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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Definite Integration

If f(x) satisfies the relation f(x)=e^ x +∫_ 0 ^ 1 (y+x e^ x ) f(y)...

If f(x)f(x) satisfies the relation f(x)=ex+∫01(y+xex)f(y)dyf(x)=e^{x}+\int_{0}^{1}\left(y+x e^{x}\right) f(y) d y, then e+f(0)e+f(0) is equal to ____\_\_\_\_.

Correct answer

0

Step-by-step solution

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

Splitting the integral:
f(x)=ex+∫01yf(y) dy+xex∫01f(y) dyf(x) = e^x + \displaystyle\int_0^1 yf(y)\, dy + xe^x \int_0^1 f(y)\, dy

Let A=∫01yf(y) dyA = \displaystyle\int_0^1 yf(y)\, dy and B=∫01f(y) dyB = \displaystyle\int_0^1 f(y)\, dy

⇒f(x)=(1+Bx)ex+A\Rightarrow f(x) = (1 + Bx)e^x + A

At x=0x = 0: f(0)=1+Af(0) = 1 + A

Now compute BB using the expression for f(y)f(y):
B=∫01[(1+By)ey+A]dyB = \displaystyle\int_0^1 \left[(1 + By)e^y + A\right] dy
B=∫01ey dy+B∫01yey dy+∫01A dyB = \displaystyle\int_0^1 e^y\, dy + B\int_0^1 ye^y\, dy + \int_0^1 A\, dy

Evaluating each integral:
∫01ey dy=e−1\displaystyle\int_0^1 e^y\, dy = e - 1
∫01yey dy=[yey]01−∫01ey dy=e−(e−1)=1\displaystyle\int_0^1 ye^y\, dy = [ye^y]_0^1 - \int_0^1 e^y\, dy = e - (e - 1) = 1
∫01A dy=A\displaystyle\int_0^1 A\, dy = A

⇒B=(e−1)+B(1)+A\Rightarrow B = (e - 1) + B(1) + A
⇒0=e−1+A\Rightarrow 0 = e - 1 + A
⇒A=1−e\Rightarrow A = 1 - e

f(0)=1+A=1+(1−e)=2−ef(0) = 1 + A = 1 + (1 - e) = 2 - e

e+f(0)=e+(2−e)=2e + f(0) = e + (2 - e) = 2

Answer: 22
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