Let the area of the region bounded by the curve y= x, x, lines x=0, x=(3 )/(2), and the x-…

Question

Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$, lines $x=0, x=\frac{3 \pi}{2}$, and the $x$-axis be A . Then, $\mathrm{A}+\mathrm{A}^2$ is equal to $\_\_\_\_$

Answer: 12

Accepted Answer

$\begin{gathered} f(x)=max(\sin x,\cos x) \ =\begin{gathered} \cos x,x\in (0,\frac{\pi }{4}) \ \sin x,x\in (\frac{\pi }{4},\frac{5\pi }{4}) \ \cos x,x\in (\frac{5\pi }{4},\frac{3\pi }{2}) \end{gathered} \end{gathered}$

$\begin{gathered} \Rightarrow {\int }_{\frac{3\pi }{4}}^{0}|f(x)|dx={\int }_{\frac{\pi }{4}}^{0}(\cos x)dx+{\int }_{\frac{5\pi }{4}}^{\frac{\pi }{4}}(\sin x)dx+{\int }_{\frac{3\pi }{2}}^{\frac{5\pi }{4}}(\cos x)dx \ =3\Rightarrow A=3 \ A+{A}^{2}=12 \end{gathered}$

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