -infinitely less, in fact. This infinite division of time constitutes the "trick," so to speak, behind the paradox. Time in the story never exceeds the time that it takes to reach the sum of the infinite series, thus resulting in the tortoise's apparent victory.

67. This exercise involves a research report, and thus a preset solution is not applicable. It should be noted, however, that the series of the reciprocal of primes (i.e., the harmonic series of primes) does diverge, with multiple proofs available in the literature.

590

Integral Test

1. The function f (x) = 1 x 1.1 is continuous and decreasing on [1, ∞). Since 3. Rewriting the series, we have

PROBLEMAS 4.4

INTEGRAL TEST

597 24. Since the terms ln(1 + 3k) are increasing and positive, we see that the terms do not converge to zero. Therefore, the series diverges.

25. The function f (x) = 1 x(x + 1)

is continuous and decreasing on [1, ∞). Since

the integral converges and ∞ k=1 1 k(k + 1) converges.

26. The function f (x) = 2x + 1

x(x + 1) is continuous and decreasing on [1, ∞). Since

the integral diverges and ∞ k=1 2k + 1 k(k + 1) diverges .

27. The function f (x) = 1 (x + 1)(x + 7)

is continuous and decreasing on [1, ∞). Since converges.

28. The function f (x) = 1 x(x 2 + 1)

is continuous and decreasing on [1, ∞). Since

the integral converges and ∞ k=1 1 k(k 2 + 1) converges.

593

The function f (x) = 1 x 3/2 is continuous and decreasing on [1, ∞). Since

the integral converges and the series 1

4. Rewriting the series, we have 1 100

the integral diverges and the series 1 100

5. The function f (x) = 1 2x + 7 is continuous and decreasing on [1, ∞). Since 6. Since lim n→∞ n 3n + 1 = 1 3 = 0, the series diverges by the n-th term test.

7. The function f (x) = 1 1 + 5x 2 is continuous and decreasing on [1, ∞). Since

the integral converges and 15. The function f (x) = 10 x(ln x) 2 is continuous and decreasing on [2, ∞). Since

17. The function f (x) = arctan x 1 + x 2 is continuous and decreasing on [1, ∞). Since 18. The function f (x) = x 1 + x 4 is continues and decreasing on [1, ∞). Since

The integral converges and ∞ k=1 k 1 + k 4 converges.

596

Since lim

sin t t = 1, the series diverges by the n-th term test.

598

599

For p > 1, this limit converges to − 1 (ln 2) p−1 . For p = 1,

ln(ln 2)] = ∞. For p < 1, the integral also diverges.

For

By the same reasoning as in Exercise 35, the series converges for p > 1 and diverges for p ≤ 1.

37. For p ≥ 0, f (x) = x p ln x is not decreasing. So for p < 0, integration by parts gives

The integral converges for p < −1 and diverges for −1 ≤ p < 0 and for p ≥ 0 (that is, the integral diverges for p ≥ −1).

Since f is decreasing and f

Also, note that f (x) ≥ f (k+1) = a k+1 on . [k, k+1]. This implies = ln(n + 1).

Therefore, e 100 ≤ n + 1 or n ≥ e 100 − 1.

41. Since f is decreasing and

Since this is true for every integer p, we can let p → ∞ to get

Since this is true for every integer p, we can let p → ∞ to get

42. Letting f (x) = 1 x 2 and using the result from Problem 41, we need to find n such that 16. Using the limit comparison test with a n = 1 10 + √ n and b n = 1 √ n , we have lim n→∞ a n b n = lim

COMPARISON TESTS

605

Since ∞ n=1 b n is a geometric series with r = 1/2 < 1, it converges and the series ∞ k=1 1 2 + 1 2k k converges.

28. Using the limit comparison test with a n = n (n + 1)(n + 2) and b n = 1 n , we have lim n→∞ a n b n = lim n→∞ n/(n + 1)(n + 2) 1/n = lim n→∞ n 2 n 2 + 3n + 2 = 1.

Since ∞ n=1 1 n diverges, the series ∞ n=1 n (n + 1)(n + 2) diverges.

29. Since lim k→∞ k 100 √ k 2 + 1 = 1 100 , the series diverges by the n-th term test.

30. Using the limit comparison test with a n = 1 n + √ n and b n = 1 n , we have lim n→∞ a n b n = lim

31. Since lim k→∞ ln 5 + k 5 = ∞, the series diverges by the n-th term test.

32. Using the limit comparison test with a n = ln 1 + 1 3 n and b n = 1 3 n , we have (using L'Hôpital's Rule) Since ∞ n=1 b n is a geometric series with r = 1/3 < 1, it converges and the series ∞ k=1 ln 1 + 1 3 k converges.

