Mathematics

# $-2 \int x^2 e^{-2x} dx = e^{-2x} (ax^2 + bx + c) + d$, then

##### SOLUTION
$-2\int x^2e^{-2x}dx$

$=-2\left ( x^2\dfrac{-1}{2}e^{-2x}+\dfrac{1}{4}\left ( -2e^{-2x}x+e^{-2x} \right ) \right )+C$

$=e^{-2x}x^2+e^{-2x}x-\dfrac{1}{2}e^{-2x}+C$

$-2\int x^2e^{-2x}dx=e^{-2x}x^2-e^{-2x}x-\dfrac{1}{2}e^{-2x}+C$

$\Rightarrow -2\int x^2e^{-2x}dx=e^{-2x}\left ( x^2+x-\dfrac{1}{2} \right )+C$

$\Rightarrow$ $e^{-2x}(ax^2+bx+c)+d=e^{-2x}\left ( x^2+x-\dfrac{1}{2} \right )+C$

$\Rightarrow a=1$

$\Rightarrow b=1$

$\Rightarrow c=\dfrac{-1}{2}$

$\Rightarrow d=C(constant)$

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

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