Mathematics

# $I = \displaystyle \int _{ 0 }^{ 1 }{ { x }^{ 2 } } { e }^{ -x }dx$

$I=2-\dfrac{1}{e}$

##### SOLUTION

Consider the given integral.

$I=\int_{0}^{1}{{{x}^{2}}{{e}^{-x}}}dx$

We know that

$\int{uvdx=u\int{vdx-}}\int{\left( \dfrac{d}{dx}\left( u \right)\int{vdx} \right)}dx$

Therefore,

$I=\left[ {{x}^{2}}\left( \dfrac{{{e}^{-x}}}{-1} \right) \right]_{0}^{1}-\int_{0}^{1}{2x\left( \dfrac{{{e}^{-x}}}{-1} \right)}dx$

$I=\left[ -{{x}^{2}}{{e}^{-x}} \right]_{0}^{1}+2\int_{0}^{1}{x{{e}^{-x}}}dx$

$I=-\left( 1\left( {{e}^{-1}} \right)-0 \right)+2\left[ \left( x\left( \dfrac{{{e}^{-x}}}{-1} \right) \right)_{0}^{1}-\int_{0}^{1}{1\left( \dfrac{{{e}^{-x}}}{-1} \right)}dx \right]$

$I=-\dfrac{1}{e}+2\left[ \left( x{{e}^{-x}} \right)_{0}^{1}+\int_{0}^{1}{{{e}^{-x}}}dx \right]$

$I=-\dfrac{1}{e}+2\left[ \left( 1{{e}^{-1}}-0 \right)+\left( \dfrac{{{e}^{-x}}}{-1} \right)_{0}^{1} \right]$

$I=-\dfrac{1}{e}+2\left( \dfrac{1}{e}-{{e}^{-1}}+{{e}^{0}} \right)$

$I=-\dfrac{1}{e}+2\left( 1 \right)$

$I=2-\dfrac{1}{e}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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