Question

Let the area of the region enclosed by the curve $y = \min\{\sin x, \cos x\}$ and the x-axis between $x = -\pi$ to $x = \pi$ be $A$. Then $A^2$ is equal to _____.

Answer 1

Correct Answer: 16

Let \( y = \min \{ \sin x, \cos x \} \). The critical points are \( x = \pi, \quad x = \frac{\pi}{4}, \quad x = 0 \). The integral from \( 0 \) to \( \frac{\pi}{4} \) of \( \sin x \) with respect to \( x \) is \( (\cos x) \bigg|_0^{\frac{\pi}{4}} = 1 - \frac{1}{\sqrt{2}} \). The integral from \( -\frac{3\pi}{4} \) to \( \pi \) of \( (\sin x - \cos x) \) with respect to \( x \) is \( (-\cos x - \sin x) \bigg|_{-\frac{3\pi}{4}}^{-\pi} \). This is equal to \( (\cos x + \sin x) \bigg|_{-\frac{3\pi}{4}}^{-\frac{\pi}{4}} = (-1 + 0) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = -1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \). The integral from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \) of \( \cos x \) with respect to \( x \) is \( (\sin x) \bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 1 - \frac{1}{\sqrt{2}} \). Given \( A = 4 \), then \( A^2 = 16 \).