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Unformatted text preview: 622 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS PROJECT 10.6 Peculiar Indeterminate Forms
Sometimes the application of L’Hopital’s rule to evaluate an indeˆ terminate form leads to peculiar situations. In this project, we investigate two such cases. First, we let 2x f (x ) = √ 1 + x2 and consider lim f (x). This limit is an indeterminate of the form ∞/∞. Problem 1. a. Try to evaluate lim √
x→∞ x→∞ Problem 2. a. Try to evaluate lim what happens. b. Write g in the equivalent form g (x ) = e−1/x using L’Hopital’s rule and describe ˆ x →0 x 2 1/x e1/x2
x →0 and use L’Hopital’s rule to evalute lim g (x). ˆ 2x 1 + x2 c. Use a graphing utility to graph g and verify the limit you found in part (b). Problem 3. Deﬁne the function h by h(x) = e−1/x x = 0 0 x = 0.
2 using L Hopital’s rule and ’ˆ describe what happens. b. Use a graphing utility to graph f and determine lim f (x).
x→∞ c. Actually this limit is easy to calculate by algebraic methods. Use algebra to verify the limit that you found in part (b). Now consider the function g (x ) = e
−1/x2 a. Use the deﬁnition of the derivative to show that h is differentiable at x = 0. What is h (0)? b. Show that lim e−1/x = 0 for every positive integer n. x →0 xn
2 x and consider lim g (x). This is an indeterminate of the form 0/0.
x →0 c. Calculate h (x), h (x), · · · , for x = 0. d. Use part (b) and the results in part (c) to conclude that h(n) (0) = 0 for every positive integer n. 10.7 IMPROPER INTEGRALS
In the deﬁnition of the deﬁnite integral
b f (x) dx
a it is assumed that [a, b] is a bounded interval and f is a bounded function. In this section we use the limit process to investigate integrals in which either the interval is unbounded or the integrand is an unbounded function. Such integrals are called improper integrals. Integrals over Unbounded Intervals
y We begin with a function f continuous on an unbounded interval [a, ∞). For each number b > a we can form the deﬁnite integral
b f
a f (x) dx.
b x (Figure 10.7.1) a If, as b tends to ∞, this integral tends to a ﬁnite limit L,
b b→∞ a Figure 10.7.1 lim f (x) dx = L, then we write
∞ a f (x) dx = L 10.7 IMPROPER INTEGRALS 623 and say that
∞ the improper integral
a f (x) dx converges to L. Otherwise, we say that
∞ the improper integral
a f (x) dx diverges. In a similar manner,
b b improper integrals
−∞ f (x) dx arise as limits of the form a→−∞ a lim f (x) dx. Example 1
∞ (a)
0 ∞ e−2x dx = 1 . 2 dx = 1. x2 ∞ (b)
1 1 dx x diverges. diverges. (c)
1 (d)
−∞ cos π x dx VERIFICATION
∞ (a)
0 e−2x dx = lim dx = lim b→∞ x dx = lim b→∞ x2 b b→∞ 0 b 1 b 1 e−2x dx = lim −
b→∞ e−2x 2 b = lim
0 b→∞ 1 1 − 2b 2 2e = 1 . 2 ∞ (b)
1 ∞ dx = lim ln b = ∞. b→∞ x dx 1 = lim − 2 b→∞ x x
b (c)
1 = lim
1 b→∞ 1− 1 b = 1. (d) Note ﬁrst that
1 a cos π x dx = 1 sin π x π 1 =−
a 1 sin π a. π As a tends to −∞, sin π a oscillates between −1 and 1. Therefore the integral oscillates between 1/π and −1/π and does not converge. The usual formulas for area and volume are extended to the unbounded case by means of improper integrals.
Example 2 graph of y Let p be a positive number. If f (x ) = 1 , xp is the region between the xaxis and the
f (x) = x ≥ 1, xp (Figure 10.7.2)
1 Ω x then area of 1 , = p−1 ∞, if if p>1 p ≤ 1. Figure 10.7.2 624 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS This comes about from setting area of For p = 1,
∞ 1 = lim b b→∞ 1 dx = xp ∞ 1 dx . xp dx = lim b→∞ xp b 1 1 dx 1 , (b1−p − 1) = p − 1 = lim p b→∞ 1 − p x ∞, if if p>1 p < 1. For p = 1,
∞ 1 dx = xp ∞ 1 dx = ∞, x as you have seen already. For future reference we record the following: Let p > 0,
∞ (10.7.1)
1 dx converges for p > 1 and diverges for p ≤ 1. xp y Example 3 We know that the region below the graph of f (x) = 1/x, x ≥ 1, has inﬁnite area. Suppose that this region with inﬁnite area is revolved about the xaxis (see Figure 10.7.3). What is the volume V of the resulting solid? It may surprise you somewhat, but the volume is not inﬁnite. In fact, it is π . Using the disc method to calculate the volume (Section 6.2), we have f (x) = 1 x V=
1 ∞ π [ f (x)]2 dx = π
1 ∞ dx = π lim b→∞ x2 = π lim b 1 dx x2
b x b→∞ −1 x = π · 1 = π.
