The area bounded by the curves y-1=cos x,y=si | vedclass

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(D) The curves are $y = \cos x + 1$ and $y = \sin x$. We need to find the area bounded by these curves and the $X$-axis between $x=0$ and $x=\pi$.From

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$\frac{\pi}{2}$

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(D) The curves are $y = \cos x + 1$ and $y = \sin x$. We need to find the area bounded by these curves and the $X$-axis between $x=0$ and $x=\pi$.From the graph,the intersection point $A$ occurs where $\cos x + 1 = \sin x$. However,looking at the region bounded by the $X$-axis,the area is split at $x = \pi/2$.For $0 \le x \le \pi/2$,the region is bounded above by $y = \sin x$ and below by the $X$-axis.For $\pi/2 \le x \le \pi$,the region is bounded above by $y = \cos x + 1$ and below by the $X$-axis.Area $= \int_0^{\pi/2} \sin x \, dx + \int_{\pi/2}^{\pi} (\cos x + 1) \, dx$$= [-\cos x]_0^{\pi/2} + [\sin x + x]_{\pi/2}^{\pi}$$= (-\cos(\pi/2) - (-\cos 0)) + ((\sin \pi + \pi) - (\sin(\pi/2) + \pi/2))$$= (0 + 1) + (0 + \pi - 1 - \pi/2)$$= 1 + \pi - 1 - \pi/2 = \pi/2$.

The area bounded by the curves $y-1=\cos x$,$y=\sin x$ and the $X$-axis between $x=0$ and $x=\pi$ is

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