Question

The area of the region under the curve \( y = |\sin x - \cos x| \) in the interval \( 0 \leq x \leq \frac{\pi}{2} \), above the x-axis, is (in square units):

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

Hint:

For absolute value functions, identify points where the function changes sign and split the integral accordingly.

Answer 1

The Correct Option is C

Step 1: Understanding the given function
We have: \[ y = |\sin x - \cos x|. \] To simplify this expression, we use the transformation: \[ \sin x - \cos x = \sqrt{2} \sin \left( x - \frac{\pi}{4} \right). \] Thus, the function can be rewritten as: \[ y = \left| \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \right|. \] Step 2: Finding points where \( \sin(x - \frac{\pi}{4}) = 0 \)
\[ \sin(x - \frac{\pi}{4}) = 0 \quad \Rightarrow \quad x - \frac{\pi}{4} = 0. \] Solving for \( x \): \[ x = \frac{\pi}{4}. \] Step 3: Evaluating the area integral
The function changes sign at \( x = \frac{\pi}{4} \). Therefore, we split the integral: \[ A = \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx. \] Using standard integration: \[ \int (\cos x - \sin x) dx = \sin x + \cos x. \] Evaluating in \( [0, \pi/4] \): \[ \left[ \sin x + \cos x \right]_0^{\pi/4} = (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (0 + 1). \] \[ = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - 1 = \sqrt{2} - 1. \] Similarly, for \( [\pi/4, \pi/2] \): \[ \left[ \sin x + \cos x \right]_{\pi/4}^{\pi/2} = (1 + 0) - (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}). \] \[ = 1 - \sqrt{2} + 1. \] Adding both areas: \[ A = 2 (\sqrt{2} - 1). \] Step 4: Conclusion
Thus, the correct answer is: \[ 2(\sqrt{2} - 1). \]