0 cosh x dx e 0 1 x 30 1 dx 2 5x 6 x cos x dx 2 0 1

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Unformatted text preview: 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 th...
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