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SalasSV_10_07 - 622 CHAPTER 10 SEQUENCES INDETERMINATE...

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10.7 IMPROPER INTEGRALS In the de fi nition of the de fi nite integral b a f ( x ) dx 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 We begin with a function f continuous on an unbounded interval [ a , ). For each number b > a we can form the de fi nite integral b a f ( x ) dx . (Figure 10.7.1) If, as b tends to , this integral tends to a fi nite limit L , y f x a b Figure 10.7.1 lim b →∞ b a f ( x ) dx = L , then we write a f ( x ) dx = L
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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, improper integrals b −∞ f ( x ) dx arise as limits of the form lim a →−∞ b a f ( x ) dx . Example 1 (a) 0 e 2 x dx = 1 2 . (b) 1 dx x diverges. (c) 1 dx x 2 = 1. (d) 1 −∞ cos π x dx diverges. VERIFICATION (a) 0 e 2 x dx = lim b →∞ b 0 e 2 x dx = lim b →∞ e 2 x 2 b 0 = lim b →∞ 1 2 1 2 e 2 b = 1 2 . (b) 1 dx x = lim b →∞ b 1 dx x = lim b →∞ ln b = ∞ . (c) 1 dx x 2 = lim b →∞ b 1 dx x 2 = lim b →∞ 1 x b 1 = lim b →∞ 1 1 b = 1. (d) Note fi rst that 1 a cos π x dx = 1 π sin π x 1 a = − 1 π sin π a . As a tends to −∞ , sin π a oscillates between 1 and 1. Therefore the integral oscillates between 1 and 1 and does not converge. The usual formulas for area and volume are extended to the unbounded case by means of improper integrals. Example 2 Let p be a positive number. If is the region between the x -axis and the graph of f ( x ) = 1 x p , x 1, (Figure 10.7.2) y x 1 f ( x ) = x p Figure 10.7.2 then area of = 1 p 1 , if p > 1 , if p 1.
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624 CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS This comes about from setting area of = lim b →∞ b 1 dx x p = 1 dx x p . For p = 1, 1 dx x p = lim b →∞ b 1 dx x p = lim b →∞ 1 1 p ( b 1 p 1) = 1 p 1 , if p > 1 , if p < 1. For p = 1, 1 dx x p = 1 dx x = ∞ , as you have seen already. For future reference we record the following: Let p > 0, (10.7.1) 1 dx x p converges for p > 1 and diverges for p 1. Example 3 We know that the region below the graph of f ( x ) = 1 / x , x 1, has in fi nite area. Suppose that this region with in fi nite area is revolved about the x -axis (see Figure 10.7.3). What is the volume V of the resulting solid? It may surprise you somewhat, but the volume is not in fi nite. In fact, it is π . Using the disc method to calculate the volume (Section 6.2), we have V = 1 π [ f ( x )] 2 dx = π 1 dx x 2 = π lim b →∞ b 1 dx x 2 = π lim b →∞ 1 x b 1 = π · 1 = π .
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