Integral Calculus – Definite Integration – JEE Main 24 Jan 2026 Shift 2

Solution:

The given equation is:
$$f(x) = e^x + \int_{0}^{1} (y + x e^x) f(y) dy$$

Separate the terms inside the integral. Note that the integration is with respect to $y$, so terms with $x$ behave as constants:
$$f(x) = e^x + \int_{0}^{1} y f(y) dy + \int_{0}^{1} x e^x f(y) dy$$

$$f(x) = e^x + \int_{0}^{1} y f(y) dy + x e^x \int_{0}^{1} f(y) dy$$

Since the limits of integration ($0$ to $1$) are constants, the definite integrals result in constant values. Let us define these constants as $A$ and $B$:
Let $A = \int_{0}^{1} y f(y) dy$
Let $B = \int_{0}^{1} f(y) dy$

Now substitute these constants back into the expression for $f(x)$:
$$f(x) = e^x + A + x e^x B$$
$$f(x) = (1 + Bx)e^x + A$$

To find the value of $f(0)$, we substitute $x=0$ into this new expression:
$$f(0) = (1 + B(0))e^0 + A$$
$$f(0) = 1 + A$$

So, we only need to find the value of $A$. We can use the definition of $B$ to find a relationship between $A$ and $B$.
$$B = \int_{0}^{1} f(y) dy$$
Substitute the function $f(y) = (1+By)e^y + A$:
$$B = \int_{0}^{1} \left[ (1+By)e^y + A \right] dy$$
$$B = \int_{0}^{1} e^y dy + B \int_{0}^{1} y e^y dy + \int_{0}^{1} A dy$$

Calculate the individual integrals:
1. $\int_{0}^{1} e^y dy = [e^y]_0^1 = e – 1$
2. $\int_{0}^{1} A dy = [Ay]_0^1 = A$
3. $\int_{0}^{1} y e^y dy$: Using integration by parts ($u=y, dv=e^y dy$):
$= [y e^y]_0^1 – \int_{0}^{1} e^y dy = (1\cdot e – 0) – (e – 1) = e – e + 1 = 1$

Now put these back into the equation for $B$:
$$B = (e – 1) + B(1) + A$$
$$B = e – 1 + B + A$$

Subtract $B$ from both sides:
$$0 = e – 1 + A$$
$$A = 1 – e$$

Now we can find $f(0)$ using the relation we found earlier ($f(0) = 1 + A$):
$$f(0) = 1 + (1 – e) = 2 – e$$

Finally, find $e + f(0)$:
$$e + f(0) = e + (2 – e) = 2$$

Ans. (2)