Mathematics - Area Under Curves Question with Solution | TestHub

$\begin{array}{rl}& y=\min \{\sin x,\cos x\}\\ & x-\text{ axis }x-\pi x=\pi \end{array}$

$\begin{array}{rl}& {\int }_0^{\pi /4}\sin x=(\cos x{)}_{\pi /4}^0=1-\frac{1}{\sqrt{2}}\\ & {\int }_{-\pi }^{-3\pi /4}(\sin x-\cos x)=(-\cos x-\sin x{)}_{-\pi }^{-3\pi /4}\\ & =(\cos x+\sin x{)}_{-3\pi /4}^{-\pi }\\ & =(-1+0)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\\ & =-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ & {\int }_{\pi /4}^{\pi /2}\cos xdx=(\sin x{)}_{\pi /4}^{\pi /2}=1-\frac{1}{\sqrt{2}}\\ & A=4\\ & A^2=16\end{array}$