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Coefficients in ∫x²e^(−2x)dx = e^(−2x)(ax²+bx+c)+d | JEE

JEE Maths question with a full step-by-step solution.

Question

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

, then

a=-\dfrac{1}{2},\ b=-\dfrac{1}{2},\ c=-\dfrac{1}{4}

correct

a=-\dfrac{1}{2},\ b=-\dfrac{1}{2},\ c=\dfrac{1}{4}

a=-\dfrac{1}{2},\ b=-1,\ c=-\dfrac{1}{2}

a=1,\ b=1,\ c=-\dfrac{1}{2}

Solution

Step 1: Substitute

t=-2x

, so

x=-\dfrac{t}{2}

dx=-\dfrac{dt}{2}

x^{2}=\dfrac{t^{2}}{4}

and

e^{-2x}=e^{t}

I=\int\dfrac{t^{2}}{4}\,e^{t}\!\left(-\dfrac{dt}{2}\right)=-\dfrac{1}{8}\int t^{2}e^{t}\,dt

Step 2: Use

\displaystyle\int t^{2}e^{t}\,dt=(t^{2}-2t+2)e^{t}

(integration by parts twice):

I=-\dfrac{1}{8}(t^{2}-2t+2)e^{t}+d

Step 3: Back-substitute

t=-2x

, so

t^{2}=4x^{2}

and

-2t=4x

I=-\dfrac{1}{8}(4x^{2}+4x+2)e^{-2x}+d

Step 4: Distribute

-\dfrac{1}{8}

I=e^{-2x}\!\left(-\dfrac{x^{2}}{2}-\dfrac{x}{2}-\dfrac{1}{4}\right)+d

Step 5: Match with

e^{-2x}(ax^{2}+bx+c)+d

a=-\dfrac{1}{2},\quad b=-\dfrac{1}{2},\quad c=-\dfrac{1}{4}

Correct answer: (1)

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