The area of the region bounded by the curve $y=\sin x$ between $x=-\pi$ and $x=\frac{3 \pi}{2}$ is
The area of the region bounded by the curve $y=\sin x$ between $x=-\pi$ and
$x=\frac{3 \pi}{2}$ is
2 (unit) $^{2}$
5 (unit) $^{2}$
3 (unit) $^{2}$
1 (unit) $^{2}$
Solution
Required area is shaded :
$\begin{aligned}
A &=2 \int_{0}^{\pi} \sin x d x+\int_{\pi}^{3 \pi / 2} \sin x d x \\
&=2\left[-\left.\cos x\right|_{0} ^{\pi}+[-\cos x]_{\pi}^{\frac{3 \pi}{2}}\right.
\end{aligned}$
$=|2[-\cos \pi+\cos 0]|+\left[-\cos \left(\frac{3 \pi}{2}\right)+\cos \pi\right]$
$=2[-(-1)+1]+[0+(-1)]$
$=2(2)-(1)=5(\text { unit })^{2}$
