Question

The area of the region \(\{(x, y): y \le \pi - |x|, y \le |x \sin x|, y \ge 0\}\) is:

• A. \(1 + \frac{\pi^2}{8}\)
• B. \(2 + \frac{\pi^2}{4}\)
• C. \(\frac{\pi^2}{8} - 1\)
• D. \(4 + \frac{\pi^2}{2}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The region is bounded by a symmetric "tent" shape $y = \pi - |x|$ and the wave $y = |x \sin x|$. Since both functions are even, the area is twice the area in the first quadrant ($x \ge 0$).

Step 2: Key Formula or Approach:
Area $= 2 \int_0^\pi \min(\pi - x, x \sin x) dx$. We need to find where $x \sin x$ intersects $\pi - x$. At $x = \pi/2$: $\pi/2 \sin(\pi/2) = \pi/2$ and $\pi - \pi/2 = \pi/2$. They intersect at $\pi/2$.

Step 3: Detailed Explanation:
1. From $0$ to $\pi/2$, $x \sin x \le \pi - x$. 2. From $\pi/2$ to $\pi$, $\pi - x \le x \sin x$. 3. Total Area $= 2 \left[ \int_0^{\pi/2} x \sin x \, dx + \int_{\pi/2}^\pi (\pi - x) \, dx \right]$. 4. Integral 1: $\int x \sin x \, dx = -x \cos x + \sin x$. Evaluating from $0$ to $\pi/2$: $(0 + 1) - (0 + 0) = 1$. 5. Integral 2: $\int_{\pi/2}^\pi (\pi - x) \, dx$ is the area of a triangle with base $\pi/2$ and height $\pi/2$. Area $= \frac{1}{2} \times \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{8}$. 6. Total: $2(1 + \pi^2/8) = 2 + \pi^2/4$.

Step 4: Final Answer:
The area is \(2 + \frac{\pi^2}{4}\).