How do you find the integral int_1^2e^(1/x)/x^2dx ?

1 Answer
Sep 11, 2014

By using the substitution $u = \frac{1}{x}$, we can find
${\int}_{1}^{2} {e}^{\frac{1}{x}} / {x}^{2} \mathrm{dx} = e - \sqrt{e}$.

Let $u = \frac{1}{x}$. $R i g h t a r r o w \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{x} ^ 2 R i g h t a r r o w \mathrm{dx} = - {x}^{2} \mathrm{du}$
Since $x$ goes from 1 to 2, $u$ goes from 1 to 1/2.

Let us look at some details,
${\int}_{1}^{2} {e}^{\frac{1}{x}} / {x}^{2} \mathrm{dx}$

by Substitution,
$= {\int}_{1}^{\frac{1}{2}} {e}^{u} / {x}^{2} \cdot \left(- {x}^{2}\right) \mathrm{du}$

by cancelling ${x}^{2}$,
$= - {\int}_{1}^{\frac{1}{2}} {e}^{u} \mathrm{du}$

by using the negative sign to switch the lower and upper limits,
$= {\int}_{\frac{1}{2}}^{1} {e}^{u} \mathrm{du}$

by taking the antiderivative,
$= {\left[{e}^{u}\right]}_{\frac{1}{2}}^{1} = e - {e}^{\frac{1}{2}} = e - \sqrt{e}$