JEE Main 2025 Jan 28 Shift 2, Mathematics Q9: Area as Definite Integral

JEE Main2025Jan 28, Shift 2Mathematics

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is :

A
$2 (\frac{π}{2} - \frac{1}{3})$
B
$\frac{π}{4} - \frac{1}{3}$
C
$\frac{π}{2} - \frac{1}{3}$
D
$\frac{1}{2} (\frac{π}{2} - \frac{1}{3})$

Solution

Intersection: substituting $y^{2} = 2 x$ into $x (1 + y^{2}) = 1$ gives $x (1 + 2 x) = 1 \Rightarrow 2 x^{2} + x - 1 = 0$ , so $x = 1/2$ (the negative root is rejected since $y^{2} = 2 x \geq 0$ ). Then $y = \pm 1$ .

Both curves are symmetric in $y$ . Express each as $x = h (y)$ : the first gives $x = \frac{1}{1 + y ^{2}}$ and the second $x = \frac{y ^{2}}{2}$ . For $∣ y ∣ \leq 1$ , $\frac{1}{1 + y ^{2}} \geq \frac{y ^{2}}{2}$ .

Area $= \int_{- 1}^{1} (\frac{1}{1 + y ^{2}} - \frac{y ^{2}}{2}) d y = [tan^{- 1} y - \frac{y ^{3}}{6}]_{- 1}^{1} = 2 (\frac{π}{4} - \frac{1}{6}) = \frac{π}{2} - \frac{1}{3}$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q9: Area as Definite Integral

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