Mathematics

Find the area of one of the curvilinear triangles formed by $$y=\sin{x},y=\cos{x}$$ and $$x$$ axis


SOLUTION
The point of intersection of  two curves $$y=\sin{x}$$ and $$y=\cos{x}$$ is at $$\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)$$

Required Area$$=\displaystyle\int_{0}^{\tfrac{\pi}{4}}{\sin{x}dx}+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}{\cos{x}dx}$$

$$=\displaystyle\left[-\cos{x}\right]_{0}^{\frac{\pi}{4}}+\left[\sin{x}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$

$$=-\left[1-\dfrac{1}{\sqrt{2}}\right]+\left[\dfrac{1}{\sqrt{2}}-1\right]$$

$$=2\left[\dfrac{1}{\sqrt{2}}-1\right]$$

$$=\dfrac{2}{\sqrt{2}}-2$$

$$=\dfrac{2}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}-2$$

$$=\sqrt{2}-2$$sq.unit
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Subjective Hard Published on 17th 09, 2020
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