Question

The value of $\underset{1}{\overset{2}{\int }}\frac{1}{x^2}{e}^{-1/x}dx$ is

• A. $\frac{1}{\sqrt{e}}+\frac{1}{e}$
• B. $\frac{1}{e}-\frac{1}{\sqrt{e}}$
• C. $\frac{1}{\sqrt{e}}-\frac{1}{e}$
• D. 0

Answer 1

The correct option is C $\frac{1}{\sqrt{e}}-\frac{1}{e}$

Consider ${\int }_1^2\frac{1}{x^2}{e}^{-1/x}dx$

Let $u=-\frac{1}{x}$

$\Rightarrow du=\frac{1}{x^2}dx$

When $x=1$, $u=-1$ and when $x=2,u=-\frac{1}{2}$

${\int }_1^2\frac{1}{x^2}{e}^{-1/x}dx={\int }_{-1}^{-1/2}{e}^{u}du$

$=-{\int }_{-1/2}^{-1}{e}^{u}du$

$=-[e^u{]}_{-1/2}^{-1}$

$=-[e^{-1}-e^{-1/2}]$

$=-[\frac{1}{e}-\frac{1}{\sqrt{e}}]$

$I={\int }_1^2\frac{1}{x^2}{e}^{-1/x}dx$$=\frac{1}{\sqrt{e}}-\frac{1}{e}$