1 Figure 10.7.3 It may surprise you further to learn that the surface area of this solid is inﬁnite. See Exercise 41. This surface is known as Gabriel’s horn. It is often difﬁcult to determine the convergence or divergence of a given improper integral by direct methods, but we can usually gain some information by comparison with integrals of known behavior. y g f (A comparison test ) Suppose that f and g are continuous and 0 ≤ f (x) ≤ g (x) for allx ∈ [a, ∞). (Figure 10.7.4)
x a Figure 10.7.4 (10.7.2) ∞ ∞ (i) If
a g (x) dx
∞ converges, diverges , then
a ∞ f (x) dxconverges. g (x) dx diverges.
a (ii) If
a f (x) dx then 10.7 IMPROPER INTEGRALS 625 A similar result holds for integrals form −∞ to b. The proof of the result is left to you as an exercise.
∞ Example 4 The improper integral
1 √ dx 1 + x3 and converges since
∞ 1 √ 1 1 + x3 < 1 x 3 /2 for x ∈ [1, ∞) dx converges. x 3 /2 In contrast, if we tried to evaluate
b b→∞ 1 lim dx √ 1 + x3 directly, we would ﬁrst have to calculate the integral dx , √ 1 + x3 and this cannot be done by any of the methods we have developed so far.
∞ Example 5 The improper integral
1 dx diverges since √ 1 + x2
∞ 1 1 ≤√ 1+x 1 + x2 for x ∈ [1, ∞) and
1 dx diverges. 1+x This result can be obtained by evaluating
b 1 √ dx 1 + x2 and then calculating the limit as b → ∞. Try it. Suppose now that f is continuous on (−∞, ∞). The improper integral
∞ f (x) dx
−∞ is said to converge if
0 ∞ f (x) dx
−∞ and
0 f (x) dx both converge. We then set
∞ −∞ f (x ) = L + M ,
∞ where 0 −∞ f (x) dx = L and
0 f (x) dx = M . 626 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS
y Example 6 Let r > 0. Determine whether the improper integral
∞ −∞ r2 r dx + x2 (Figure 10.7.5) x converges or diverges. If it converges, gives its value.
SOLUTION Figure 10.7.5 According to the deﬁnition, we need to determine the convergence or divergence of each of the improper integrals
0 −∞ r dx 2 + x2 r ∞ and
0 r2 r dx. + x2 For the ﬁrst integral:
0 −∞ r dx = lim a→−∞ r 2 + x2 r x dx = lim tan−1 a→−∞ r 2 + x2 r a a π π = − lim tan−1 =− − =. a→−∞ r 2 2 0 0 a For the second integral:
∞ 0 r dx = lim b→∞ r 2 + x2
b→∞ b 0 r x dx = lim tan−1 b→∞ r 2 + x2 r b r = π . 2 b 0 = lim tan−1 Since both of these integrals converge, the improper integral
∞ −∞ r2 r dx + x2 converges. Its value is 1 π + 1 π = π . 2 2 Remark Note that we did not deﬁne
∞ (1)
−∞ f (x) dx as
y
b (2) b→∞ −b lim f (x) dx. f It is easy to show that if (1) exists, then (2) exists and (1) = (2). However, the existence of (2) does not imply the existence of (1). For example, (2) exists and is 0 for every odd function f , but this is certainly not the case for (1). You are asked to verify these statements in Exercise 57. x=b a c x Integrals of Unbounded Functions
Improper integrals can also arise on bounded intervals. Suppose that f is continuous on the halfopen interval [a, b) but is unbounded there. See Figure 10.7.6. For each number c < b, we can form the deﬁnite integral
c Figure 10.7.6 f (x) dx.
a 10.7 IMPROPER INTEGRALS 627 If, as c → b− , the integral tends to a ﬁnite limit L,
c c→b− lim f (x) dx = L, a then we write
b a f (x) dx = L and say that
b the improper integral
a f (x) dx converges to L. Otherwise, we say that the improper integral diverges. Similarly, functions continuous but unbounded on halfopen intervals of the form (a, b] lead to limits of the form
b c→a+ lim f (x) dx.
c As above,
b the improper integral
a f (x) dx converges to L b if c→a+ lim f (x) dx = L. c Otherwise, the improper integral diverges.
