The area bounded by the curves y = cos x and | vedclass

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(C) The area $A$ bounded by the curves $y = \cos x$ and $y = \sin x$ between $x = 0$ and $x = \frac{\pi}{4}$ is given by the integral of the differenc

Vedclass · Accepted Answer

$\sqrt{2} - 1$

Vedclass · Answer

(C) The area $A$ bounded by the curves $y = \cos x$ and $y = \sin x$ between $x = 0$ and $x = \frac{\pi}{4}$ is given by the integral of the difference between the upper curve and the lower curve.In the interval $[0, \frac{\pi}{4}]$,$\cos x \ge \sin x$.Therefore,the area is:$A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx$$A = [\sin x]_0^{\frac{\pi}{4}} - [-\cos x]_0^{\frac{\pi}{4}}$$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$.

The area bounded by the curves $y = \cos x$ and $y = \sin x$ and the ordinates $x = 0$ and $x = \frac{\pi}{4}$ is:

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