Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 94 pps

29.4 Improper Integrals 911 y x y = e –x 2 y = e –x 1 Figure 29.13 The first summand is proper. We’ll concern ourselves with the second integral and compare with the convergent integral  ∞ 1 e −x dx.  ∞ 1 e −x 2 dx = lim b→∞  b 1 e −x 2 dx We will show that this limit exists and is finite. We know e −x 2 > 0, so for b>1,  b 1 e −x 2 dx increases with b. Therefore, as b →∞,  b 1 e −x dx either grows without bound or is finite. 0 <  b 1 e −x 2 dx <  b 1 e −x dx because e −x 2 ≤ e −x on [1, b] 0 ≤ lim b→∞  b 1 e −x 2 dx ≤ lim b→∞  b 1 e −x dx lim b→∞  b 1 e −x dx = lim b→∞ −e −x     b 1 = lim b→∞ − 1 e b + 1 e = 1 e so 0 ≤  ∞ 1 e −x 2 dx ≤ 1 e . Therefore  ∞ 1 e −x 2 dx is convergent. We conclude that  ∞ 0 e −x 2 dx converges. ◆ y x y = e –x 2 1 Figure 29.14 REMARK  ∞ −∞ e −x 2 dx is an interesting integral. We’ve concluded that it is convergent; using more advanced methods it can be shown that its value is √ π . If we wanted to approximate  ∞ −∞ e −x 2 dx, we could proceed as follows. i.  ∞ −∞ e −x 2 dx =2  ∞ 0 e −x 2 dx ii. Cut the tail off of  ∞ 0 e −x 2 dx and bound it. For instance,  ∞ 6 e −x 2 dx <  ∞ 6 e −x dx < 0.0025. Even better, bound  ∞ c e −x 2 dx by  ∞ c xe −x 2 dx. For instance, 912 CHAPTER 29 Computing Integrals  ∞ 5 e −x 2 dx <  ∞ 5 xe −x 2 dx = 1 2e 25 < 7 × 10 −12 The interested reader has many details to fill in here. The claim is that xe −x 2 is a much better bound, a tighter fit, than is e −x . y x y = e –x 2 y = xe –x 2 1 Figure 29.15 iii. Use numerical methods to approximate the proper integral  k 0 e −x 2 dx after the tail,  ∞ k e −x 2 dx, has been amputated. Notice that the graph of e −x 2 is bell-shaped. As stands, e −x 2 cannot be a probability density function because the area under such a function must be exactly 1. A bit of tinkering takes care of this. A standard normal distribution in statistics is described mathematically by the formula p(x) = 1 √ 2π e − x 2 2 . The Method of Comparison We’ve used the method of comparison in several instances. We state the comparison theorem below. We omit the formal proof, but the statements should seem quite reasonable; an informal argument was provided earlier in this section. Comparison Theorem Let f and g be continuous functions with 0 ≤ g(x) ≤f(x)for x ≥a. If  ∞ a f(x)dx converges, then  ∞ a g(x) dx converges. If  ∞ a g(x) dx diverges, then  ∞ a f(x)dx diverges. Suppose h(x) is positive and continuous. To show  ∞ a h(x) dx converges, we must produce a larger function whose improper integral converges. To show  ∞ a h(x) dx diverges, we must produce a smaller function whose improper integral diverges. 29.4 Improper Integrals 913 Naturally, we can’t produce both, so we begin by taking a guess about whether or not  ∞ a h(x) converges. The integral  ∞ 1 1 x p dx (and constant multiples of this integral) can be useful for comparison. Recall that this integral converges for p>1and diverges for p ≤ 1. ◆ EXAMPLE 29.29 Is  ∞ 1 √ 1+sin 2 x x dx convergent? y x y = √1 + sin 2 x x Figure 29.16 SOLUTION 0 ≤ 1 x ≤ √ 1+sin 2 x x  ∞ 1 1 x dx diverges, so  ∞ 1 √ 1+sin 2 x x dx diverges by comparison. ◆ REMARK Not only can we compare improper integrals with one another but we can compare improper integrals with infinite series. See Figure 29.17. x y 12345 y = 1 x area = 1 area = area = 1 2 1 3 area = 1 4 Figure 29.17 The shaded circumscribed rectangles have areas corresponding to the terms of the harmonic series. The rectangles lie above 1 x . We can argue that the harmonic series 1 + 1 2 + 1 3 + 1 4 + ···diverges because  n k=1 1 k is larger than  n+1 1 1 x dx and  ∞ 1 1 x dx is a divergent improper integral. We can argue that the infinite series  ∞ n=2 1 n 2 = 1 2 + 1 3 + 1 4 +···converges by com- paring it to the convergent improper integral  ∞ 1 1 x 2 dx. 914 CHAPTER 29 Computing Integrals x y 12345 y = 1 x 2 area = area = 1 4 1 9 area = 1 16 Figure 29.18 Similarly, after completing Exercise 29.11 we can argue that  ∞ n=1 1 n p converges for p>1 and diverges for p ≤ 1. We will formalize the comparison between improper integrals and infinite series by the end of the next chapter. PROBLEMS FOR SECTION 29.4 In Problems 1 through 5, pinpoint all the improprieties in the integral. If necessary rewrite the integral as a sum of integrals so that each impropriety occurs at an endpoint and there is only one impropriety per integral. 