472 Chapter 6 Further Topics in Integration 4 This section extends the concept of the definite integral to integrals of the form ϩϱ Improper ͵ f(x) dx Integrals a in which the upper limit of integration is not a finite number. Such integrals are known as improper integrals and arise in a variety of practical situations. ϩϱ GEOMETRIC INTERPRETATION If f is nonnegative, the improper integral ͵ f(x) dx can be interpreted as the area of a the region under the graph of f to the right of x ϭ a (Figure 6.21a). Although this region has infinite extent, its area may be either finite or infinite, depending on how quickly f(x) approaches zero as x increases. yy y = f(x) y = f(x) yyy xx a aN (a) (b) yyy ϱ N FIGURE 6.21 Area ͵ f(x) dx ϭ lim ͵ f(x) dx. ϩϱ a Nfi a A reasonable strategy for finding the area of such a region is to first use a defi- nite integral to compute the area from x ϭ a to some finite number x ϭ N and then to let N approach infinity in the resulting expression. That is, N Total area ϭ lim (area from a to N) ϭ lim ͵ f(x) dx ϩ ϩ Nfi ϱ Nfi ϱ a This strategy is illustrated in Figure 6.21b and motivates the following definition of the improper integral. The Improper Integral I If f(x) is continuous for x Ն a, then ϩϱ N ͵ ϭ ͵ f(x) dx lim f(x) dx ϩ a Nfi ϱ a Chapter 6 I Section 4 Improper Integrals 473 If the limit defining the improper integral is a finite number, the integral is said to converge. Otherwise the integral is said to diverge. Here are some examples. EXAMPLE 4.1 ϩϱ ͵ 1 Evaluate 2 dx. 1 x Solution First compute the integral from 1 to N and then let N approach infinity. Arrange your work as follows: ϩϱ N N ͵ 1 ϭ ͵ 1 ϭ Ϫ1 ϭ Ϫ 1 ϩ ϭ dx lim dx lim lim 1 1 2 ϩϱ 2 ϩϱ ΂ ΃ ϩϱ ΂ ΃ 1 x Nfi 1 x Nfi x Խ1 Nfi N Refer to Example 4.1. Place EXAMPLE 4.2 1 ϩϱ f(x) ϭ into Y1 of the 1 x2 Evaluate ͵ dx. x equation editor and store the 1 numerical integral into Y2, using Solution the interval from x ϭ 1 to X. Set ϩϱ 1 N 1 N the table feature to start at x ϭ ͵ dx ϭ lim ͵ dx ϭ lim ln ԽxԽ ϭ lim ln N ϭϩϱ ϩϱ ϩϱ ΂ ΃ ϩϱ 1 x Nfi 1 x Nfi Խ1 Nfi 500 increasing in increments of 500. Explain what you observe. 1 Notice that the improper integral of the function f(x) ϭ in Example 4.1 x2 1 converged, while that of the function f(x) ϭ in Example 4.2 diverged. In geometric x 1 terms, this says that the area to the right of x ϭ 1 under the curve y ϭ is finite, x2 1 while the corresponding area under the curve y ϭ is infinite. The reason for the x 1 1 difference is that, as x increases, approaches zero more quickly than does (see x2 x Figure 6.22). 474 Chapter 6 Further Topics in Integration y 3 2 1 1 y = x Finite area 1 Infinite area y = x2 x 1 1 1 FIGURE 6.22 The areas under y ϭ and y ϭ . x x2 It can be shown that lim xm eϪkx ϭ 0 xfi ϩϱ for all m Ͼ 0 and k Ͼ 0. This kind of limit occurs frequently in practical problems involving improper integrals. Here is an example. EXAMPLE 4.3 ϩϱ Evaluate ͵ xeϪ2x dx. 0 Solution ϩϱ N ͵ Ϫ2x ϭ ͵ Ϫ2x xe dx lim xe dx ϩ 0 Nfi ϱ 0 N N 1 Ϫ 1 Ϫ ϭ lim Ϫ xe 2x ϩ ͵ e 2x dx (integration by parts) ϩϱ ΂ ΃ Nfi 2 Խ0 2 0 N 1 Ϫ2x 1 Ϫ2x ϭ lim Ϫ xe Ϫ e ϩϱ ΂ ΃ Nfi 2 4 Խ0 Chapter 6 I Section 4 Improper Integrals 475 1 1 1 ϭ lim ΂Ϫ NeϪ2N Ϫ eϪ2N ϩ 0 ϩ ΃ Nfi ϩϱ 2 4 4 1 ϭ (since NeϪ2N fi 0) 4 APPLICATIONS OF THE The following applications of the improper integral generalize applications of the def- IMPROPER INTEGRAL inite integral that you saw in Sections 2 and 3. In each, the strategy is to use the char- acterization of the definite integral as the limit of a sum to construct an appropriate definite integral and then to let the upper limit of integration increase without bound. As you read these examples, you may want to refer back to the corresponding exam- ples in Sections 2 and 3. PRESENT VALUE In Example 2.2 of Section 2, you saw that the present value of an investment that generates income