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The area of the region bounded by the graph of $y=\sin x$ and $y=\cos x $ between $x=0$ and $x=\large\frac{\pi}{4}$ is

\[\begin{array}{1 1}(1)\sqrt{2}+1&(2)\sqrt{2}-1\\(3)2\sqrt{2}-2&(4)2\sqrt{2}+2\end{array}\]

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Area $A= \int \limits _0^{\pi/4} (y_1-y_2)$
$\qquad= \int \limits _0^{\pi/4} (\cos x - \sin x )dx$
$\qquad= [ \sin x -(- \cos x ) ] _0^{\pi/4}$
$\qquad= (\sin \large\frac{\pi}{4} -$$ \sin 0 ) + (\cos \large\frac{\pi}{4} $$- \cos 0)$
$\qquad= (\large\frac{1}{\sqrt 2} -0)+ (\frac{1}{\sqrt 2} -1)$
$\qquad = \large\frac{2}{\sqrt 2}$$ -1$
$\qquad= \sqrt 2-1$
Hence 2 is the correct answer.
answered May 21, 2014 by meena.p

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