The area bonded by the curve $y=\sin ^{2} x, x$ -axis and the lines $x=0$ and $x=\frac{\pi}{2}$ is
The area bonded by the curve $y=\sin ^{2} x, x$ -axis and the lines $x=0$ and $x=\frac{\pi}{2}$ is
1 sq. units
$\frac{\pi}{8}$ sq. units
$\frac{\pi}{4}$ sq. units
$\frac{\pi}{2}$ sq. units
Solution
Required area is shaded.
$\begin{aligned}
A &=\int_{0}^{\pi / 2} \sin ^{2} x d x \\
&=\int_{0}^{\pi / 2} \frac{1-\cos 2 x}{2} d x \\
&=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}-\frac{1}{4}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\
&=\frac{\pi}{4}-0=\frac{\pi}{4}
\end{aligned}$