Key Concepts and Formulas

• Area Between Curves: The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by $A = \int_a^b |f(x) - g(x)| dx$.
• Intersection Points: To find where the curves intersect, solve $f(x) = g(x)$. These points are crucial for determining the intervals where one function is above the other.
• Trigonometric Values: Knowing the values of $\sin x$ and $\cos x$ at key angles is essential. Specifically, $\sin(\pi/4) = \cos(\pi/4) = \frac{1}{\sqrt{2}}$.

Step-by-Step Solution

Step 1: Find the intersection points of the curves

We need to find where $\cos x = \sin x$ in the interval $[0, \frac{3\pi}{2}]$. $\cos x = \sin x$ $\tan x = 1$ The solutions in the given interval are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. WHY: The intersection points define the regions where one function is above the other, and we need these points to correctly evaluate the integral.

Step 2: Divide the interval into subintervals based on intersection points

The interval $[0, \frac{3\pi}{2}]$ is divided into three subintervals by the intersection points: $[0, \frac{\pi}{4}]$, $[\frac{\pi}{4}, \frac{5\pi}{4}]$, and $[\frac{5\pi}{4}, \frac{3\pi}{2}]$.

WHY: We need to integrate the absolute difference between the two functions. Since the relative positions of the curves change at the intersection points, we split the integral into multiple integrals over the subintervals.

Step 3: Determine which function is greater in each subinterval

• In $[0, \frac{\pi}{4}]$, $\cos x \ge \sin x$. For example, at $x=0$, $\cos 0 = 1 > \sin 0 = 0$.
• In $[\frac{\pi}{4}, \frac{5\pi}{4}]$, $\sin x \ge \cos x$. For example, at $x=\pi/2$, $\sin (\pi/2) = 1 > \cos(\pi/2) = 0$.
• In $[\frac{5\pi}{4}, \frac{3\pi}{2}]$, $\cos x \ge \sin x$. For example, at $x=\pi$, $\cos \pi = -1 > \sin \pi = 0$ is false, but note that we need a value in the range $(\frac{5\pi}{4}, \frac{3\pi}{2})$. Consider $x = \frac{11\pi}{6}$ which is in this interval. Then $\cos(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{11\pi}{6}) = -\frac{1}{2}$. Then $\cos(\frac{11\pi}{6}) > \sin(\frac{11\pi}{6})$.

WHY: This step is crucial for setting up the correct integrals. We need to know which function to subtract from which in each subinterval to ensure we're integrating a positive difference.

Step 4: Set up the definite integrals

The total area is given by: $A = \int_0^{\pi/4} (\cos x - \sin x) dx + \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) dx + \int_{5\pi/4}^{3\pi/2} (\cos x - \sin x) dx$

WHY: This integral represents the sum of the areas in each subinterval where we take the absolute value of the difference in the curves.

Step 5: Evaluate the integrals

• $\int_0^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_0^{\pi/4} = (\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0)) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$
• $\int_{\pi/4}^{5\pi/4} (\sin x - \cos x) dx = [-\cos x - \sin x]_{\pi/4}^{5\pi/4} = (-\cos(5\pi/4) - \sin(5\pi/4)) - (-\cos(\pi/4) - \sin(\pi/4)) = (-(-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}})) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$
• $\int_{5\pi/4}^{3\pi/2} (\cos x - \sin x) dx = [\sin x + \cos x]_{5\pi/4}^{3\pi/2} = (\sin(3\pi/2) + \cos(3\pi/2)) - (\sin(5\pi/4) + \cos(5\pi/4)) = (-1 + 0) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 - (-\frac{2}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$

WHY: We evaluate the definite integrals to obtain the numerical value of the area in each subinterval.

Step 6: Sum the areas

$A = (\sqrt{2} - 1) + 2\sqrt{2} + (\sqrt{2} - 1) = 4\sqrt{2} - 2$

WHY: We sum the areas from each subinterval to find the total area bounded by the curves.

Common Mistakes & Tips

• Forgetting the Absolute Value: Always remember to use the absolute value $|f(x) - g(x)|$ in the integral. This ensures you're always integrating a positive quantity representing the area. If you forget, you might get a negative area, which is incorrect.
• Incorrectly Identifying the Upper and Lower Functions: Make sure you correctly identify which function is greater than the other in each subinterval. A quick way to do this is to plug in a test value within the interval.
• Sign Errors: Be extremely careful with signs when evaluating the integrals of trigonometric functions, especially when dealing with negative angles or angles in different quadrants.

Summary

To find the area bounded by $y=\cos x$ and $y=\sin x$ from $x=0$ to $x=\frac{3\pi}{2}$, we first found the intersection points of the curves, which are $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$. Then, we split the integral into three subintervals based on these intersection points and determined which function was greater in each subinterval. Finally, we evaluated the definite integrals in each subinterval and summed them to find the total area.

The final answer is $4\sqrt{2} - 2$.

Final Answer

The final answer is \boxed{4\sqrt 2 - 2}, which corresponds to option (D).