over a finite period of time is given by a definite integral. The pres- ent value of an investment that generates income in perpetuity is given by an improper integral. Here is an example. EXAMPLE 4.4 A donor wishes to make a gift to a private college from which the college will draw $7,000 per year in perpetuity to support the operation of its computer center. Assum- ing that the prevailing annual interest rate will remain fixed at 10% compounded con- tinuously, how much should the donor give the college? That is, what is the present value of the endowment? Solution To find the present value of a gift that generates $7,000 per year for N years, divide the N-year time interval 0 Յ t Յ N into n subintervals of length ⌬nt years and let tj denote the beginning of the jth subinterval (Figure 6.23). –0.1tj ∆ ∆ 7,000e nt 7,000 nt t 0 tj N ∆ nt FIGURE 6.23 The present value of the money generated during the jth subinterval. 476 Chapter 6 Further Topics in Integration Then Amount generated during 7,000 ⌬ t jth subinterval n Present value of amount Ϫ0.1tj and generated during 7,000e ⌬nt jth subinterval n Present value of Ϫ0.1tj Hence, ϭ lim ͚ 7,000e ⌬nt ϩ N-year gift nfi ϱ jϭ1 N ϭ ͵ 7,000eϪ0.1t dt 0 To find the present value of the total gift, take the limit of this integral as N approaches infinity. That is, N Present value of ϭ ͵ Ϫ0.1t lim 7,000e dt ϩ total gift Nfi ϱ 0 N Ϫ0.1t ϭ lim Ϫ70,000e ϩϱ ΂ ΃ Nfi Խ0 Ϫ0.1N ϭ lim Ϫ70,000(e Ϫ 1) Nfi ϩϱ ϭ $70,000 NUCLEAR WASTE The next example is similar in structure to the problem involving survival and renewal functions in Example 3.2 of Section 3. EXAMPLE 4.5 It is estimated that t years from now, a certain nuclear power plant will be producing radioactive waste at the rate of f(t) ϭ 400t pounds per year. The waste decays exponentially at the rate of 2% per year. What will happen to the accumulation of radioactive waste from the plant in the long run? Solution To find the amount of radioactive waste present after N years, divide the N-year inter- val 0 Յ t Յ N into n equal subintervals of length ⌬nt years and let tj denote the begin- ning of the jth subinterval (Figure 6.24). Then, Amount of waste produced 400t ⌬ t during jth subinterval j n Chapter 6 I Section 4 Improper Integrals 477 N – tj years ∆ –0.02(N – t ) ∆ 400tj nt 400tj e j nt t 0 tj N ∆ nt FIGURE 6.24 Radioactive waste generated during the jth subinterval. Since the waste decays exponentially at the rate of 2% per year, and since there are N Ϫ tj years between times t ϭ tj and t ϭ N, it follows that Ϫ0.02(NϪtj) Amount of waste produced Ӎ 400tje ⌬nt during jth subinterval still present at t ϭ N Thus, n Amount of waste present Ϫ0.02(NϪtj) ϭ lim ͚ 400tje ⌬nt ϩ after N years nfi ϱ jϭ1 N ϭ ͵ Ϫ0.02(NϪt) 400te dt 0 N ϭ Ϫ0.02N ͵ 0.02t 400e te dt 0 The amount of radioactive waste present in the long run is the limit of this expres- sion as N approaches infinity. That is, Amount of waste N ϭ Ϫ0.02N ͵ 0.02t present in the lim 400e te dt ϩϱ long run Nfi 0 N Ϫ0.02N 0.02t 0.02t ϭ lim 400e (50te Ϫ 2,500e ) ϩϱ Nfi Խ0 Ϫ0.02N 0.02N 0.02N ϭ lim 400e (50Ne Ϫ 2,500e ϩ 2,500) Nfi ϩϱ Ϫ0.02N ϭ lim 400(50N Ϫ 2,500 ϩ 2,500e ) Nfi ϩϱ ϭϩϱ That is, in the long run, the accumulation of radioactive waste from the plant will increase without bound. 478 Chapter 6 Further Topics in Integration PROBABILITY DENSITY Improper integrals also appear in the study of probability and statistics, which is an FUNCTIONS important tool in certain areas of the social, managerial, and life sciences. We close this section with a brief introduction to this application. For example, the life span of a lightbulb selected at random from a manufac- turer’s stock is a quantity that cannot be predicted with certainty. The process of selecting a bulb is called a random experiment, and the life span X of the bulb is a continuous random variable.