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

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:

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