If $f(x)=max\{\sin x,\cos x,\frac{1}{2}\}$, then the area of the region bounded by the curves $y = f (x)$, x-axis, y-axis and $x=\frac{5π}{3}$, is

$\sqrt{2}+\frac{\sqrt{3}}{2}+\frac{5π}{12}$

We have,

$f(x)=max\{\sin x,\cos x,\frac{1}{2}\}=\left\{\begin{matrix}\cos x,&0 ≤x≤\frac{π}{4}\\\sin x,&\frac{π}{4}≤x≤\frac{5π}{6}\\\frac{1}{2},&\frac{5π}{6}≤x≤\frac{5π}{3}\end{matrix}\right.$

Let A be the required area. Then,

$A=\int\limits_{0}^{5π/3}f(x)dx$

$⇒A=\int\limits_{0}^{π/4}\cos xdx+\int\limits_{π/4}^{5π/6}\sin xdx+\int\limits_{5π/6}^{5π/3}\frac{1}{2}dx$

$⇒A=\left[\sin x\right]_{0}^{π/4}-\left[\cos x\right]_{π/4}^{5π/6}+\frac{1}{2}\left(\frac{5π}{3}-\frac{5π}{6}\right)$

$⇒A=\frac{1}{\sqrt{2}}-(-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}})+\frac{5π}{12}=\frac{5π}{12}+\frac{\sqrt{3}}{2}+\sqrt{2}$