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Q.
Area of the region bounded by the curve $y=\cos x$ between $x=0$ and $x=\pi$ is
Application of Integrals
Solution:
We have, $y=\cos x$ and lines $x=0, x=\pi$
Draw the rough sketch of the function $\cos x$.
$ \therefore $ Required area $ =\int\limits_0^{\frac{\pi}{2}} \cos x d x+\left|\int\limits_{\frac{\pi}{2}}^\pi \cos x d x\right| $
$ =[\sin x]_0^{\pi / 2}+\left|[\sin x]_{\pi / 2}^\pi\right| $
$ =\left[\sin \frac{\pi}{2}-\sin 0\right]+\left|\left[\sin \pi-\sin \frac{\pi}{2}\right]\right| $
$ =[1-0]+|0-1| $
$ =1+1 $
$ =2 $ sq units
