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

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

Question

Let

f(x)

satisfy

f'(x)=f(x)

with

f(0)=1

, and let

g

satisfy

f(x)+g(x)=x^{2}

. The value of

\displaystyle\int_{0}^{1}f(x)\,g(x)\,dx

e-\dfrac12e^{2}-\dfrac52

e-e^{2}-3

\dfrac12(e-3)

e-\dfrac12e^{2}-\dfrac32

correct

Solution

Step 1: Solve for

f

. From

\dfrac{f'(x)}{f(x)}=1

, integrating gives

\ln f(x)=x+c

, so

f(x)=ke^{x}

. The condition

f(0)=1

forces

k=1

, hence

f(x)=e^{x}

. Step 2: Determine

g

. From

f(x)+g(x)=x^{2}

g(x)=x^{2}-e^{x}.

Therefore

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,

\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

\int_{0}^{1}e^{2x}\,dx=\dfrac12\big(e^{2}-1\big).

Step 4: Combine:

\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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