Integral Calculus – Definite Integration – JEE Main 02 April 2025 Shift 2

Solution:

Given curve: $x^2 = 2y$
Given line: $y = 2x + 6$

Substitute the value of $y$ from the line equation into the curve equation:
$$x^2 = 2(2x + 6)$$
$$x^2 = 4x + 12$$
$$x^2 – 4x – 12 = 0$$
$$(x – 6)(x + 2) = 0$$

$$x = 6 \quad \text{or} \quad x = -2$$

For $x = 6$, $y = 2(6) + 6 = 18 \Rightarrow (6, 18)$ is in the first quadrant.
For $x = -2$, $y = 2(-2) + 6 = 2 \Rightarrow (-2, 2)$ is in the second quadrant.

Since $(a, b)$ lies in the second quadrant, we have $(a, b) = (-2, 2)$.
Thus, $a = -2$ and $b = 2$.

Now, evaluate the integral:
$$I = \int_{-2}^{2} \frac{9x^2}{1+5^x} dx \quad \dots (i)$$

Using the property $\int_{-A}^{A} f(x) dx = \int_{-A}^{A} f(-x) dx$:
$$I = \int_{-2}^{2} \frac{9(-x)^2}{1+5^{-x}} dx$$
$$I = \int_{-2}^{2} \frac{9x^2}{1 + \frac{1}{5^x}} dx$$
$$I = \int_{-2}^{2} \frac{9x^2 \cdot 5^x}{5^x + 1} dx \quad \dots (ii)$$

Adding equations $(i)$ and $(ii)$:
$$2I = \int_{-2}^{2} \left( \frac{9x^2}{1+5^x} + \frac{9x^2 \cdot 5^x}{1+5^x} \right) dx$$
$$2I = \int_{-2}^{2} \frac{9x^2(1+5^x)}{1+5^x} dx$$
$$2I = \int_{-2}^{2} 9x^2 dx$$

Since $9x^2$ is an even function:
$$2I = 2 \int_{0}^{2} 9x^2 dx$$
$$I = 9 \left[ \frac{x^3}{3} \right]_0^2$$
$$I = 3 [x^3]_0^2$$
$$I = 3 (8 – 0)$$
$$I = 24$$

Ans. (1)