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 $A = \int_a^b |f(x) - g(x)| \, dx$.
• Trigonometric Identity: $\sin(a+b) = \sin a \cos b + \cos a \sin b$.
• Trigonometric Values: $\sin(\pi/4) = \cos(\pi/4) = \frac{1}{\sqrt{2}}$.

Step-by-Step Solution

Step 1: Define the functions and the interval

We are given the functions $y_1 = \sin x + \cos x$ and $y_2 = |\cos x - \sin x|$, and the interval $x \in [0, \frac{\pi}{2}]$. Our goal is to find the area enclosed by these curves and the given lines.

Step 2: Simplify $y_1 = \sin x + \cos x$

We can rewrite $y_1$ using the identity $R\sin(x+\alpha) = \sin x + \cos x$ where $R = \sqrt{1^2+1^2} = \sqrt{2}$ and $\alpha = \arctan(1) = \frac{\pi}{4}$. This simplifies the analysis of the function's behavior.

$y_1 = \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \right) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$

Since $x \in [0, \frac{\pi}{2}]$, then $x+\frac{\pi}{4} \in [\frac{\pi}{4}, \frac{3\pi}{4}]$. Since $\sin(x)$ is positive in the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]$, $y_1$ is always non-negative in the given interval.

Step 3: Simplify $y_2 = |\cos x - \sin x|$

To handle the absolute value, we need to determine when $\cos x - \sin x$ is positive or negative. This involves finding where $\cos x = \sin x$, which occurs at $x = \frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.

• For $x \in [0, \frac{\pi}{4}]$, $\cos x \ge \sin x$, so $y_2 = \cos x - \sin x$.
• For $x \in [\frac{\pi}{4}, \frac{\pi}{2}]$, $\sin x \ge \cos x$, so $y_2 = -(\cos x - \sin x) = \sin x - \cos x$.

Step 4: Determine which function is greater in each subinterval

We need to compare $y_1$ and $y_2$ in the two subintervals $[0, \frac{\pi}{4}]$ and $[\frac{\pi}{4}, \frac{\pi}{2}]$.

• For $x \in [0, \frac{\pi}{4}]$: $y_1 - y_2 = (\sin x + \cos x) - (\cos x - \sin x) = 2 \sin x$. Since $\sin x \ge 0$ in this interval, $y_1 \ge y_2$.

• For $x \in [\frac{\pi}{4}, \frac{\pi}{2}]$: $y_1 - y_2 = (\sin x + \cos x) - (\sin x - \cos x) = 2 \cos x$. Since $\cos x \ge 0$ in this interval, $y_1 \ge y_2$.

Therefore, $y_1 \ge y_2$ throughout the interval $[0, \frac{\pi}{2}]$.

Step 5: Set up the integral for the area

Since $y_1 \ge y_2$ in the entire interval, we can write the area as: $A = \int_0^{\pi/2} (y_1 - y_2) \, dx = \int_0^{\pi/4} (y_1 - y_2) \, dx + \int_{\pi/4}^{\pi/2} (y_1 - y_2) \, dx$ Substituting the expressions for $y_1$ and $y_2$ in each subinterval: $A = \int_0^{\pi/4} [(\sin x + \cos x) - (\cos x - \sin x)] \, dx + \int_{\pi/4}^{\pi/2} [(\sin x + \cos x) - (\sin x - \cos x)] \, dx$ $A = \int_0^{\pi/4} 2 \sin x \, dx + \int_{\pi/4}^{\pi/2} 2 \cos x \, dx$

Step 6: Evaluate the integrals

$A = \left[ -2 \cos x \right]_0^{\pi/4} + \left[ 2 \sin x \right]_{\pi/4}^{\pi/2}$ $A = \left( -2 \cos \frac{\pi}{4} - (-2 \cos 0) \right) + \left( 2 \sin \frac{\pi}{2} - 2 \sin \frac{\pi}{4} \right)$ $A = \left( -2 \cdot \frac{1}{\sqrt{2}} + 2 \right) + \left( 2 \cdot 1 - 2 \cdot \frac{1}{\sqrt{2}} \right)$ $A = \left( -\sqrt{2} + 2 \right) + \left( 2 - \sqrt{2} \right)$ $A = 4 - 2\sqrt{2}$

Step 7: Rewrite the answer in the desired format

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

Common Mistakes & Tips

• Absolute Value: Forgetting to consider the different cases when dealing with absolute values is a common error. Always analyze the sign of the expression inside the absolute value.
• Integration Errors: Double-check your integration and the evaluation of the limits. Sign errors are common.
• Identifying the Upper Curve: Always determine which curve is above the other in the interval of integration to correctly set up the integral.

Summary

We found the area enclosed by the curves $y = \sin x + \cos x$ and $y = |\cos x - \sin x|$ and the lines $x = 0$ and $x = \frac{\pi}{2}$ by first simplifying the given functions, then determining the intervals where each function was greater, setting up the definite integrals, and finally evaluating the integrals. The area is $2\sqrt{2}(\sqrt{2} - 1)$.

Final Answer

The final answer is \boxed{2\sqrt 2 (\sqrt 2 - 1)}, which corresponds to option (A).