The area enclosed between the curves $y=x^2$ and $x=y^2$ is |
1 $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{4}$ |
$\frac{1}{3}$ |
So $y=x^2$, and $x=y^2$ finding area along x-axis uppor cueve $y^2=x$ $\Rightarrow y=\sqrt{x}$ lower corve $y=x^2$ intersecting points $x^2=\sqrt{x}$ So x = 0, 1 y = 0, 1 So area $=\int\limits_0^1 \sqrt{x}-x^2 d x$ area $=\int\limits_0^1 \sqrt{x}-x^2 d x= \left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{2+1}}{2+1}\right]_0^1$ $=\left[\frac{2}{3} x^{3 / 2}-\frac{x^3}{3}\right]_0^1$ $=\frac{2}{3}-\frac{1}{3}$ $=\frac{1}{3}$ |