1. (a)  ∞ 0 1 x 2 +4 dx (b)  ∞ 0 1 x 2 −4 dx 2. (a)  ∞ 0 1 x 2 dx (b)  ∞ −∞ 1 x 2 dx 3. (a)  ∞ −∞ 1 x 2 +4 dx (b)  ∞ −∞ 1 x 2 −4 dx 4. (a)  π 2 −π 2 tan xdx (b)  π 0 tan xdx 5. (a)  ∞ 0 tan −1 xdx (b)  ∞ −∞ tan −1 xdx 6. Show that  ∞ 1 1 x p dx converges for p>1and diverges for p ≤ 1. 7. Show that  1 0 1 x p dx converges for p<1and diverges for p ≥ 1. 8. Show that  ∞ −1 1 x 4 dx diverges. 9. (a) Evaluate  ∞ 0 xe −x 2 dx. (b) Evaluate  ∞ −∞ xe −x 2 dx. 10. Show  ∞ 4 e −x 2 dx < 0.0000001. Hint: Compare it to  ∞ 4 xe −x 2 dx. In Problems 11 through 36, determine whether the integral is convergent or divergent. Evaluate all convergent integrals. Be efficient. If lim x→∞ = 0, then  ∞ a f(x)dx is divergent. 29.4 Improper Integrals 915 11.  ∞ 0 e −2x dx 12.  ∞ 0 xdx 13.  ∞ 0 cos x + 1 dx 14.  ∞ 0 cos xdx 15.  ∞ 0 xe −x dx 16.  1 −1 1 5x 2 dx 17.  ∞ −∞ 1 x 3 dx 18.  1 0 ln xdx 19.  ∞ 1 ln xdx 20.  ∞ 1 1 x(x+1) dx 21.  ∞ 0 1 x(x+1) dx 22.  ∞ e 1 x ln x dx 23.  ∞ e 2 1 x(ln x) 2 dx 24.  ∞ 0 x 2+x 2 dx 25.  ∞ 0 1 √ x+1 dx 26.  1 −1 1 √ x+1 dx 27.  5 1 1 x−3 dx 28.  ∞ 1 ln xdx 29.  ∞ 1 x √ 3+x 2 dx 30.  ∞ 1 arctan x 1+x 2 dx 31.  2 1 1 x ln x dx 32.  ∞ 1 arctan xdx 33.  π 0 tan xdx 916 CHAPTER 29 Computing Integrals 34.  ∞ 0 x 2 +3 x+1 dx 35.  ∞ 1 1 (x+1) 3 dx 36.  e 2 1 dx x √ ln x Use the comparison theorem to determine whether the integral is convergent or diver- gent. 37.  ∞ 1 (sin x) 2 x 2 dx 38.  ∞ 2 1 x(x+1) dx 39.  ∞ 2 2 x 2 ln x dx 40.  ∞ 1 1 √ x 7 +1 dx 41.  ∞ 0 sin xe −x dx 42.  ∞ 1 cos x x 2 dx 43. (a) Show that  ∞ 1 1 1+x 4 dx converges. (b) Approximate  ∞ 1 1 1+x 4 dx with error < 0.01. This involves making some choices, but the gist should be as follows. i. Snip off the tail,  ∞ c 1 1+x 4 dx, for some constant c. Bound it using  ∞ c 1 x 4 dx. ii. Approximate  c 1 1 1+x 4 dx using numerical methods. iii. Be sure the sum of the bound in part (i) and the error in part (ii) is less than 0.01. 44. The surface formed by revolving the graph of y = 1 x on [1, ∞) about the x-axis is known as Gabriel’s horn. Find the volume of the horn. Curiously, you will find that the volume is finite even though the area under y = 1 x on [1, ∞) is infinite. Probability Density Functions (Problems 45 through 48) As mentioned at the beginning of this section, statisticians use probability density functions to determine the probability of a random variable falling in a certain interval. If p(x) is a probability density function, then p(x) ≥0 for all x and  ∞ −∞ p(x) dx = 1. 45. A probability density function of the form p(x) =  λe −λx for x ≥ 0, 0 for x<0 where λ is a positive constant describes what is known as an exponential distribution. Verify that  ∞ −∞ p(x) dx =1. 29.4 Improper Integrals 917 46. A cumulative density function, C(x), gives the probability of a random variable taking on a value less than or equal to x.Itisgivenby C(x) =  x −∞ p(y) dy. Show that for an exponential distribution (refer to Problem 45), the cumulative density function is given by C(x) =  1 − e −λx for x ≥ 0, 0 for x<0. Find lim x→∞ C(x). 47. The mean of a probability distribution is given by µ =  ∞ −∞ xp(x) dx, where p(x) is the probability density function. Think of this as p(x) giving a fractional weight to each value of x. Show that the mean of an exponential distribution (see Problem 45) is 1 λ .(Note: You will need to use integration by parts to find µ.) 48. Suppose the number of minutes a caller spends on hold when calling a health clinic can be modeled using the probability density function. p(x) =  10e −10x for x ≥ 0, 0 for x<0. The probability that a random caller will wait at least 5 minutes on hold is given by  ∞ 5 p(x) dx. Find this probability. Note: it is not necessary to compute an improper integral in order to answer this question. 49. Essay Question. Two of your classmates are having some trouble with improper integrals. Todd believes that improper integrals ought to diverge. He reasons that if f is positive, then the accumulated area keeps increasing, even if only by a little bit, so how can we get anything other than infinity? Dylan, on the other hand, is convinced that if lim x→∞ f(x)=0,then  ∞ 0 f(x)dx ought to converge. After all, he reasons, the rate at which area is accumulating is going to zero. Why isn’t that enough to assure convergence? Write an essay responding to Todd and Dylan’s misconceptions. Your essay should be designed to help your classmates see the errors in their reasoning. PART X Series 30 CHAPTER Series 30.1 APPROXIMATING A FUNCTION BY A