JEE Main 2025 Apr 2 Shift 2, Mathematics Q7: Some Properties of Definite Integrals

JEE Main2025Apr 2, Shift 2Mathematics

Let $(a, b)$ be the point of intersection of the curve $x^{2} = 2 y$ and the straight line $y - 2 x - 6 = 0$ in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x ^{2}}{1 + 5 ^{x}} d x$ is equal to :

A
$24$
B
$27$
C
$18$
D
$21$

Solution

Solve $x^{2} = 2 (2 x + 6) \Rightarrow x^{2} - 4 x - 12 = 0 \Rightarrow (x - 6) (x + 2) = 0$ , giving intersection points $(6, 18)$ and $(- 2, 2)$ . The second-quadrant point is $(- 2, 2)$ , so $a = - 2$ and $b = 2$ .

Use the symmetry trick. Let $I = \int_{- 2}^{2} \frac{9 x ^{2}}{1 + 5 ^{x}} d x$ . Substituting $x \to - x$ gives $I = \int_{- 2}^{2} \frac{9 x ^{2} \cdot 5 ^{x}}{1 + 5 ^{x}} d x$ .

Adding the two forms: $2 I = \int_{- 2}^{2} 9 x^{2} d x = 18 [\frac{x ^{3}}{3}]_{0}^{2} = 18 \cdot \frac{8}{3} = 48$ .

$∴ I = 24$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q7: Some Properties of Definite Integrals

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