33. The function f (x) =

x (x 2 + 1) 2 is continuous and decreasing on [1, ∞). Since

the integral converges and ∞ k=1 k (k 2 + 1) 2 converges.

(The direct comparison test and limit comparison test can also be used.)

606

Ratio and Root Tests

12. Since lim n→∞ a n+1 a n = lim n→∞ (n + 1)! n! · e n 2 e (n+1) 2 = lim n→∞ (n + 1) · e n 2 e n 2 ·e 2n · e = lim n→∞ n + 1 e 2n+1 = 0 < 1, the series converges by the ratio test. 14. Since

the series diverges by the ratio test.

15. Since lim k→∞ a k+1 k n = lim k→∞ 1 · 3 · 5 · · · (2k + 1)/(k + 1)! 1 · 3 · 5 · · · (2k − 1)/k! = lim

Thus, the integral converges and ∞ k=2 ln k k p converges for p > 1.

RATIO AND ROOT TESTS

Dividing numerator and denominator by (1 + √ 5) n we have 3. Since lim k→∞ k k + 1 = 1 = 0, the series diverges.

4. Let f (x) = x x 2 + 1 . Then f (x) = 1 − x 2 (x 2 + 1) 2 < 0 for x > 1 and a k+1 < a k . Since lim k→∞ k k 2 + 1 = lim k→∞ 1 k + 1/k = 0, the series converges.

x 2 + 6 x 4 < 0 for x ≥ 1 and a k+1 < a k . Since lim k→∞ k 2 + 2 k 3 = lim k→∞ 1/k + 2/k 3 1 = 0, the series converges.

6. Since lim k→∞ 3k − 1 k + 5 = 3 = 0, the series diverges.

7. Since a k+1 = 1 k + 1 + 1 3 k+1 < 10. Let f (x) = x 1/3 x + 1 . Then f (x) = 1 − 2x 3x 2/3 (x + 1) 2 < 0 for x ≥ 1 and a n+1 < a n . Since lim n→∞ 3 √ n n + 1 = lim n→∞ 1 n 2/3 + n −1/3 = 0, the series converges.

Note that cos

and a n+1 < a n . Since lim n→∞ √ n + 1 n + 2 = lim n→∞ 1 + 1/n √ n + 2/ √ n = 0, the series converges.

12 14. Since a k+1 = 1 ln(k + 1) < 1 ln k = a k and lim k→∞ 1 ln k = 0, the series converges.

15. Apply the limit comparison test to ∞ k=1 1 2k + 1 with a k = 1 2k + 1 and b k = 1 k :

Since ∞ k=1 b k diverges, the given series is not absolutely convergent. Since a k+1 = 1 2k + 3 < 1 2k + 1 = a k and lim k→∞ 1 2k + 1 = 0, the series is conditionally convergent.

Apply the limit comparison test to

b k diverges, the given series is not absolutely 27. Since cos kπ = (−1) k and lim k→∞ (−1) k is not 0, the series diverges by the n-th term test.

28. Since sin 2k + 1 2 π = sin kπ + π 2 = (−1) k , the series is

The harmonic series

The series is absolutely convergent for 2|x| < 1 or |x| < 1/2. At x = −1/2, the series ∞ k=1 (−1) k k converges by the alternating series test. At x = 1/2, the series ∞ k=1 1 k is the harmonic series which diverges. Thus, the given series converges on [−1/2, 1/2).

31.

∞ k=1 1 x k is a geometric series with common ratio r = 1/x. It converges for |1/x| < 1 or

x > 1. 33. This is a geometric series with common ratio r = x + 1 x and will converge for x + 1 x < 1 or |x + 1| < |x|. If x ≥ 0, this means that x + 1 < x which has no solution. For x < 0, |x| = −x and the inequality can be written as |x + 1| < −x or x < x + 1 < −x. Since x < x + 1 is valid for all x, we have x + 1 < −x, 2x < −1, or x < −1/2. Thus, the given series converges on (−∞, −1/2).

34. This is a geometric series with common ratio r = x 2x + 4 and will converge for x 2x + 4 < 1 or |x| < |2x + 4|. If x ≥ 0, this means x < 2x + 4 which is true for all x ≥ 0. Thus, the series converges on [0, ∞). If x < 0, |x| = −x and the inequality can be written as −x < |2x + 4|. This is equivalent to 2x + 4 < −x(−x) or 2x + 4 > −x. Solving for x, we obtain x < −4 or x > −4/3. Therefore, the given series converges on (−∞, −4) (−4/3, ∞).

