Question

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

• A. $a=-\frac{1}{2}$
• B. $b=2$
• C. $c=-\frac{1}{4}$
• D. $d\in R$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $a=-\frac{1}{2}$
C $c=-\frac{1}{4}$
D $d\in R$

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