Example 7
1 (a)
0 (1 − x)−2/3 dx = 3. 2 (b)
0 dx diverges . x VERIFICATION
1 (a )
0 (1 − x)−2/3 dx = lim c 0 c→1− (1 − x)−2/3 dx
c 0 = lim
2 c→1− − 3(1 − x)1/3
2 c = lim [−3(1 − c)1/3 + 3] = 3.
c→1− (b)
0 dx = lim x c→0+ 2 c dx = lim ln x x c→0+ = lim [ln 2 − ln c] = ∞.
c→0+ Now suppose that f is continuous on an interval [a, b] except at some point c in (a, b) where f (x) → ±∞ as x → c− or as x → c+ . We say that the improper integral
b f (x) dx
a 628 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS converges iff both of the integrals
c b f (x) dx
a and
c f (x) dx converge. If
c a f (x) dx = L b and
c f (x) dx = M , we then set
b a f (x) dx = L + M . Example 8 To evaluate
y (∗)
f (x) =
1
(x – 2)2 4 dx , (x − 2)2 we need to calculate
c x=2 1 2 3 4 x c→2− lim 1 dx (x − 2)2 4 and c→2+ lim c dx (x − 2)2 (the integrand blows up at x = 2; see Figure 10.7.7). For c < 2,
c 1 Figure 10.7.7 −1 dx = (x − 2)2 (x − 2) c =−
1 1 −1→∞ c−2
c as c → 2− . Since
1 dx (x − 2)2 does not converge, improper integral (∗) does not converge. (As you can verify, the second limit does not exist either.) Notice that if we ignore the fact that integral (∗) is improper, then we are led to the incorrect conclusion that
4 1 −1 dx = 2 (x − 2) (x − 2)
1 4 1 3 =− . 2 Example 9 Evaluate
−2 dx . x 4 /5 SOLUTION Since 1/x4/5 → ∞ as x → 0− and as x → 0+ , the given integral is improper. We need to calculate
0 −2 dx x 4 /5 1 and
0 dx . x 4 /5 Now dx = lim x4/5 c→0− −2
0 c −2 dx = lim 5x1/5 x4/5 c→0− c −2 = lim [5c1/5 − 5(−2)1/5 ] = 5(21/5 )
c→0− 10.7 IMPROPER INTEGRALS
1 629 and
0 dx = lim x4/5 c→0+ 1 c dx = lim 5x1/5 x4/5 c→0+ 1 c = lim [5 − 5c1/5 ] = 5.
c→0+ The improper integral converges and
1 −2 dx = 5 + 5(21/5 ) ∼ 10. 74. = x 4 /5 EXERCISES 10.7
Evaluate the improper integrals that converge.
∞ 1.
1 ∞ dx . x2 dx . 4 + x2 e dx,
px ∞ 2.
0 ∞ dx . 1 + x2 e
−px c In Exercises 35–36, use a graphing utility to draw the graph of the integrand in each of the improper integrals. Then use a CAS to determine whether the integral converges or diverges.
∞ 3.
0 ∞ 4.
0 1 dx, p > 0. 35. (a)
0 ∞ x dx. (16 + x2 )2 x dx. 16 + x4 x3 2−x x 2−x dx. dx. ∞ (b)
0 ∞ x2 dx. (16 + x2 )2 x dx. 16 + x2 1 2−x 1 2x − x 2 dx. dx. 5.
0 8 p > 0. 6.
0 1 dx √. x dx . x2 √ dx 1−x dx a2 − x 2 . . (c)
0 2 (d)
0 2 7.
0 1 dx . x2/3 √ dx 1 − x2 x 4 − x2 . dx. 36. (a)
0 2 8.
0 1 √ 3 √ (b)
0 2 √ (c)
0 (d)
0 √ 9.
0 2 10.
0 a 37. Evaluate
1 0 11.
0 √
∞ 12.
0 √
∞ sin−1 x dx 13.
e 1 ln x dx. x 14.
e ∞ dx . x ln x dx . x(ln x)2 dx . x2 − 1 15.
