The area enclosed between the curves y = sin | vedclass

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(B) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$.For $0 \le x \le \frac{\pi}{4}$,the area is bounded by $y = \sin x$ and

Vedclass · Accepted Answer

$2 - \sqrt{2}$

Vedclass · Answer

(B) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$.For $0 \le x \le \frac{\pi}{4}$,the area is bounded by $y = \sin x$ and the $x$-axis.For $\frac{\pi}{4} \le x \le \frac{\pi}{2}$,the area is bounded by $y = \cos x$ and the $x$-axis.The total area $A$ is given by:$A = \int_{0}^{\frac{\pi}{4}} \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx$$A = [-\cos x]_{0}^{\frac{\pi}{4}} + [\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$A = \left(-\cos \frac{\pi}{4} - (-\cos 0)\right) + \left(\sin \frac{\pi}{2} - \sin \frac{\pi}{4}\right)$$A = \left(-\frac{1}{\sqrt{2}} + 1\right) + \left(1 - \frac{1}{\sqrt{2}}\right)$$A = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$

The area enclosed between the curves $y = \sin x$,$y = \cos x$ and the $x$-axis for $0 \le x \le \frac{\pi}{2}$ is:

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