Question

Evaluate: $\underset{1}{\overset{2}{\int }}\frac{1}{x^2}{e}^{\frac{-1}{x}}dx=$

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

Answer 1

Answer: (B)

$\frac{\sqrt{e}-1}{e}$

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

Let $t=-\frac{1}{x}\Rightarrow dt=-(-\frac{1}{x^2})dx=\frac{1}{x^2}dx$

When $x=1\Rightarrow t=-1$

When $x=2\Rightarrow t=-\frac{1}{2}$

$={\int }_{-1}^{-\frac{1}{2}}{e}^{t}dt$

$={\left[e^t\right]}_{-1}^{-\frac{1}{2}}$

$=e^{-\frac{1}{2}}-e^{-1}$

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

$=\frac{e-\sqrt{e}}{e\sqrt{e}}$

$=\frac{\sqrt{e}(\sqrt{e}-1)}{e\sqrt{e}}$

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