Mathematics

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

$a=-\displaystyle \frac {1}{2}$

$c=-\displaystyle \frac{1}{4}$

$d\in R$

##### SOLUTION
$\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d$
Differentiating both sides, we get
$x^{2}e^{-2x}=e^{-2x}(2ax+b)+(ax^{2}+bx+c)(-2e^{-2x})$
$x^2e^{-2x}=e^{-2x}(-2ax^{2}+2(a-b)x+b-2c)$
On comparing the coefficients, we get
$-2a=1,\:2(a-b)=0,\:b-2c=0$
$\displaystyle a=\frac{-1}{2}, b=-\frac{1}{2},\:c=-\frac{1}{4}$
Also, $d\in R$

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

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