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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
If ∫x2e−2x dx=e−2x(ax2+bx+c)+d\displaystyle\int x^{2}e^{-2x}\,dx=e^{-2x}(ax^{2}+bx+c)+d, then
Aa=−12, b=−12, c=−14a=-\dfrac{1}{2},\ b=-\dfrac{1}{2},\ c=-\dfrac{1}{4}correct
Ba=−12, b=−12, c=14a=-\dfrac{1}{2},\ b=-\dfrac{1}{2},\ c=\dfrac{1}{4}
Ca=−12, b=−1, c=−12a=-\dfrac{1}{2},\ b=-1,\ c=-\dfrac{1}{2}
Da=1, b=1, c=−12a=1,\ b=1,\ c=-\dfrac{1}{2}
Solution
Step 1: Substitute t=−2xt=-2x, so x=−t2x=-\dfrac{t}{2}, dx=−dt2dx=-\dfrac{dt}{2}, x2=t24x^{2}=\dfrac{t^{2}}{4} and e−2x=ete^{-2x}=e^{t}:
I=∫t24 et ⁣(−dt2)=−18∫t2et dtI=\int\dfrac{t^{2}}{4}\,e^{t}\!\left(-\dfrac{dt}{2}\right)=-\dfrac{1}{8}\int t^{2}e^{t}\,dt
 Step 2: Use ∫t2et dt=(t2−2t+2)et\displaystyle\int t^{2}e^{t}\,dt=(t^{2}-2t+2)e^{t} (integration by parts twice):
I=−18(t2−2t+2)et+dI=-\dfrac{1}{8}(t^{2}-2t+2)e^{t}+d
 Step 3: Back-substitute t=−2xt=-2x, so t2=4x2t^{2}=4x^{2} and −2t=4x-2t=4x:
I=−18(4x2+4x+2)e−2x+dI=-\dfrac{1}{8}(4x^{2}+4x+2)e^{-2x}+d
 Step 4: Distribute −18-\dfrac{1}{8}:
I=e−2x ⁣(−x22−x2−14)+dI=e^{-2x}\!\left(-\dfrac{x^{2}}{2}-\dfrac{x}{2}-\dfrac{1}{4}\right)+d
 Step 5: Match with e−2x(ax2+bx+c)+de^{-2x}(ax^{2}+bx+c)+d:
a=−12,b=−12,c=−14a=-\dfrac{1}{2},\quad b=-\dfrac{1}{2},\quad c=-\dfrac{1}{4}
 Correct answer: (1)
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