MCQ
The area bounded by $y=\sin ^2 x, x=\frac{\pi}{2}$ and $x=\pi$ is
- A$\frac{\pi}{2}$
- ✓$\frac{\pi}{4}$
- C$\frac{\pi}{8}$
- D$\frac{\pi}{16}$
✓
Answer
Correct option: B.
$\frac{\pi}{4}$
(B)
Required area $=\int_{\frac{\pi}{2}}^\pi \sin ^2 x d x=\int_{\frac{\pi}{2}}^\pi \frac{1-\cos 2 x}{2} d x$
$\begin{array}{l}=\left[\frac{1}{2} x-\frac{\sin 2 x}{4}\right]_{\frac{\pi}{2}}^\pi \\ =\frac{\pi}{2}-\frac{\pi}{4} \\ =\frac{\pi}{4} \text { sq. units }\end{array}$
Required area $=\int_{\frac{\pi}{2}}^\pi \sin ^2 x d x=\int_{\frac{\pi}{2}}^\pi \frac{1-\cos 2 x}{2} d x$
$\begin{array}{l}=\left[\frac{1}{2} x-\frac{\sin 2 x}{4}\right]_{\frac{\pi}{2}}^\pi \\ =\frac{\pi}{2}-\frac{\pi}{4} \\ =\frac{\pi}{4} \text { sq. units }\end{array}$
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