0 x ln x dx.
∞ 16.
e ∞ by using integration by parts even though the technique leads to an improper integral. 38. (a) For what values of r is
∞ 0 17.
−∞ ∞ dx . 1 + x2 dx . x2 dx . x(x + 1) x x2 − 9 dx. 18.
2 3 xr e−x dx 19.
−∞ ∞ 20. 22. dx . √ 3 3x − 1 1/ 3
0 convergent? (b) Use mathematical induction to show that
∞ 0 21.
1 5 x ex dx.
−∞ 4 xn e−x dx = n!, n = 1, 2, 3, . . . . 23.
3 3 √ 24.
1 dx . 2−4 x x dx. (1 + x2 )2 1 dx. ex + e−x 39. The integral
∞ 0 √ 25.
−3 1 dx . x(x + 1) x2 dx . −4 ∞ 1 dx x (1 + x) 26.
1 ∞ 27.
−3 ∞ 28.
−∞ 4 is improper in two distinct ways: the interval of integration is unbounded and the integrand is unbounded. If we rewrite the integral as
1 29.
0 ∞ cosh x dx. e
0 1 −x 30.
1 dx . 2 − 5x + 6 x cos x dx.
2 √
0 1 dx + x (1 + x) ∞ 1 √ 1 dx, x(1 + x) ∞ 31. 33.
0 sin x dx. 32.
0 ex √ dx. x √ π /2 34.
0 cos x dx. √ sin x then we have two improper integrals, the ﬁrst having an unbounded integrand and the second deﬁned on an unbounded interval. If each of these integrals converges with values L1 and L2 , respectively, then the original integral converges and has the value L1 + L2 . Evaluate the original integral. 630 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS 40. Evaluate
∞ 1 57. (a) Show that the improper integral 1 dx √ x x2 − 1
∞ 0 2x dx 1 + x2 using the method given in Exercise 39. 41. Show that if the region below the graph of f (x) = 1/x, x ≥ 1, is revolved about the xaxis, then the surface area of the resulting solid is inﬁnite (see Example 3). 42. Sketch the graphs of y = sec x and y = tan x for 0 ≤ x < π/2. Calculate the area of the region between the two curves. 43. Let be the region bounded by the coordinate axes, the √ graph of y = 1/ x, and the line x = 1. (a) Sketch . (b) Show that has ﬁnite area and ﬁnd it. (c) Show that if is revolved about the xaxis, the solid obtained does not have ﬁnite volume. 44. Let be the region between the graph of y = 1/(1 + x2 ) and the xaxis, x ≥ 0. (a) Sketch . (b) Find the area of . (c) Find the volume of the solid obtained by revolving about the xaxis. (d) Find the volume of the solid obtained by revolving about the yaxis. 45. Let be the region bounded by the curve y = e−x and the xaxis, x ≥ 0. (a) Sketch . (b) Find the area of . (c) Find the volume of the solid obtained by revolving about the xaxis. (d) Find the volume obtained by revolving about the yaxis. (e) Find the lateral surface area of the solid in part (c). 46. What point would you call the centroid of the region in Exercise 45? Does Pappus’s theorem work in this instance? 47. Let be the region bounded by the curve y = e−x and the xaxis, x ≥ 0. (a) Show that has ﬁnite area. (The area is √ actually 1 π , as you will see in Chapter 16.) (b) Calculate 2 the volume generated by revolving about the yaxis. 48. Let be the region bounded below by y(x2 + 1) = x, above by xy = 1, and to the left by x = 1. (a) Find the area of . (b) Show that the solid generated by revolving about the xaxis has ﬁnite volume. (c) Calculate the volume generated by revolving about the yaxis. 49. Let be the region bounded by the curve y = x−1/4 and the xaxis, 0 < x ≤ 1. (a) Sketch . (b) Find the area of . (c) Find the volume of the solid obtained by revolving about the xaxis. (d) Find the volume of the solid obtained by revolving about the yaxis. 50. Prove the validity of the comparison test (10.7.2). In Exercises 51–56, use the comparison test (10.7.2) to determine whether the integral converges.
