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NVAMediumJEE 2026Indefinite Integrals

JEE Mathematics 2026 Question with Solution

If f(x)f(x) satisfies the relation

f(x)=ex+∫01(y+xex)f(y) dy,f(x)=e^x+\int_0^1 (y+x e^x)f(y)\,dy,

then e+f(0)e+f(0) is equal to

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given:

f(x)=ex+∫01(y+xex)f(y) dyf(x)=e^x+\int_0^1 (y+x e^x)f(y)\,dy

Find: e+f(0)e+f(0)

Let

∫01f(y) dy=Aand∫01yf(y) dy=B\int_0^1 f(y)\,dy=A \quad \text{and} \quad \int_0^1 y f(y)\,dy=B

Then,

f(x)=ex+B+xexAf(x)=e^x+B+x e^x A

Substitute x=0x=0:

f(0)=1+Bf(0)=1+B

Integrate both sides from 00 to 11:

A=∫01ex dx+B+∫01xex dx⋅AA=\int_0^1 e^x \, dx+B+\int_0^1 x e^x \, dx\cdot A

So,

A=(e−1)+B+(e−2)AA=(e-1)+B+(e-2)A

From the working, solving gives

A=1,B=2A=1,\quad B=2

Hence,

f(0)=1+2=3f(0)=1+2=3

Therefore,

e+f(0)=e+3=4e+f(0)=e+3=4

So the required answer is 44.

Using constant integrals

Given: the integral equation contains integration with respect to yy, so the integrals can be treated as constants with respect to xx.

Find: e+f(0)e+f(0)

Define

A=∫01f(y) dy,B=∫01yf(y) dyA=\int_0^1 f(y)\,dy, \qquad B=\int_0^1 y f(y)\,dy

Then the given relation becomes

f(x)=ex+B+xexAf(x)=e^x+B+x e^x A

Now put x=0x=0 to get

f(0)=1+Bf(0)=1+B

Next, integrate the expression for f(x)f(x) from 00 to 11:

A=∫01ex dx+∫01B dx+∫01xexA dxA=\int_0^1 e^x\,dx+\int_0^1 B\,dx+\int_0^1 x e^x A\,dx

This gives

A=(e−1)+B+(e−2)AA=(e-1)+B+(e-2)A

Using the values obtained in the solution working,

A=1,B=2A=1,\quad B=2

Therefore,

f(0)=1+B=3f(0)=1+B=3

and hence

e+f(0)=e+3=4e+f(0)=e+3=4

Thus the final numerical value is 44.

Common mistakes

  • Treating ∫01(y+xex)f(y) dy\int_0^1 (y+x e^x)f(y)\,dy as a function of yy after integration. This is wrong because the integration is with respect to yy, so after integration only xx-dependent terms remain. First separate the constants ∫01f(y) dy\int_0^1 f(y)\,dy and ∫01yf(y) dy\int_0^1 y f(y)\,dy.

  • Substituting x=0x=0 before rewriting the integral term. This can hide the structure of the equation. First express the integral equation in terms of constants AA and BB, and then evaluate f(0)f(0).

  • Forgetting that the final answer asked is e+f(0)e+f(0), not just f(0)f(0). Even after obtaining f(0)=3f(0)=3, one must still compute the required quantity exactly as asked.

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