8. Application and integration — Maths STD 12 Science — Question

We have, $y=x^2$ and $y=1-x^2$

Intersection point of $y=x^2$ and $y=1-x^2$ is $A\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)$ and $C\left(\frac{-1}{\sqrt{2}}, \frac{1}{2}\right)$

Area of shaded region

$=2 \int \limits_0^{1 / \sqrt{2}}\left[\left(1-x^2\right)-\left(x^2\right)\right] d x$

$=2 \int \limits_0^{1 / \sqrt{2}}\left(1-2 x^2\right) d x$

$=2\left[x-\frac{2 x^3}{3}\right]_0^{1 / \sqrt{2}}$

$=2\left[\frac{1}{\sqrt{2}}-\frac{1}{3 \sqrt{2}}\right]$

$=\frac{2}{\sqrt{2}}\left[\frac{3-1}{3}\right]=\frac{2 \sqrt{2}}{3}$ sq units