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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area bounded by the curve \(y = \sin\: x\) between \(x = 0\) and \(x = 2π\).

$\begin{array}{1 1} 4 \\ 2 \\ 6 \\ 8 \end{array}$

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The required area is
A=area of OAB+area of BCD
$\;\;=\int_0^\pi ydx+\int_\pi^{2\pi}-ydx$
The curve between the limits 0 to $\pi$ is on the positive side of y-axis
The curve between the limits $\pi$ to $2\pi$ is on the negative side of y-axis
Hence $A=\int_0^\pi\sin xdx+\int_\pi^{2\pi}\sin xdx$
$\;\;\;\;\;\;\;\;\;A=[-\cos x]_0^\pi+[-\cos x]_\pi^{2\pi}$
$\;\;\;\;\;\;\;\;\;A=2+2=4sq.units$
Hence the required area is 4 sq,units.
answered Dec 21, 2013 by yamini.v
 

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