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:
\[ A = 2 \cdot \left( 1 - \frac{1}{\sqrt{2}} \right). \]
\[ A = 4. \]
\[ A^2 = 16. \]
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
Find \( P(0<X<5) \).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32