Question

The integral \( \int_{0}^{\pi} |x \cos 2x| dx \) is equal to:

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

Hint:

For integrals of \( |f(x)| \), determine the sign of \( f(x) \) in every sub-interval to remove the modulus correctly[cite: 1017].

Answer 1

The Correct Option is C

Concept: The absolute value function \( |x \cos 2x| \) changes sign where \( \cos 2x = 0 \). We split the integral at these points in the interval \([0, \pi]\)[cite: 1017].

Step 1: Identify the split points.
\( \cos 2x = 0 \) at \( 2x = \frac{\pi}{2}, \frac{3\pi}{2} \), which gives \( x = \frac{\pi}{4}, \frac{3\pi}{4} \). $$ I = \int_{0}^{\pi/4} x \cos 2x \, dx - \int_{\pi/4}^{3\pi/4} x \cos 2x \, dx + \int_{3\pi/4}^{\pi} x \cos 2x \, dx $$

Step 2: Evaluate the integral by parts.
The general integral is \( \int x \cos 2x \, dx = \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \)[cite: 1017].

Step 3: Apply the limits.
Evaluating at \( [0, \pi/4] \): $$ \left[ \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right]_{0}^{\pi/4} = \left( \frac{\pi}{8} \cdot 1 + 0 \right) - \left( 0 + \frac{1}{4} \cdot 1 \right) = \frac{\pi}{8} - \frac{1}{4} $$ Evaluating at \( [\pi/4, 3\pi/4] \): $$ \left[ \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right]_{\pi/4}^{3\pi/4} = \left( -\frac{3\pi}{8} - \frac{1}{4} \right) - \left( \frac{\pi}{8} - \frac{1}{4} \right) = -\frac{4\pi}{8} = -\frac{\pi}{2} $$ Evaluating at \( [3\pi/4, \pi] \): $$ \left[ \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right]_{3\pi/4}^{\pi} = \left( 0 + \frac{1}{4} \right) - \left( -\frac{3\pi}{8} - \frac{1}{4} \right) = \frac{3\pi}{8} + \frac{1}{2} $$

Step 4: Combine the results.
$$ I = \left( \frac{\pi}{8} - \frac{1}{4} \right) - \left( -\frac{\pi}{2} \right) + \left( \frac{3\pi}{8} + \frac{1}{2} \right) = \pi + \frac{1}{4} $$ $$\boxed{\pi + \frac{1}{4}}$$