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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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The area of the region bounded by the curve $y=sin x$ between the ordinates $x=0$ and $x=\large\frac{\pi}{2}$ and the x-axis 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 $\int_a^b ydx.$
  • $\int \sin xdx=-\cos x+c$
Step 1:
Given $y=\sin x$ bounded between $0$ and $\large\frac{\pi}{2}$.
Area of the region is the shaded portion shown in the fig.
$A=\int_0^{\Large\frac{\pi}{2}}$$\sin xdx.$
Step 2:
On integrating we get,
$A=[-\cos x]_0^{\Large\frac{\pi}{2}}$
Applying the limits we get,
$A=-[\cos\large\frac{\pi}{2}$$-\cos 0]$
But $\cos\large\frac{\pi}{2}$=0 and $\cos 0$=1
Therefore A=1 sq.units.
Hence D is the correct option.
answered May 7, 2013 by sreemathi.v
edited May 7, 2013 by sreemathi.v

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