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

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(A) The area $A$ of the region bounded by the $y$-axis $(x=0)$,$y=\cos x$,and $y=\sin x$ for $0 \leq x \leq \frac{\pi}{4}$ is given by the integral of

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$(\sqrt{2}-1)$ sq. units

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(A) The area $A$ of the region bounded by the $y$-axis $(x=0)$,$y=\cos x$,and $y=\sin x$ for $0 \leq x \leq \frac{\pi}{4}$ is given by the integral of the upper curve minus the lower curve.In the interval $[0, \frac{\pi}{4}]$,$\cos x \geq \sin x$.Therefore,the area $A$ is:$A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx$$A = [\sin x - (-\cos x)]_{0}^{\frac{\pi}{4}}$$A = [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$$A = (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)$$A = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$$A = \frac{2}{\sqrt{2}} - 1$$A = \sqrt{2} - 1$ sq. units.

The area of the region bounded by the $y$-axis,$y=\cos x$,and $y=\sin x$,when $0 \leq x \leq \frac{\pi}{4}$,is

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