35. Applying the root test, we have lim n→∞ |a n | 1/n = lim n→∞ x 2 + 2 6 n 2 1/n = lim n→∞ x 2 + 2 6 n < 1.

x 2 + x = (x + 2) − 2 x + 2 = 1 − 2 x + 2 = 1 − 2 1 + (x + 1)

which converges for |x + 1| < 1. Since the series diverges at x = −2 and x = 0, the interval of convergence is (−2, 0).

Setting

x 2 + 2 6 n < 1, we obtain x 2 + 2 6 < 1 or x 2 < 4. Thus, the series converges on (−2, 2). At x = ±2, the series ∞ k=0 1 k 2 diverges by the n-th term test. Therefore, the given series converges on (−2, 2).

36. Applying the ratio test, we have lim n→∞ a n+1 a n = lim n→∞ (n + 1)!/[(n + 1)x] n+1 n!/(nx) n = lim n→∞ (n + 1)n n (n + 1) n+1 x = lim n→∞ n n + 1 n 1 x = lim n→∞ 1 (1 + 1/n) n · 1 |x| = 1 e|x| .

Setting 1/e|x| < 1, we obtain |x| > 1/e. Thus, the series converges on (∞, −1/e) (1/e, ∞).

At x = −1/e and 1/e we obtain the series

Letting a k = k!e k k k , we will show that a k+1 ≥ a k . First, compute a k+1 a k = (k + 1)!e k+1 /(k + 1) k+1 k!e k /k k = (k + 1)k k e (k + 1) k+1 = e k + 1 k k .

628

38.

Applying the ratio test, we have lim n→∞ a n+1 a n = lim n→∞ (n + 1)!e −(n+1)x 2 n!e −nx 2 = lim n→∞ (n+1)e −x 2 = ∞. Thus, the series diverges for all x.

39. This is a geometric series with r = 2 √ 3 sin x and will converge for 2 √ 3 sin x < 1 or | sin x| < √ 3/2. On [0, 2π] this will be on [0, π/3) (2π/3, 4π/3) (5π/3, 2π].

40. Since | sin kx k 2 | ≤ 1 k 2 , we see that

Representing Functions by Power Series

In this section we will use the fact that the geometric series ∞ k=0 ar k converges to a 1 − r for r < 1. (−3, 3). At x = −3 and x = 3, the series diverges by the n-th term test. Thus, the interval of convergence is (−3, 3). 1 − x 2 (1 + x 2 ) 2 = d dx

The integral of convergence remains (−1, 1) .

15. Using Problem 5,

2k + 1 for x in (−1, 1). At x = −1 and x = 1, the series converges by the alternating series test. Thus, the interval of convergence is [−1, 1].

16. tan −1 x 2 = 2 1 4 + x 2 dx = 2

Thus ln(4 + x) = ln 4 + ∞ k=0 (−1) k (k + 1)4 k+1 x k+1 . This series converges for |x/4| < 1, |x| < 4, or on (−4, 4). At x = −4, the series diverges since it is the negative harmonic series. At x = 4, the series converges by the alternating series test. Thus, the interval of convergence is (−4, 4].

ln

3 + x 3 − x = ln(3 + x) − ln(3 − x)

(1 + (−1) k ) x 3 k+1 k + 1 which converges for x in (−3, 3). At x = 3, we have the series (1 + (−1) k ) k + 1 which diverges.

REPRESENTING FUNCTIONS BY POWER SERIES

633 At x = −3, we have the series ∞ k=0

(1 + (−1) k )(−1) k+1 k + 1 which also diverges. Thus, the interval of convergence is (−3, 3).

21

.

which converges for x in − 1 2 , 1 2 and diverges at x = − 1 2 and x = 1 2 . Thus, the interval of

2x k which converges for x in (−1, 1) and diverges at x = −1 and x = 1. Thus, the interval of convergence is (−1, 1) .

23.

x 2 (1 − x) 3 = x 2 · 1 (1 + x) 3 = x 2 · 1 2

k(k − 1)(−1) k x k which converges for x in (−1, 1) and diverges at x = −1 and x = 1. Thus, the interval of convergence is (−1, 1) .

x k+3 4 k which converges for x in (−4, 4) and diverges at x = −4 and x = 4. Thus, the interval of convergence is (−4, 4) .

634

s ln(1 +

which converges for x in (−1, 1). At x = −1, we have the series 26.

which converges for x in (−1, 1). At x = −1, we have the series (−1) k t 2k+2 (2k + 1)(2k + 2)

x 0 = ∞ k=0 (−1) k x 2k+2 (2k + 1)(2k + 2) which converges for x in (−1, 1). At x = −1, we have the series ∞ k=0 (−1) 3k+2 (2k + 1)(2k + 2) which converges. At x = 1, we have the series ∞ k=0 (−1) k (2k + 1)(2k + 2) which converges. Thus, the interval of convergence is [−1, 1].

