Area of Bounded Region with Absolute Values

To find the area of the region bounded by $\mathcal{R} = \{(x, y) : y \le \pi - |x|, y \le |x\sin x|, y \ge 0\}$

we analyze the symmetry and the boundaries of the region.

1. Symmetry of the Region

Observe that replacing $x$ with $-x$ leaves the inequalities unchanged: $\pi - |-x| = \pi - |x|$ $|-x \sin(-x)| = |x \sin x|$

Since the region is symmetric about the $y$-axis ($x = 0$), the total area $A$ is twice the area $A_1$ of the region in the first quadrant ($x \ge 0$): $A = 2 A_1$

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2. Region in the First Quadrant ($x \ge 0$)

For $x \ge 0$ and $y \ge 0$:

1. $y \le \pi - x \implies \pi - x \ge 0 \implies 0 \le x \le \pi$.
2. For $x \in [0, \pi]$, we have $\sin x \ge 0$, which means $|x \sin x| = x \sin x$.

Thus, for $x \in [0, \pi]$, the upper boundary for $y$ at a given $x$ is: $y(x) = \min(x \sin x, \pi - x)$

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3. Finding the Point of Intersection

Let us compare $f(x) = x \sin x$ and $g(x) = \pi - x$ in the interval $[0, \pi]$:

• At $x = \frac{\pi}{2}$: $f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$ $g\left(\frac{\pi}{2}\right) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$ Hence, $x = \frac{\pi}{2}$ is an intersection point.

• For $x \in \left[0, \frac{\pi}{2}\right]$: Since $\sin x \le 1$ and $x \le \pi - x$, we have: $x \sin x \le x \le \pi - x \implies \min(x \sin x, \pi - x) = x \sin x$

• For $x \in \left[\frac{\pi}{2}, \pi\right]$: Consider the function $k(x) = x \sin x - (\pi - x)$.

  • $k\left(\frac{\pi}{2}\right) = 0$ and $k(\pi) = 0$.
  • $k''(x) = 2\cos x - x\sin x < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$ because $\cos x < 0$ and $\sin x > 0$.

  Since $k(x)$ is strictly concave on $\left[\frac{\pi}{2}, \pi\right]$ and vanishes at the endpoints, $k(x) \ge 0$ for all $x \in \left[\frac{\pi}{2}, \pi\right]$. Therefore, $x \sin x \ge \pi - x$ on this interval, giving: $\min(x \sin x, \pi - x) = \pi - x$

Thus, the function $y(x)$ is defined piecewise as: $y(x) = \begin{cases} x \sin x, & 0 \le x \le \frac{\pi}{2} \\ \pi - x, & \frac{\pi}{2} \le x \le \pi \end{cases}$

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4. Integration for Area $A_1$

The area in the first quadrant is given by: $A_1 = \int_0^{\pi/2} x \sin x \, dx + \int_{\pi/2}^\pi (\pi - x) \, dx$

Evaluating the First Integral:

Using integration by parts ($\int u \, dv = uv - \int v \, du$ with $u = x$ and $dv = \sin x \, dx$): $\int_0^{\pi/2} x \sin x \, dx = \left[ -x \cos x \right]_0^{\pi/2} + \int_0^{\pi/2} \cos x \, dx$ $= (0 - 0) + \left[ \sin x \right]_0^{\pi/2} = 1 - 0 = 1$

Evaluating the Second Integral:

$\int_{\pi/2}^\pi (\pi - x) \, dx = \left[ \pi x - \frac{x^2}{2} \right]_{\pi/2}^\pi$ $= \left( \pi^2 - \frac{\pi^2}{2} \right) - \left( \frac{\pi^2}{2} - \frac{\pi^2}{8} \right) = \frac{\pi^2}{2} - \frac{3\pi^2}{8} = \frac{\pi^2}{8}$

Adding both parts gives: $A_1 = 1 + \frac{\pi^2}{8}$

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5. Total Bounded Area

The total area $A$ of the region is: $A = 2 A_1 = 2 \left( 1 + \frac{\pi^2}{8} \right) = 2 + \frac{\pi^2}{4}$

Thus, the correct option is B.