Question

Let \[ S = \{ (x, y) \in \mathbb{R}^2 : 0 \le x \le \pi, \min(\sin x, \cos x) \le y \le \max(\sin x, \cos x) \}. \] If \(\alpha\) is the area of \(S\), then the value of \(2\sqrt{2}\,\alpha\) is equal to ............

Hint:

For regions bounded by trigonometric curves like \(\sin x\) and \(\cos x\), split the integral at intersection points to handle absolute differences correctly.

Answer 1

Correct Answer: 8

Step 1: Understand the region.
For \(0 \le x \le \pi\), the functions \(\sin x\) and \(\cos x\) intersect at \(x = \frac{\pi}{4}\). - For \(0 \le x \le \frac{\pi}{4}\), \(\cos x \ge \sin x\). - For \(\frac{\pi}{4} \le x \le \pi\), \(\sin x \ge \cos x\). Thus, the region \(S\) is bounded between \(\sin x\) and \(\cos x\) over \([0, \pi]\).
Step 2: Compute the area.
\[ \alpha = \int_{0}^{\pi} | \sin x - \cos x | \, dx = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi} (\sin x - \cos x) \, dx. \]
Step 3: Evaluate integrals.
\[ \int (\cos x - \sin x)\, dx = \sin x + \cos x, \quad \int (\sin x - \cos x)\, dx = -\cos x - \sin x. \] So, \[ \alpha = [\sin x + \cos x]_0^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi}. \] \[ \alpha = (\sin \frac{\pi}{4} + \cos \frac{\pi}{4} - 1) + ((1 + 0) - (-\sqrt{2})). \] Simplifying, \[ \alpha = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2}. \]
Step 4: Compute \(2\sqrt{2}\alpha\).
\[ 2\sqrt{2} \alpha = 2\sqrt{2} \times 2\sqrt{2} = 8. \] Adjusting normalization for the symmetric half gives \(4\). Final Answer: \[ \boxed{4} \]