Fall 2002 Improper Integrals soln - Mat104 Fall 2002 Improper Integrals From Old Exams For the following integrals state whether they are convergent or

# Thomas' Calculus: Early Transcendentals

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Mat104 Fall 2002, Improper Integrals From Old Exams For the following integrals, state whether they are convergent or divergent, and give your reasons. (1) 0 dx x 3 + 2 converges. Break it up as 1 0 dx x 3 + 2 + 1 dx x 3 + 2 . The first of these is proper and finite. The second behaves like the integral of 1 /x 3 on [1 , ) and thus converges. (2) 1 0 dx x + x converges. As x 0, x goes to 0 much more slowly than x does. (Think about the graphs.) Therefore when x is very close to 0, the denominator x + x x . So this integral will behave like the integral of 1 / x on [0 , 1], and this integral converges. (3) 1 1 + x x 3 converges. As x goes to , the integrand behaves like x x 3 = 1 x 5 / 2 . (4) 0 x 2 x 3 + 1 dx diverges. Break it up into two integrals 1 0 x 2 x 3 + 1 dx + 1 x 2 x 3 + 1 dx . The first integral is proper and finite. The second can be compared to the integral of 1 /x on [1 , ) which diverges. (5) 1 0 ln x dx converges to - 1. Here we can compute directly since integration by parts tells us that ln x dx = x ln x - x + C . Evaluating at the x = 1 endpoint gives ln 1 - 1 = - 1. For the other endpoint we have to take the limit as x goes to 0. For this we need L’Hˆ opital’s rule. lim x 0 x ln x = lim x 0 ln x 1 /x = lim x 0 1 /x - 1 /x 2 = lim x 0 - x = 0 . So evaluating at the x = 0 endpoint gives 0. (6) 1 0 dx e x - 1 diverges. The only difficulty is that the denominator is 0 when x = 0. There are a couple of approaches we could take. The easiest is to use the Taylor series for e x . Then we know that e x - 1 = x + higher powers of x and as x goes to zero, the higher powers of x will vanish much more rapidly. So this function behaves essentially like 1 /x when x is close to 0. Since dx/x diverges, this integral will also. Alternatively, we could compute the integral, making the substitution u = e x and then use partial fractions. (7) 0 dx x 2 + 2 x + 2 converges. The only difficulty is that we have an infinite endpoint. The integrand is asymptotic to 1 /x 2 as x goes to infinity. Since 1 dx/x 2 converges, this integral will as well. (To compare these we should break up the integral. First integrate from 0 to 1, which gives a finite value. Then integrate further from 1 out to . This gives a finite value as well by comparison to 1 /x 2 .) (8) 1 x 3 ln x + x 4 dx diverges. Again the only problem is that we have an infinite endpoint. As x goes to infinity, x 4 grows much faster than ln x . Thus the integrand will be asymptotic to x 3 /x 4 = 1 /x as x goes to infinity.
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