The area of the region bounded by the curves | vedclass

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(A) The required area is bounded by $y = \cos x$ (upper curve) and $y = \sin x$ (lower curve) from $x = 0$ to their point of intersection.To find the

Vedclass · Accepted Answer

$\sqrt{2} - 1$

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(A) The required area is bounded by $y = \cos x$ (upper curve) and $y = \sin x$ (lower curve) from $x = 0$ to their point of intersection.To find the point of intersection,set $\sin x = \cos x$,which gives $	an x = 1$,so $x = \frac{\pi}{4}$.The required area $A$ is given by the integral:$A = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx$$A = [\sin x + \cos x]_{0}^{\pi/4}$$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 = \sqrt{2} - 1$.

The area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and $x = 0$ is

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