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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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The area of the region bounded by the curve $y=\cos x$ between $x=0$ and $x=\pi$ is \begin{array}{1 1}(A)\;2 sq.units & (B)\;4 sq.units\\(C)\;3 sq.units & (D)\;1sq.units \end{array}

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  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int\cos xdx=\sin x+c$
Step 1:
Given $y=\cos x$ bounded between $x=0$ and $x=\pi$
Required area is the shaded portion shown in the fig.
$A=\int_0^{\Large\frac{\pi}{2}}$$\cos xdx-\int_{\Large\frac{\pi}{2}}^\pi\cos xdx.$
Since the area of the curve between $\large\frac{\pi}{2}$ and $\pi$ is on the negative side.
Step 2:
On integrating we get,
$A=\begin{bmatrix}\sin x\end{bmatrix}_0^{\Large\frac{\pi}{2}}-\begin{bmatrix}\sin x\end{bmatrix}_{\Large\frac{\pi}{2}}^\pi$
On applying limits we get,
$A=[\sin\large\frac{\pi}{2}-$$\sin 0]-[\sin\pi-\sin\large\frac{\pi}{2}]$
But $\sin 0=\sin \pi=0$ and $\sin\large\frac{\pi}{2}$$=1$
Hence $A=[1-0]-[0-1]=2$ sq.units.
Hence A is the correct option.
answered May 7, 2013 by sreemathi.v

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