We start with the given integral: ? f(x) dx = ? [(8x)/(sqrt(9x^4+x^3))] dx, boun

Question

We start with the given integral: ? f(x) dx = ? [(8x)/(sqrt(9x^4+x^3))] dx, bounds of: 7 < x < infinity (1) Choose the best comparison F(x) for f(x): a) (8x)/(9x^4) b) 8/(x^(1/2)) c) (8x)/(x^3) d) 8/(3x) (2) What is the limit as x goes to infinity of f(x)/F(x)? (3) Is it convergent or divergent?

Explanation / Answer

1)c) (8x)/(x^3) 2)(1) Let f( x ) = x^3 e^(-x^4) for short. Its antiderivative ( indefinite integral ) is g( x ) = (- 1/4 ) *e^(-x^4).This can be found by using the substitution u = - x^4, du = - 4 x^3 dx. The improper integral of f from - infinity to + infinity is defined to be the sum of two limits, if both limits exist. We first integrate f from 0 to b ( an unspecified positive number ). Using the antiderivative g, we get g( b ) - g( 0 ). The limit of this as b --> infinity is 0 - g( 0 ) = - g( 0 ) = 1/4. This is the improper integral of f from 0 to infinity. Next, we integrate f from c ( an unspecified negative number ) to 0. This gives g( 0 ) - g( c ). Take the limit as c --> - infinity to get g( 0 ) - 0 = g( 0 ) = - 1/4. This is the improper integral of f from - infinity to 0. Since both limits exist, we conclude that the improper integral of f from - infinity to + infinity converges and equals 1/4 + ( - 1/4 ) = 0. ( 2 ) Let f( y ) = 1/( 4y-1). Note that f has an infinite discontinuity at y = 1/4 , which is in the interval of integration. The improper integral of f from 0 to 1 is defined to be the sum of two limits, if both exist. To compute the first one, integrate f from 0 to b , a number slightly less than 1/4. The antiderivative of f is g( y ) = (1/4)*ln| 4y-1|. Hence the integral equals g(b)-g(0). The improper integral of f from 0 to 1/4 is defined to be the limit of g(b)-g(0) as b --> 1/4 from the left. However this limit does not exist. Thus the improper integral diverges. There is a similar problem with the improper integral of f from 1/4 to 1. To compute it, we first integrate f from c , a number slightly larger than 1/4 , to 1. This gives g(1)-g(c). We then take the limit of this as c --> 1/4 from the right. We find that the limit does not exist, hence the improper integral diverges. The given improper integral of f from 0 to 1 is the sum of the two improper integrals above, provided that both of them converge. Since neither one converges, we conclude that the given integral diverges.

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