Area Calculation for $y = \max\{\sin x, \cos x\}$

The problem asks for the area $A$ bounded by the curve $y = \max\{\sin x, \cos x\}$, the vertical lines $x = 0$ and $x = \frac{3\pi}{2}$, and the x-axis ($y=0$). This requires integrating the absolute value of the function, i.e., $A = \int_{0}^{3\pi/2} |\max\{\sin x, \cos x\}| \, dx$. We need to find where $\sin x = \cos x$ and where the function $\max\{\sin x, \cos x\}$ is positive or negative.

Determining Function Intervals and Values

First, find points where $\sin x = \cos x$ in the interval $[0, \frac{3\pi}{2}]$. This occurs when $\tan x = 1$, giving $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.

We analyze the function $y = \max\{\sin x, \cos x\}$ over the interval $[0, \frac{3\pi}{2}]$:

• Interval $[0, \frac{\pi}{4}]$: $\cos x \ge \sin x$. Since both are positive here, $y = \cos x$.
• Interval $[\frac{\pi}{4}, \frac{5\pi}{4}]$: $\sin x \ge \cos x$. $y = \sin x$. Note that $\sin x$ is positive in $[\frac{\pi}{4}, \pi]$ and negative in $(\pi, \frac{5\pi}{4}]$.
• Interval $[\frac{5\pi}{4}, \frac{3\pi}{2}]$: $\cos x \ge \sin x$. Since both are negative here, $y = \cos x$.

The area calculation requires integrating $|\max\{\sin x, \cos x\}|$. We split the integral based on the intervals and the sign of the function:

Step 1: Calculate Area in Each Sub-interval

• Region 1: $x \in [0, \frac{\pi}{4}]$ Here, $y = \cos x$, and $\cos x \ge 0$. Area$_1 = \int_{0}^{\pi/4} \cos x \, dx = [\sin x]_{0}^{\pi/4} = \sin(\frac{\pi}{4}) - \sin(0) = \frac{\sqrt{2}}{2}$.
• Region 2: $x \in [\frac{\pi}{4}, \pi]$ Here, $y = \sin x$, and $\sin x \ge 0$. Area$_2 = \int_{\pi/4}^{\pi} \sin x \, dx = [-\cos x]_{\pi/4}^{\pi} = -\cos(\pi) - (-\cos(\frac{\pi}{4})) = -(-1) - (-\frac{\sqrt{2}}{2}) = 1 + \frac{\sqrt{2}}{2}$.
• Region 3: $x \in [\pi, \frac{5\pi}{4}]$ Here, $y = \sin x$, and $\sin x < 0$. We need the area, so we integrate $|\sin x| = -\sin x$. Area$_3 = \int_{\pi}^{5\pi/4} |\sin x| \, dx = \int_{\pi}^{5\pi/4} -\sin x \, dx = [\cos x]_{\pi}^{5\pi/4} = \cos(\frac{5\pi}{4}) - \cos(\pi) = -\frac{\sqrt{2}}{2} - (-1) = 1 - \frac{\sqrt{2}}{2}$.
• Region 4: $x \in [\frac{5\pi}{4}, \frac{3\pi}{2}]$ Here, $y = \cos x$, and $\cos x < 0$. We need the area, so we integrate $|\cos x| = -\cos x$. Area$_4 = \int_{5\pi/4}^{3\pi/2} |\cos x| \, dx = \int_{5\pi/4}^{3\pi/2} -\cos x \, dx = [-\sin x]_{5\pi/4}^{3\pi/2} = -\sin(\frac{3\pi}{2}) - (-\sin(\frac{5\pi}{4})) = -(-1) - (-(-\frac{\sqrt{2}}{2})) = 1 - \frac{\sqrt{2}}{2}$.

Step 2: Calculate Total Area A

The total area $A$ is the sum of the areas from the four regions:

$A = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 + \text{Area}_4$ $A = \frac{\sqrt{2}}{2} + (1 + \frac{\sqrt{2}}{2}) + (1 - \frac{\sqrt{2}}{2}) + (1 - \frac{\sqrt{2}}{2})$ $A = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) + (1 + 1 + 1)$ $A = 0 + 3 = 3$

Step 3: Calculate $A + A^2$

Given $A = 3$, we calculate $A + A^2$:

$A + A^2 = 3 + 3^2 = 3 + 9 = 12$

The value of $A + A^2$ is 12.