improp_ints - (8.8 Improper Integrals t Infinite integrals...

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(8.8) Improper Integrals Infinite integrals. The definition of a f ( x ) dx is a f ( x ) dx = lim t →∞ t a f ( x ) dx. Put differently, suppose F ( x ) is an antiderivative of f ( x ), so f ( x ) dx = F ( x ) + C . So t a f ( x ) dx = F ( x ) t a = F ( t ) - F ( a ). Then a f ( x ) dx = lim t →∞ ( F ( t ) - F ( a ) ) . Example. 0 e - 5 x dx . Here, t 0 e - 5 x dx = - 1 5 e - 5 x t 0 = 1 5 - 1 5 e - 5 t . So 0 e - 5 x dx = lim t →∞ t 0 e - 5 x dx = lim t →∞ ( 1 5 - 1 5 e - 5 t ) = 1 5 . (Note lim t →∞ e - 5 t = lim t →∞ 1 e 5 t = 0; this is because 1 large = small.) Example. 0 x 2 e - x dx . Here, by two integrations by parts, x 2 e - x dx = - ( x 2 + 2 x + 2) e - x + C. Therefore, 0 x 2 e - x dx = lim t →∞ t 0 x 2 e - x dx = lim t →∞ - ( x 2 + 2 x + 2) e - x t 0 = lim t →∞ - ( t 2 + 2 t + 2) e - t + 2 . We must compute this limit. Now lim t →∞ t 2 e - t = lim t →∞ t 2 e t = lim t →∞ 2 t e t = lim t →∞ 2 e t = 0 by two applications of L’Hospital’s rule. In the same way, we see 0 x 2 e - x dx = lim t →∞ - ( t 2 + 2 t + 2) e - t + 2 = 2 . Note, it may be useful to remember that if n is any positive integer, then lim t →∞ t n e t = 0. (This requires n applications of L’Hospital’s rule. This says that e t grows much faster for large t than does t n .) 1
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Example. 1 1 x dx . Now t 1 1 x dx = 2 x t 1 = 2 t - 2. Therefore, 1 1 x dx = lim t →∞ t 1 1 x dx = lim t →∞ 2 t - 2 = .
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