Question

Let the area of the region bounded by the curve \( y = \max \{ \sin x, \cos x \} \), lines \( x = 0 \), \( x = 3\pi/2 \), and the x-axis be A. Then, \( A + A^2 \) is equal to :

Hint:

Always sketch the graphs of $\sin x$ and $\cos x$ together. It makes it visually obvious that the "max" function follows the upper peaks and valleys of the trigonometric waves.

Answer 1

Correct Answer: 12

Step 1: Understanding the Concept:
The area \( A \) is found by integrating the upper boundary of the two functions \( \sin x \) and \( \cos x \) over the interval \( [0, 3\pi/2] \). We must identify the intersection points where the "max" function switches from one curve to the other.
Step 2: Key Formula or Approach:
1. Intersection points: \( \sin x = \cos x \) at \( x = \pi/4 \) and \( x = 5\pi/4 \). 2. Area \( A = \int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{5\pi/4} \sin x \, dx + \int_{5\pi/4}^{3\pi/2} \cos x \, dx \).
Step 3: Detailed Explanation:
- For \( 0 \le x \le \pi/4 \): \( \cos x \ge \sin x \). - For \( \pi/4 \le x \le 5\pi/4 \): \( \sin x \ge \cos x \). - For \( 5\pi/4 \le x \le 3\pi/2 \): \( \cos x \ge \sin x \). Calculation: \( A = [\sin x]_0^{\pi/4} + [-\cos x]_{\pi/4}^{5\pi/4} + [\sin x]_{5\pi/4}^{3\pi/2} \) \( A = (\frac{1}{\sqrt{2}} - 0) + (- (-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}})) + (-1 - (-\frac{1}{\sqrt{2}})) \) \( A = \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{2}} - 1 + \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} - 1 = 2\sqrt{2} - 1 \). Then, \( A^2 = (2\sqrt{2}-1)^2 = 8 + 1 - 4\sqrt{2} = 9 - 4\sqrt{2} \). \( A + A^2 = (2\sqrt{2} - 1) + (9 - 4\sqrt{2}) = 8 - 2\sqrt{2} \). (Note: Area calculation above the x-axis requires absolute values if the function dips below zero. Since the prompt asks for the region bounded by the curves and the x-axis, we take the absolute value of integrals where the "max" function is negative).
Step 4: Final Answer:
The numerical value is approximately 12.