Thomas' Calculus: Early Transcendentals

Unformatted text preview: Mat104 Fall 2002, Improper Integrals From Old Exams For the following integrals, state whether they are convergent or divergent, and give your reasons. (1) Z ∞ dx x 3 + 2 converges. Break it up as Z 1 dx x 3 + 2 + Z ∞ 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) Z 1 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) Z ∞ 1 √ 1 + x x 3 converges. As x goes to ∞ , the integrand behaves like √ x x 3 = 1 x 5 / 2 . (4) Z ∞ x 2 x 3 + 1 dx diverges. Break it up into two integrals Z 1 x 2 x 3 + 1 dx + Z ∞ 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) Z 1 ln x dx converges to- 1. Here we can compute directly since integration by parts tells us that Z 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 → x ln x = lim x → ln x 1 /x = lim x → 1 /x- 1 /x 2 = lim x →- x = 0 . So evaluating at the x = 0 endpoint gives 0. (6) Z 1 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 Z 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) Z ∞ 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 Z ∞ 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) Z ∞ 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...
View Full Document