The area bounded by the y-axis,y=cos x and y= | vedclass

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Question

(C) The given curves are $y=\cos x$ and $y=\sin x$. They intersect at $x = \frac{\pi}{4}$,where $y = \frac{1}{\sqrt{2}}$.The required area is bounded

Vedclass · Accepted Answer

$\sqrt{2}-1$

Vedclass · Answer

(C) The given curves are $y=\cos x$ and $y=\sin x$. They intersect at $x = \frac{\pi}{4}$,where $y = \frac{1}{\sqrt{2}}$.The required area is bounded by the $y-$axis $(x=0)$,$y=\cos x$,and $y=\sin x$ for $0 \leq x \leq \frac{\pi}{4}$.Area $= \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx$$= [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$$= (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)$$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$$= \frac{2}{\sqrt{2}} - 1$$= \sqrt{2} - 1 	ext{ sq. units}$.Therefore,the correct answer is option $C$.

The area bounded by the $y-$axis,$y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$ is

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