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# Find the area bounded by the curve y=sinx between $x=0$ and $x=2\pi$.

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## 1 Answer

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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 :$y=\sin x$ between $x=0$ and $x=2\pi$
The sketch of the curve $y=\sin x$ between $x=0$ and $x=2\pi$ is shown in the fig.
The shaded portion is the required region.
The required area A=area of $OAB$+area of $BA'B'$.
$\Rightarrow A=\int_0^{\pi}y dx+\int_{\pi}^{2\pi} -y dx.$
$\quad\quad=\int_0^{\pi}\sin xdx-\int_{\pi}^{2\pi}\sin xdx.$
Step 2:
On integrating we get,
$\quad\quad=[-\cos x]_0^{\pi}-[-\cos x]_{\pi}^{2\pi}$
On applying limits we get,
$\quad\quad=-[\cos \pi-\cos 0]+[\cos 2\pi-\cos \pi]$
But $\cos 0=\cos 2\pi=1$ and $\cos\pi=-1.$
$\quad\quad=-[-1-1]+[1-(-1)]=2+2=4$
Hence the required area is 4 sq.units.
answered Apr 29, 2013
edited Apr 29, 2013

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