28. Since ln(1 + t 2 ) = ∞ k=0 (−1) k t 2k+2 k + 1 , we have

which converges for x in (−1, 1). At x = −1, we have the series (−1) k x − 6 5 k which converges for x − 6 5 < 1 or |x − 6| < 5. Since the series diverges at x = 11 and x = 1, the interval of convergence is (1, 11).

30

.

x + 2 2 k which converges for x + 2 2 < 1 or |x + 2| < 2. Since the series diverges at x = −4 and x = 0, the interval of convergence is (−4, 0).

32.

x − 2

which converges for |x − 2| < 1. Since the series diverges at x = 1 and x = 3, the interval of convergence is (1, 3).

33

. 7x

3 k x k which converges for (−3, 3). Since the series diverges at x = −3 and x = 3, the interval of convergence is (−3, 3).

34.

3

−1 2 k+1 + (−1) k x k which converges for (−1, 1). Since the series diverges at x = −1 and x = 1, the interval of convergence is (−1, 1).

35.

1

x k = 1 2 1 + x 2 + x 2 4 + x 3 8 + · · · · [1 + x + x 2 + x 3 + · · · ] = 1 2 + 3 4

x + 7 8

x 2 + 15 16

x 3 + · · · 36.

x (1 + 2x)(1 + x 2 ) = 1 1 + 2x

37. Writing f (x) = ∞ k=1 (−1) k+1 x k k3 k and applying the ratio test, we have lim n→∞ a n+1 a n = lim n→∞ x n+1 /(n + 1)3 n+1 x n /n3 n = lim n→∞ 1 3 n n + 1 |x| = 1 3 |x|.

The series is absolutely convergent for 1 3 |x| < 1, |x| < 3, or on (−3, 3). At x = −3, the series is divergent since it is the negative harmonic series. At x = 3, the series converges by the alternating series test. Thus, the domain of the function is (−3, 3].

38. Writing f (x) = ∞ k=0 2 k k! x k and applying the ratio test, we have lim n→∞ a n+1 a n = lim n→∞ 2 n+1 x n+1 /(n + 1)! 2 n x n /n! = lim n→∞ 2 n + 1 |x| = 0.

Thus, the series is absolutely convergent for all x and the domain of the function is (−∞, ∞).

Using

4k + 2 .

Letting x = 1/3, we have (1/2) 4k+3 4k + 3 = ∞ 0 (−1) k 1 2 4k+3 (2k + 1)(4k + 3) = 1 2 3 · 1 · 3 − 1 2 7 · 3 · 7 + 1 2 1 1 · 5 · 11 − · ≈ 0.04167 − 0.00037 + 0.00001 − · ≈ 0.0413.

Using tan

46. Using Theorem 9.7.2 in the text, we must have a n+1 = 1 2(n + 1) + 1 = 1 2n + 3 < 0.00005 or 2n + 3 > 20, 000. This means n > 9998.5. Thus, it will require S 9999 to approximate π/4 to four decimal places.

x 2 2! + x 3 3! + x 4 4! + · · · · 1 + x + x 2 + x 3 + x 4 + · · · = 1 + 2x + 5 2

x 2 + 8 3

x 3 + 65 24

x 4 + · · · 38. e x sin x = 1 + x +

39. e x cos x = 1 + x + x 2 2 + x 3 3! + · · · 1 − x 2 2 + x 4 4! − · · · = 1 + x + x 2 + 2x 3 3 + x 4 2 + · · · 40. sec x = 1 cos x = 1 1 − x 2 2 + x 4 4! − x 5 5! + · · · = 1 + x 2 2 + 5x 4 24 + 61x 6 720 + 277x 8 8064 + · · · 652 CHAPTER 9. SEQUENCES AND SERIES 41.

44. Using e x = 1 + x + x 2 2! + x 3 3! + · · · , we have e −1 = 1 − 1 + 1 2! − 1 3! + 1 4! − 1 5! + · · · = 1 2! − 1 3! + 1 4! − 1 5! + · · · 45. Using cos x = 1 − x 2 2! + x 4 4! − x 6 6! + · · · , we have cos π = 1 − π 2 2! + π 4 4! − π 6 6! + · · · 46. Using sin x = x − x 3 3! + x 5 5! − x 7 7! + · · · , we have sin π = π − π 3 3! + π 5 5! − π 7 7! Therefore, sec x = 1 + x 2 2 + · · ·.

Approximating sec x by 1 +

− x)x 2 dx = 2π x 3