Integral Calculus – Area Under Curves – JEE Main 28 Jan 2025 Shift 2

Solution:

The equations of the curves are:
1) $x = \frac{1}{1+y^2}$
2) $x = \frac{y^2}{2}$

Find the intersection points by equating $x$:
$$\frac{1}{1+y^2} = \frac{y^2}{2}$$
$$2 = y^2(1+y^2)$$
$$y^4 + y^2 – 2 = 0$$
$$(y^2+2)(y^2-1) = 0$$
Since $y^2$ cannot be negative, $y^2 = 1 \Rightarrow y = \pm 1$.


At $y=0$, curve (1) gives $x=1$ and curve (2) gives $x=0$. Thus, $x = \frac{1}{1+y^2}$ is the right curve.

The area is given by the integral with respect to $y$ from $-1$ to $1$:
$$Area = \int_{-1}^{1} \left( x_{\text{right}} – x_{\text{left}} \right) dy$$
$$Area = \int_{-1}^{1} \left( \frac{1}{1+y^2} – \frac{y^2}{2} \right) dy$$

Since both functions are even, we can write:
$$Area = 2 \int_{0}^{1} \left( \frac{1}{1+y^2} – \frac{y^2}{2} \right) dy$$
$$Area = 2 \left[ \tan^{-1}y – \frac{y^3}{6} \right]_{0}^{1}$$
$$Area = 2 \left[ \left( \tan^{-1}(1) – \frac{1}{6} \right) – (0 – 0) \right]$$
$$Area = 2 \left[ \frac{\pi}{4} – \frac{1}{6} \right]$$
$$Area = \frac{\pi}{2} – \frac{1}{3}$$

Ans. (3)