∞
2 diverges. Thus, the improper integral 2x dx 1 + x2 −∞ diverges. (b) Show that lim 58. Show that
∞ b→∞ ∞ 2x dx = 0. 2 −b 1 + x and b (a)
−∞ sin x dx diverges
b (b) lim b→∞ −b sin x dx = 0. 59. Calculate the arc distance from the origin to the point (x(θ1 ), y(θ1 )) along the exponential spiral r = a ecθ . (Take a > 0, c > 0.) 60. The function 1 f (x ) = √ 2π
x −∞ e−t 2 /2 dt is important in statistics. Prove that the integral on the right converges for all real x. Exercises 59–62: Laplace transforms. Let f be continuous on [0, ∞). The Laplace transform of f is the function F deﬁned by F (s) =
0 ∞ e−sx f (x) dx. The domain of F is the set of all real numbers s such that the improper integral converges. Find the Laplace transform F of each of the following functions and give the domain of F . 61. f (x) = 1. 63. f (x) = cos 2x. 62. f (x) = x. 64. f (x) = eax . Exercises 63–66: Probability density functions. A nonnegative function f deﬁned on ( − ∞, ∞) is called a probability density function if
∞ −∞ f (x) dx = 1. 65. Show that the function f deﬁned by f (x ) = 6x/(1 + 3x2 )2 0 x≥0 x<0 51.
1 ∞ dx. √ 1 + x5 (1 + x ) ln x dx. x2
5 −1/6 x ∞ 52.
1 ∞ 2 −x2 dx. 53.
0 ∞ dx. 54.
π ∞ sin2 2x dx. x2 √ dx x + 1 ln x . is a probability density function. 66. Let k > 0. Show that the function f (x ) = ke−kx x ≥ 0 0 x < 0, 55.
1 56.
e CHAPTER HIGHLIGHTS 631 is a probability density function. It is called the exponential density function. 67. The mean of a probability density function f is deﬁned as the number µ=
∞ where µ is the mean. Calculate the standard deviation for the exponential density function. 69. (Useful later) Let f be a continuous, positive, decreasing function on [1, ∞). Show that
∞ x f (x) dx.
−∞ 1 f (x) dx converges iff the sequence
n Calculate the mean for the exponential density function. 68. The standard deviation of a probability density function f is deﬁned as the number σ=
∞ −∞ 1/ 2 an =
1 f (x) dx (x − µ) f (x) dx
2 converges. CHAPTER HIGHLIGHTS
10.1 The Least Upper Bound Axiom for each α > 0, lim upper bound, bounded above, least upper bound (p. 585) least upper bound axiom (p. 586) lower bound, bounded below, greatest lower bound (p. 587)
10.2 Sequences of Real Numbers 1 =0 nα n x for each real x, lim =0 n→∞ n! ln n lim =0 lim n1/n = 1 n→∞ n n→∞ xn for each real x, lim 1 − = ex n→∞ n Cauchy sequence (p. 610)
n→∞ sequence (p. 590) bounded above, bounded below, bounded (p. 591) increasing, nondecreasing, decreasing, nonincreasing (p. 591) recurrence relation (p. 594) It is sometimes possible to obtain useful information about a sequence yn = f (n) by applying the techniques of calculus to the function y = f (x).
10.3 Limit of a Sequence 10.5 The Indeterminate Form (0/0) L Hopital’s rule (0/0) (p. 611) ’ˆ Cauchy meanvalue theorem (p. 613)
10.6 The Indeterminate Form (∞/∞) ; other Indeterminate Forms limit of a sequence (p. 595) uniqueness of the limit (p. 597) convergent, divergent (p. 597) pinching theorem (p. 600) Every convergent sequence is bounded (p. 597); thus, every unbounded sequence is divergent. A bounded, monotonic sequence converges. (p. 598) Suppose that cn → c as n → ∞, and all the cn are in the domain of f . If f is continuous at c, then f (cn ) → f (c). (p. 601)
10.4 Some Important Limits L Hopital’s rule (∞/∞) (p. 616) ’ˆ ln x xk lim α = 0 lim x = 0 lim xx = 1 x→∞ x x→∞ e x →0 + other indeterminate forms: 0 · ∞, ∞ − ∞, 00 , 1∞ , ∞ ∞ (p. 617)
10.7 Improper Integrals integrals over inﬁnite intervals (p. 622) convergent, divergent (p. 623)
∞ 1 dx converges for p > 1 and diverges for p ≤ 1. xp for x > 0, n→∞ lim x1/n = 1; for x < 1 n→∞ lim xn = 0 a comparison test (p. 624) integrals of unbounded functions (p. 626) convergent, divergent (p. 627) ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston  Downtown.
 Spring '10
 SMITH
 Improper Integrals, Integrals

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