Question

The area enclosed by the curves \( y = \sin x + \cos x \) { and } \( y = | \cos x - \sin x | \) { over the interval} \( \left[ 0, \frac{\pi}{2} \right] \) { is:}

• A. \( 4(\sqrt{2} - 1) \)
• B. \( 2\sqrt{2}(\sqrt{2} - 1) \)
• C. \( 2(\sqrt{2} + 1) \)
• D. \( 2\sqrt{2}(\sqrt{2} + 1) \)

Hint:

When dealing with absolute value functions, split the function into cases based on the sign of the expression inside the absolute value. Then integrate over the appropriate intervals to find the enclosed area.

Answer 1

The Correct Option is B

We are asked to find the area enclosed by the curves \( y = \sin x + \cos x \) and \( y = | \cos x - \sin x | \) over the interval \( \left[ 0, \frac{\pi}{2} \right] \). Step 1: Analyze the functions - The curve \( y = \sin x + \cos x \) is a smooth continuous curve. - The curve \( y = | \cos x - \sin x | \) involves the absolute value function, so we need to split it into two cases based on the value of \( \cos x - \sin x \). For \( x \in [0, \frac{\pi}{2}] \), we know: - \( \cos x - \sin x \geq 0 \) for \( x \in [0, \frac{\pi}{4}] \) - \( \cos x - \sin x \leq 0 \) for \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \) Thus, the function \( y = | \cos x - \sin x | \) becomes: - \( y = \cos x - \sin x \) for \( x \in [0, \frac{\pi}{4}] \) - \( y = \sin x - \cos x \) for \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \) Step 2: Set up the integrals The area enclosed by the curves is the sum of two areas: 1. The area between \( y = \sin x + \cos x \) and \( y = \cos x - \sin x \) over the interval \( [0, \frac{\pi}{4}] \). 2. The area between \( y = \sin x + \cos x \) and \( y = \sin x - \cos x \) over the interval \( [\frac{\pi}{4}, \frac{\pi}{2}] \). Area for \( x \in [0, \frac{\pi}{4}] \): \[ A_1 = \int_0^{\frac{\pi}{4}} \left[ (\sin x + \cos x) - (\cos x - \sin x) \right] dx = \int_0^{\frac{\pi}{4}} 2\sin x \, dx \] Integrating: \[ A_1 = 2 \left[ -\cos x \right]_0^{\frac{\pi}{4}} = 2 \left( -\cos \frac{\pi}{4} + \cos 0 \right) = 2 \left( -\frac{\sqrt{2}}{2} + 1 \right) \] \[ A_1 = 2 \left( 1 - \frac{\sqrt{2}}{2} \right) = 2\left( \sqrt{2} - 1 \right) \] Area for \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \): \[ A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left[ (\sin x + \cos x) - (\sin x - \cos x) \right] dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos x \, dx \] Integrating: \[ A_2 = 2 \left[ \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 2 \left( \sin \frac{\pi}{2} - \sin \frac{\pi}{4} \right) = 2 \left( 1 - \frac{\sqrt{2}}{2} \right) \] \[ A_2 = 2 \left( \sqrt{2} - 1 \right) \] Step 3: Total area The total area enclosed by the curves is: \[ A = A_1 + A_2 = 2\left( \sqrt{2} - 1 \right) + 2\left( \sqrt{2} - 1 \right) = 2\sqrt{2}(\sqrt{2} - 1) \] Thus, the correct answer is Option B.