Mathematics

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


ANSWER

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


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,

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

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

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

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

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

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

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

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

 

Hence, this is the answer.

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