Question

If $\int_0^{2\pi} |x \sin x| dx = k\pi$, then $k=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Hint:

Handling Absolute Values in Integration.

• Break the domain at points where the function inside absolute changes sign.
• Compute definite integrals over sub-intervals by adjusting sign accordingly.

Answer 1

The Correct Option is D

Since $x \ge 0$ in $[0,2\pi]$, $|x \sin x| = x |\sin x|$. Then: \[ \int_0^{2\pi} x |\sin x| dx = \int_0^\pi x \sin x dx + \int_\pi^{2\pi} x (-\sin x) dx = \pi + 3\pi = 4\pi \] So $k = 4$.

Answer 2

Step 1: Understand the integral
We need to evaluate:
\[ \int_0^{2\pi} |x \sin x| \, dx = k \pi \]
and find the value of \( k \).

Step 2: Analyze the integrand \( |x \sin x| \)
Since \( x \) is positive in \( [0, 2\pi] \), the absolute value comes from \( \sin x \).
We know that \( \sin x \) is positive in \( (0, \pi) \) and negative in \( (\pi, 2\pi) \).

Step 3: Split the integral at \( x = \pi \)
\[ \int_0^{2\pi} |x \sin x| dx = \int_0^{\pi} x \sin x \, dx + \int_{\pi}^{2\pi} |x \sin x| dx \]
For \( x \in [\pi, 2\pi] \), \( \sin x \leq 0 \), so:
\[ |x \sin x| = -x \sin x \]
Hence,
\[ \int_0^{2\pi} |x \sin x| dx = \int_0^{\pi} x \sin x \, dx - \int_{\pi}^{2\pi} x \sin x \, dx \]

Step 4: Evaluate \( \int x \sin x \, dx \)
Use integration by parts:
Let \( u = x \), \( dv = \sin x \, dx \)
Then, \( du = dx \), \( v = -\cos x \)
So,
\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C \]

Step 5: Calculate the definite integrals
\[ \int_0^{\pi} x \sin x \, dx = \left[-x \cos x + \sin x\right]_0^{\pi} \]
Evaluate at limits:
At \( x = \pi \): \( -\pi \cos \pi + \sin \pi = -\pi (-1) + 0 = \pi \)
At \( x = 0 \): \( 0 + 0 = 0 \)
So, \( \int_0^{\pi} x \sin x \, dx = \pi \)

Similarly, for \( \int_{\pi}^{2\pi} x \sin x \, dx \):
\[ \left[-x \cos x + \sin x\right]_{\pi}^{2\pi} \]
At \( x = 2\pi \): \( -2\pi \cos 2\pi + \sin 2\pi = -2\pi (1) + 0 = -2\pi \)
At \( x = \pi \): \( -\pi \cos \pi + \sin \pi = -\pi (-1) + 0 = \pi \)
So,
\[ \int_{\pi}^{2\pi} x \sin x \, dx = (-2\pi) - (\pi) = -3\pi \]

Step 6: Combine results
\[ \int_0^{2\pi} |x \sin x| dx = \pi - (-3\pi) = \pi + 3\pi = 4\pi \]
Hence, \( k \pi = 4 \pi \implies k = 4 \).

Step 7: Final answer
\[ k = 4 \]