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

\[ y = \min \{ \sin x, \cos x \} \] \[ x = \pi, \quad x = \frac{\pi}{4}, \quad x = 0 \] \[ \int_{0}^{\frac{\pi}{4}} \sin x \, dx = (\cos x) \bigg|_0^{\frac{\pi}{4}} = 1 - \frac{1}{\sqrt{2}} \] \[ \int_{-\frac{3\pi}{4}}^{\pi} (\sin x - \cos x) \, dx = (-\cos x - \sin x) \bigg|_{-\frac{3\pi}{4}}^{-\pi} \] \[ = (\cos x + \sin x) \bigg|_{-\frac{3\pi}{4}}^{-\frac{\pi}{4}} = (-1 + 0) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \] \[ = -1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \] \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx = (\sin x) \bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 1 - \frac{1}{\sqrt{2}} \] \[ A = 4 \] \[ A^2 = 16 \] 

Answer 2

The curve \( y = \min\{\sin x, \cos x\} \) means that the curve follows the smaller of \(\sin x\) and \(\cos x\) for each \(x\). Over \([-\pi, \pi]\), the following intervals apply:
\[ y = \sin x \quad \text{for} \quad \left[ -\frac{\pi}{4}, \frac{3\pi}{4} \right], \] \[ y = \cos x \quad \text{for} \quad \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right]. \]
The total area is:
\[ A = 2 \int_{-\pi/4}^{\pi/4} \sin x \, dx. \]
Compute:
\[ \int_{-\pi/4}^{\pi/4} \sin x \, dx = \left[-\cos x \right]_{-\pi/4}^{\pi/4} = -\cos(\pi/4) - (-\cos(-\pi/4)). \]
\[ \cos(\pi/4) = \cos(-\pi/4) = \frac{1}{\sqrt{2}}. \]
Thus:

\[ A = 2 \cdot \left( 1 - \frac{1}{\sqrt{2}} \right). \]
\[ A = 4. \]
\[ A^2 = 16. \]