The area of the region bounded by the y-axis, | vedclass

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Question

(C) The region is bounded by the $y$-axis $(x=0)$,$y = \cos x$,and $y = \sin x$. These curves intersect when $\cos x = \sin x$,which occurs at $x = \f

Vedclass · Accepted Answer

$ \sqrt{2} - 1 $ Sq.units

Vedclass · Answer

(C) The region is bounded by the $y$-axis $(x=0)$,$y = \cos x$,and $y = \sin x$. These curves intersect when $\cos x = \sin x$,which occurs at $x = \frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.In the interval $[0, \frac{\pi}{4}]$,$\cos x \geq \sin x$.Therefore,the required area is given by:$	ext{Area} = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx$$= [\sin x - (-\cos x)]_{0}^{\frac{\pi}{4}}$$= [\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}$

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

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