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 \(y = 0\), \(y = \pi - |x|\), and \(y = |x \sin x|\). Both bounding functions \(y = \pi - |x|\) and \(y = |x \sin x|\) are even functions, meaning they are symmetric about the y-axis.
We can find the area in the first quadrant (\(x \ge 0\)) and multiply by 2.
Step 2: Key Formula or Approach:
For \(x \ge 0\):
\(y_1 = \pi - x\)
\(y_2 = x \sin x\)
We need to find the intersection of these two curves to determine which acts as the upper bound in different intervals.
Setting \(y_1 = y_2\):
\[ \pi - x = x \sin x \implies \sin x = \frac{\pi}{x} - 1 \] At \(x = \frac{\pi}{2}\), LHS is \(\sin(\frac{\pi}{2}) = 1\), and RHS is \(\frac{\pi}{\pi/2} - 1 = 2 - 1 = 1\).
Thus, the curves intersect at \(x = \frac{\pi}{2}\).
Step 3: Detailed Explanation:
For \(0 \le x \le \frac{\pi}{2}\), \(x \sin x \le \pi - x\), so the upper boundary is \(y = x \sin x\).
For \(\frac{\pi}{2} \le x \le \pi\), \(\pi - x \le x \sin x\), so the upper boundary is \(y = \pi - x\).
The area \(A_1\) in the first quadrant is given by:
\[ A_1 = \int_0^{\pi/2} x \sin x \, dx + \int_{\pi/2}^{\pi} (\pi - x) \, dx \] Evaluate the first integral using integration by parts:
\[ \int x \sin x \, dx = -x \cos x - \int (-\cos x) dx = -x \cos x + \sin x \] Evaluating from 0 to \(\pi/2\):
\[ [\sin x - x \cos x]_0^{\pi/2} = \left(\sin\left(\frac{\pi}{2}\right) - \frac{\pi}{2}\cos\left(\frac{\pi}{2}\right)\right) - (\sin 0 - 0) = (1 - 0) - 0 = 1 \] Evaluate the second integral:
\[ \int_{\pi/2}^{\pi} (\pi - x) \, dx = \left[ -\frac{(\pi - x)^2}{2} \right]_{\pi/2}^{\pi} = 0 - \left( -\frac{(\pi - \pi/2)^2}{2} \right) = \frac{(\pi/2)^2}{2} = \frac{\pi^2}{8} \] So, \(A_1 = 1 + \frac{\pi^2}{8}\).
Step 4: Final Answer:
Due to the symmetry about the y-axis, the total area is:
\[ \text{Total Area} = 2 \times A_1 = 2 \left( 1 + \frac{\pi^2}{8} \right) = 2 + \frac{\pi^2}{4} \]