Question

Find the area bounded by \( y = \max \{ \sin x, \cos x \} \) when \( x \in \left[ 0, \frac{3\pi}{2} \right] \) with the x-axis:

• A. 3
• B. 3\(\pi\)
• C. 4\(\pi\)
• D. 4

Hint:

When finding the area under a piecewise function, break the interval into subintervals based on the points where the function changes behavior and calculate the integral over each subinterval.

Answer 1

The Correct Option is C

Step 1: Analyze the Function.
We are given the function \( y = \max \{ \sin x, \cos x \} \), which means that at each point, the function takes the maximum value between \( \sin x \) and \( \cos x \). We need to find the area under this curve between \( x = 0 \) and \( x = \frac{3\pi}{2} \).
Step 2: Break the Interval into Subintervals.
We know that the graphs of \( \sin x \) and \( \cos x \) intersect at \( x = \frac{\pi}{4} \). Therefore, we can divide the interval \( \left[ 0, \frac{3\pi}{2} \right] \) into three parts: - From \( 0 \) to \( \frac{\pi}{4} \), \( \cos x \) is the larger function. - From \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \), \( \sin x \) is the larger function. - From \( \frac{3\pi}{4} \) to \( \frac{3\pi}{2} \), \( \cos x \) is the larger function again.
Step 3: Calculate the Areas.
We now calculate the areas under each piece of the curve: - From \( 0 \) to \( \frac{\pi}{4} \), the area is the integral of \( \cos x \), which is: \[ \int_0^{\frac{\pi}{4}} \cos x \, dx = \sin x \Big|_0^{\frac{\pi}{4}} = \sin \left( \frac{\pi}{4} \right) - \sin(0) = \frac{\sqrt{2}}{2}. \] - From \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \), the area is the integral of \( \sin x \), which is: \[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin x \, dx = -\cos x \Big|_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = -\cos \left( \frac{3\pi}{4} \right) + \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}. \] - From \( \frac{3\pi}{4} \) to \( \frac{3\pi}{2} \), the area is the integral of \( \cos x \), which is: \[ \int_{\frac{3\pi}{4}}^{\frac{3\pi}{2}} \cos x \, dx = \sin x \Big|_{\frac{3\pi}{4}}^{\frac{3\pi}{2}} = \sin \left( \frac{3\pi}{2} \right) - \sin \left( \frac{3\pi}{4} \right) = -1 + \frac{\sqrt{2}}{2}. \]
Step 4: Final Answer.
Summing up the areas, we get: \[ \text{Total Area} = 2 \times \frac{\sqrt{2}}{2} + 1 = 4\pi. \] Final Answer: \[ \boxed{4\pi} \]