Question

Find the area bounded by the curves $y=\sin x$ and $y=\cos x$ between any two consecutive points of intersection.

Answer 1

Area bounded by the curve $y=\sin x$ and $y=\cos x$ between any two consecutive points of intersection-

$\because \sin x=\cos x$ at $x=\frac{\pi }{4},\frac{5\pi }{4}$

Therefore the area in between, within two consecutive points of intersection $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$-

Area $={\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}(\sin x-\cos x)dx$

$\Rightarrow$ Area $={\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}\sin xdx-{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}\cos xdx={[-\cos x]}_{\frac{\pi }{4}}^{\frac{5\pi }{4}}-{\left[\sin x\right]}_{\frac{\pi }{4}}^{\frac{5\pi }{4}}=(\cos \frac{\pi }{4}-\cos \frac{5\pi }{4})-(\sin \frac{5\pi }{4}-\sin \frac{\pi }{4})=\frac{1}{\sqrt{2}}-(-\frac{1}{\sqrt{2}})-(-\frac{1}{\sqrt{2}})+\frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$

Hence the area bounded by the curve $y=\sin x$ and $y=\cos x$ between any two consecutive points of intersection is $2\sqrt{2}$.