Integral Calculus – Area Under Curves – JEE Main 22 Jan 2026 Shift 2

Solution:

The region is bounded by the parabola $y^2 = 4x$ and the ellipse $4x^2 + y^2 = 8$.
First, we find the points of intersection by solving the equations simultaneously:
$$ 4x^2 + 4x = 8 \Rightarrow x^2 + x – 2 = 0 $$
$$ (x+2)(x-1) = 0 $$
Since $y^2 = 4x$, we must have $x \ge 0$, so $x = 1$.
The intersection points are $(1, 2)$ and $(1, -2)$.

The region is symmetric about the x-axis. The total area is twice the area in the first quadrant.
The area is bounded by the parabola from $x=0$ to $x=1$ and by the ellipse from $x=1$ to $x=\sqrt{2}$ (since $x$-intercept of ellipse is $\sqrt{2}$).
$$ A = 2 \left( \int_{0}^{1} y_{parabola} \, dx + \int_{1}^{\sqrt{2}} y_{ellipse} \, dx \right) $$
$$ A = 2 \left( \int_{0}^{1} 2\sqrt{x} \, dx + \int_{1}^{\sqrt{2}} \sqrt{8-4x^2} \, dx \right) $$
$$ A = 4 \int_{0}^{1} \sqrt{x} \, dx + 4 \int_{1}^{\sqrt{2}} \sqrt{2-x^2} \, dx $$

Calculating the first integral:
$$ \int_{0}^{1} \sqrt{x} \, dx = \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1} = \frac{2}{3} $$

Calculating the second integral using $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$:
$$ \int_{1}^{\sqrt{2}} \sqrt{2-x^2} \, dx = \left[ \frac{x}{2}\sqrt{2-x^2} + \sin^{-1}\left(\frac{x}{\sqrt{2}}\right) \right]_{1}^{\sqrt{2}} $$
Substituting limits:
$$ = \left( 0 + \frac{\pi}{2} \right) – \left( \frac{1}{2} + \frac{\pi}{4} \right) = \frac{\pi}{4} – \frac{1}{2} $$

Substituting these values back into the area formula:
$$ A = 4\left(\frac{2}{3}\right) + 4\left(\frac{\pi}{4} – \frac{1}{2}\right) $$
$$ A = \frac{8}{3} + \pi – 2 $$
$$ A = \pi + \frac{2}{3} $$

Ans. (2)