POLYNOMIAL Preview Addition and multiplication—these are our fundamental computational tools. A high- powered computer, for all its computational sophistication, ultimately relies on these basic operations. How then can a computer numerically approximate values of transcendental functions? How are values of exponential, logarithmic, and trigonometric functions com- puted? Consider the sine function, for example. A calculator can approximate sin 0.1 with a high degree of accuracy, accuracy not readily accessible from unit circle or right triangle definitions of sin x. How can such a good approximation be obtained? If we know the value of a differentiable function f at the point x = b, then we can use the tangent line to f at x = b to approximate the function’s values near x = b. The tangent line is the best linear approximation of f near x = b; higher degree polynomials offer the possibility of staying even closer to the values of f near x = b and following the shape of f over a larger interval around b. In this section we will improve upon the tangent line approximation, obtaining quadratic, cubic, and higher degree polynomial approximations of f around x = b. We will generally find the “fit” improving with the degree of the polynomial. 919 920 CHAPTER 30 Series Such polynomial approximations are convenient because they involve only the operations of addition and multiplication; they are easily evaluated, easily differentiated, and easily integrated. The process of approximating an elusive quantity, successively refining the approxi- mation, and using a limiting process to nail it down is at the heart of theoretical calculus. In this chapter we obtain successively better polynomial approximations of a function about a point by computing increasingly higher degree polynomial approximations. By computing the limit as the degree of the polynomial increases without bound, we will discover that, under certain conditions, we can represent a function as an infinite “polynomial” known as a power series. The fact that sin x, cos x, and e x have representations as power series is remarkable in its own right. In addition, this alternative representation turns out to be com- putationally very useful. Power series representation of functions was known to Newton who used it as a computational aid, particularly for integrating functions lacking elemen- tary antiderivatives. It was the subject of work published by the English mathematician Brook Taylor in 1712 and was popularized by the Scottish mathematician Colin Maclaurin in a textbook published in 1742. Although mathematicians had been using the ideas as early as the 1660s, the names of Taylor and Maclaurin have been associated with power series representations of functions. Polynomial Approximations of sin x around x = 0 In this section we will use polynomials to numerically estimate values of some transcen- dental functions. ◆ EXAMPLE 30.1 A calculator or computer gives sin 0.1 to ten decimal places, displaying 0.0998334166. Obtain this result by using a polynomial to approximate sin x near x = 0 and evaluating this polynomial at x = 0.1. SOLUTION We will approach this problem via a sequence of polynomial approximations to f(x)=sin x for x near zero until we arrive at the desired result. We denote by P k (x) the kth degree polynomial approximation. P k (x) is of the form a 0 + a 1 x + a 2 x 2 + ···+a k x k , where a 0 , a 1 , ,a k are constants. We must determine the values of these constants so the P k (x) “fits” the graph of f well around x = 0. Constant Approximation Because sin x is continuous and 0.1 is near 0, we know sin 0.1 ≈ sin 0 = 0. P 0 (x) = 0; sin 0.1 ≈ P 0 (0.1) = 0 Tangent Line Approximation The tangent line passes through (0, 0) and has a slope of f  (0) = cos 0 = 1. P 1 (x) = x; sin 0.1 ≈ P 1 (0.1) = 0.1 . lim x→∞ f(x)=0,then  ∞ 0 f(x)dx ought to converge. After all, he reasons, the rate at which area is accumulating is going to zero. Why isn’t that enough to assure convergence? Write an essay responding to Todd and Dylan’s misconceptions involve only the operations of addition and multiplication; they are easily evaluated, easily differentiated, and easily integrated. The process of approximating an elusive quantity, successively refining. graph of e −x 2 is bell-shaped. As stands, e −x 2 cannot be a probability density function because the area under such a function must be exactly 1. A bit of tinkering takes care of this. A standard