use integration, the direct comparison test, or the limit comparison test to test the integral for…

use integration, the direct comparison test, or the limit comparison test to test the integral for convergence. if more than one method applies, use whatever method you pre\n\nselect the correct choice below and fill in the answer box to complete your choice.\n(type an exact answer.)\n\na. the integral converges because (int_{0}^{ln 2} 4 x^{-2} e^{-4 / x} d x=)\n\nb. the integral diverges because (int_{0}^{ln 2} 4 x^{-2} e^{-4 / x} d x=)

use integration, the direct comparison test, or the limit comparison test to test the integral for convergence. if more than one method applies, use whatever method you pre\n\nselect the correct choice below and fill in the answer box to complete your choice.\n(type an exact answer.)\n\na. the integral converges because (int_{0}^{ln 2} 4 x^{-2} e^{-4 / x} d x=)\n\nb. the integral diverges because (int_{0}^{ln 2} 4 x^{-2} e^{-4 / x} d x=)

Answer

Explanation:

Step1: Substitute ( t=\frac{4}{x} ), then ( x = \frac{4}{t} ) and ( dx=-\frac{4}{t^{2}}dt )

When ( x = 0 ), ( t=\infty ); when ( x=\ln2 ), ( t=\frac{4}{\ln2} ). The integral ( \int_{0}^{\ln2}4x^{-2}e^{-\frac{4}{x}}dx=\int_{\infty}^{\frac{4}{\ln2}}4\times(\frac{t}{4})^{2}e^{-t}\times(-\frac{4}{t^{2}})dt ) Simplify the integrand: ( 4\times\frac{t^{2}}{16}\times(-\frac{4}{t^{2}})e^{-t}=-e^{-t} ) So the integral becomes ( \int_{\frac{4}{\ln2}}^{\infty}e^{-t}dt )

Step2: Integrate ( e^{-t} )

The antiderivative of ( e^{-t} ) is ( -e^{-t} ). Evaluate the definite - integral: ( \lim_{b\rightarrow\infty}\int_{\frac{4}{\ln2}}^{b}e^{-t}dt=\lim_{b\rightarrow\infty}\left[-e^{-t}\right]{\frac{4}{\ln2}}^{b} ) (=\lim{b\rightarrow\infty}(-e^{-b}+e^{-\frac{4}{\ln2}})) Since ( \lim_{b\rightarrow\infty}e^{-b}=0 )

Answer:

A. The integral converges because ( \int_{0}^{\ln2}4x^{-2}e^{-\frac{4}{x}}dx = e^{-\frac{4}{\ln2}} )