Question

Determine whether the integral is convergent or divergent. [infinity] 7 e−1/x x2 dx convergent divergent Changed: Your submitted answer was incorrect. Your current answer has not been submitted. If it is convergent, evaluate it. (If the quantity diverges, enter DIVERGES.)

Answer 1

I suppose the integral could be

$\displaystyle\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx$

In that case, since $-\frac1x\to0$ as $x\to\infty$, we know $e^{-1/x}\to1$. We also have $\left(e^{-1/x}\right)'=\frac{e^{-1/x}}{x^2}>0$, so the integral is approach +1 from below. This tells us that, by comparison,

$\displaystyle\frac{e^{-1/x}}{x^2}\le\frac1{x^2}\implies\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx\le\int_7^\infty\frac{\mathrm dx}{x^2}$

and the latter integral is convergent, so this integral must converge.

To find its value, let $u=-\frac1x$, so that $\mathrm du=\frac{\mathrm dx}{x^2}$. Then the integral is equal to

$\displaystyle\int_{-1/7}^0e^u\,\mathrm du=e^0-e^{-1/7}=1-\frac1{\sqrt[7]{e}}$