Mathematics

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


ANSWER

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

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

$$d\in R$$


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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