Find the area bounded by the curve $y=2\cos x$ and the $x$-axis from $x=0$ to x=2$\pi$.

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int \sin xdx=-\cos x+c$
• $\int \cos xdx=\sin x+c.$
Step :1
Given $y=2\cos x$ and $x=0$ to $x=2\pi$
It is clear from the graph that the required area is the three regions which are shaded.
Clearly the curve ranges between $0$ to $\large\frac{\pi}{2}$ in the I region,$\large\frac{\pi}{2}$ to $\large\frac{3\pi}{2}$ in the second region which is in the negative side of $y$-axis,and $\large\frac{3\pi}{2}$ to $2\pi$ in the III region.
Hence the area of the shaded region is $A=\int_0^{\Large\frac{\pi}{2}}y_1dx+\int_{\Large\frac{\pi}{2}}^{\Large\frac{3\pi}{2}}(-y_2)dx+\int_{\Large\frac{3\pi}{2}}^{2\pi}y_3dx.$
$A=2\int_0^{\Large\frac{\pi}{2}}\cos xdx-2\int_{\Large\frac{\pi}{2}}^{\Large\frac{3\pi}{2}}\cos xdx+2\int_{\Large\frac{3\pi}{2}}^{2\pi}\cos xdx.$
Step 2:
On integrating we get,
$A=2\begin{bmatrix}\begin{bmatrix}\sin x\end{bmatrix}_0^{\Large\frac{\pi}{2}}-\begin{bmatrix}\sin x\end{bmatrix}_{\large\frac{\pi}{2}}^{\Large\frac{3\pi}{2}}+\begin{bmatrix}\sin x\end{bmatrix}_{\Large\frac{3\pi}{2}}^{2\pi}\end{bmatrix}$
Applying the limits we get,
$A=2\begin{bmatrix}(\sin\large\frac{\pi}{2}-\normalsize\sin 0)-(\sin\large\frac{3\pi}{2}-\sin\large\frac{\pi}{2})+\normalsize(\sin 2\pi-\sin\large\frac{3\pi}{2})\end{bmatrix}$
We know $\sin 0=\sin2\pi=0$ and $\sin\large\frac{\pi}{2}$$=1 and \sin\large\frac{3\pi}{2}$$=-1$
Hence $A=2[(1-0)-(-1-1)+(0-(-1))]$
$A=2[1+2+1]$
$\;\;\;=8$sq.units.
Hence the required area is $8$sq.units.
edited Dec 22, 2013