Find the area bounded by the curve $y=2 \cos x$ and the $X$-axis from $x=0$ to $x=2 \pi$.
$$\begin{aligned} \text { Required area of shaded region } & =\int_0^{2 \pi} 2 \cos x d x \\ & =\int_0^{\pi / 2} 2 \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} 2 \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} 2 \cos x d x \end{aligned}$$
$$\begin{aligned} & =2[\sin x]_0^{\pi / 2}+\left|2(\sin x)_{\pi / 2}^{3 \pi / 2}\right|+2[\sin x]_{3 \pi / 2}^{2 \pi} \\ & =2+4+2=8 \text { sq units } \end{aligned}$$
Draw a rough sketch of the given curve $y=1+|x+1|, x=-3, x=3$, $y=0$ and find the area of the region bounded by them, using integration.
We have, $y=1+|x+1|, x=-3, x=3$ and $y=0$
$\because\quad y=\left\{\begin{array}{cc}-x, & \text { if } x<-1 \\ x+2, & \text { if } x \geq-1\end{array}\right.$
$$\begin{aligned} \therefore \quad \text { Area of shaded region, } A & =\int_{-3}^{-1}-x d x+\int_{-1}^3(x+2) d x \\ & =-\left[\frac{x^2}{2}\right]_{-3}^{-1}+\left[\frac{x^2}{2}+2 x\right]_{-1}^3 \\ & =-\left[\frac{1}{2}-\frac{9}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\ & =-[-4]+[8+4] \\ & =12+4=16 \text { sq units } \end{aligned}$$
The area of the region bounded by the $Y$-axis $y=\cos x$ and $y=\sin x$, where $0 \leq x \leq \frac{\pi}{2}$, is
The area of the region bounded by the curve $x^2=4 y$ and the straight line $x=4 y-2$ is
The area of the region bounded by the curve $y=\sqrt{16-x^2}$ and $X$-axis is