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(B) The area $A$ is bounded by $y=\max\{\sin x, \cos x\}$ from $x=0$ to $x=\frac{3\pi}{2}$ and the x-axis. We split the integral based on the interval

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$12$

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(B) The area $A$ is bounded by $y=\max\{\sin x, \cos x\}$ from $x=0$ to $x=\frac{3\pi}{2}$ and the x-axis. We split the integral based on the intervals where $\sin x$ or $\cos x$ is greater: $A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} |\sin x| \, dx + \int_{5\pi/4}^{3\pi/2} |\cos x| \, dx$ Since the area is bounded by the x-axis, we take the absolute value of the functions where they are negative. $A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} (-\sin x) \, dx + \int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx$ $A = [\sin x]_{0}^{\pi/4} + [-\cos x]_{\pi/4}^{\pi} + [\cos x]_{\pi}^{5\pi/4} + [-\sin x]_{5\pi/4}^{3\pi/2}$ $A = (\frac{1}{\sqrt{2}} - 0) + (-(-1) - (-\frac{1}{\sqrt{2}})) + (-\frac{1}{\sqrt{2}} - (-1)) + (-(-1) - (-\frac{1}{\sqrt{2}}))$ $A = \frac{1}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} = 3$ Thus, $A+A^2 = 3 + 3^2 = 3 + 9 = 12$.

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, $A+A^{2}$ is equal to:

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