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∫ f·g dx with f′=f, f(0)=1, f+g=x² | JEE

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Question
Let f(x)f(x) satisfy f′(x)=f(x)f'(x)=f(x) with f(0)=1f(0)=1, and let gg satisfy f(x)+g(x)=x2f(x)+g(x)=x^{2}. The value of ∫01f(x) g(x) dx\displaystyle\int_{0}^{1}f(x)\,g(x)\,dx is
Ae−12e2−52e-\dfrac12e^{2}-\dfrac52
Be−e2−3e-e^{2}-3
C12(e−3)\dfrac12(e-3)
De−12e2−32e-\dfrac12e^{2}-\dfrac32correct
Solution
Step 1: Solve for ff. From f′(x)f(x)=1\dfrac{f'(x)}{f(x)}=1, integrating gives ln⁡f(x)=x+c\ln f(x)=x+c, so f(x)=kexf(x)=ke^{x}. The condition f(0)=1f(0)=1 forces k=1k=1, hence f(x)=exf(x)=e^{x}. Step 2: Determine gg. From f(x)+g(x)=x2f(x)+g(x)=x^{2},
g(x)=x2−ex.g(x)=x^{2}-e^{x}.
 Therefore f(x)g(x)=ex(x2−ex)=x2ex−e2x.f(x)g(x)=e^{x}\big(x^{2}-e^{x}\big)=x^{2}e^{x}-e^{2x}. Step 3: Integrate the two pieces. By parts, ∫01x2ex dx=[x2ex−2xex+2ex]01=(e−2e+2e)−(2)=e−2.\displaystyle\int_{0}^{1}x^{2}e^{x}\,dx=\big[x^{2}e^{x}-2xe^{x}+2e^{x}\big]_{0}^{1}=(e-2e+2e)-(2)=e-2. Also
∫01e2x dx=12(e2−1).\int_{0}^{1}e^{2x}\,dx=\dfrac12\big(e^{2}-1\big).
 Step 4: Combine:
∫01f(x)g(x) dx=(e−2)−12(e2−1)=e−2−e22+12=e−12e2−32.\int_{0}^{1}f(x)g(x)\,dx=(e-2)-\dfrac12\big(e^{2}-1\big)=e-2-\dfrac{e^{2}}{2}+\dfrac12=e-\dfrac12e^{2}-\dfrac32.
 Correct answer: (4)
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