Mathematics

# Evaluate: $\int\limits_1^2 {\dfrac{1}{{{x^2}}}{e^{\tfrac{{ - 1}}{x}}}dx = }$

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

##### SOLUTION
$\displaystyle\int_{1}^{2}{\dfrac{1}{{x}^{2}}{e}^{-\frac{1}{x}}dx}$

Let $t=-\dfrac{1}{x}\Rightarrow\,dt=-\left(-\dfrac{1}{{x}^{2}}\right)dx=\dfrac{1}{{x}^{2}}dx$

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

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

$=\displaystyle\int_{-1}^{-\tfrac{1}{2}}{{e}^{t}dt}$

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

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

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

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

$=\dfrac{\sqrt{e}\left(\sqrt{e}-1\right)}{e\sqrt{e}}$

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

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
Let $f$ be the continuous function on $R$ satisfying $f\left ( x+y \right )= f\left ( x \right )+f\left ( y \right )$ for all $x,\: y\: \in \: R$ with $f\left ( 1 \right )= 2$ and $g$ be a function satisfying $f\left ( x \right )\div g\left ( x \right )= e^{x}$ then the value of the integral $\displaystyle \int_{0}^{1}f\left ( x \right )\: g\left ( x \right )\: dx$ is
• A. $\dfrac{1}{e}-4$
• B. $\displaystyle \frac{1}{4}\left ( e-2 \right )$
• C. $\left (\dfrac {1}{2} \right )\left ( e-3 \right )$
• D. $\dfrac 2 3$

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Solve:
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