Question

Find the area bounded by f(x) = maximum { sin x, cos x}, x=0, x = $2\pi$ and the x-axis.

Answer 1

$f\left(x\right)=\{\cos x:0\le x\le \pi /4\sin x:\pi /4\le x\le 5\pi /4\cos x:5\pi /4\le x\le 2\pi \}A={\int }_0^{\pi /4}\cos xdx+{\int }_{\pi /4}^{5\pi /4}\sin xdx+{\int }_{5\pi /4}^{2\pi }\cos xdxA={\left[\sin x\right]}_0^{\pi /4}+{[-\cos x]}_{\pi /4}^{5\pi /4}+{\left[\sin x\right]}_{5\pi /4}^{2\pi }A=(\sin \pi /4-\sin 0)+(-\cos 5\pi /4+\cos \pi /4)+(\sin 5\pi /4-\sin 2\pi )A=(\frac{1}{\sqrt{2}}-0)+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})+(0+\frac{1}{\sqrt{2}})A=4\left(\frac{1}{\sqrt{2}}\right)squnit$