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: \[ 2(\sqrt{2} - 1). \]

Answer 2

To find the area under the curve \( y = |\sin x - \cos x| \) over the interval \( 0 \leq x \leq \frac{\pi}{2} \), we first identify where \( \sin x = \cos x \). Solving \( \sin x = \cos x \), we have \( \tan x = 1 \), which gives \( x = \frac{\pi}{4} \). Therefore, the integral becomes split at \( x = \frac{\pi}{4} \):

For \( 0 \leq x \leq \frac{\pi}{4} \), \(\sin x \geq \cos x\), so \( |\sin x - \cos x| = \sin x - \cos x \). The integral over this interval is:
\( \int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx \).
Integrating, we get:
\( [-\cos x - \sin x]_{0}^{\frac{\pi}{4}} = \left(-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right) - (-\cos(0) - \sin(0)) \).
Substitute values:
\( -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} + 1 = 1 - \sqrt{2} \).

For \( \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \), \(\cos x \geq \sin x\), so \( |\sin x - \cos x| = \cos x - \sin x \). The integral over this interval is:
\( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x - \sin x) \, dx \).
Integrating, we get:
\( [\sin x - \cos x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \left(\sin\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right)\right) - \left(\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right)\right) \).
Substitute values:
\( (1 - 0) - \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = 1 - 0 = 1 \).

Adding both integrals, we find the total area:
\( (1 - \sqrt{2}) + 1 = 2 - \sqrt{2} \).
Conclusion: The area under the curve is \( 2(\sqrt{2} - 1) \) square units.