Question

Let $f\left(x\right)=max\{sinx,cosx,\frac{1}{2}\}$ then determine the area of region bounded by curves y=f(x),x-axis,y-axis and $x=2\pi$

Answer 1

Required area $={\int }_0^{\pi /4}\cos xdx+{\int }_{\pi /4}^{5\pi /6}\sin xdx+{\int }_{5\pi /5}^{5\pi /3}\frac{1}{2}dx+{\int }_{5\pi /3}^{2\pi }\cos xdx$

Refer image.

`$=\left[\sin x\right]{|}_0^{\pi /4}-\left[\cos x\right]{|}_{\pi /4}^{5\pi /4}+\frac{1}{2}\left[x\right]{|}_{5\pi /6}^{5\pi /3}+\left[\sin x\right]{|}_{5\pi /3}^{2\pi }$

$=\sin \frac{\pi }{4}-\sin 0-\cos \frac{5\pi }{4}+\cos \frac{\pi }{4}+\frac{5\pi }{4}+\frac{5\pi }{6}-\frac{5\pi }{12}+\sin 2\pi -\sin \frac{5\pi }{3}$

$=\frac{1}{\sqrt{2}}-0+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{5\pi }{12}+0+\frac{\sqrt{3}}{2}$

$=\frac{3}{\sqrt{2}}+\frac{5\pi }{12}+\frac{\sqrt{3}}{2}$