Question:
The area bounded by \( y = x|\sin x| \) and X-axis between \( x = 0 \) and \( x = 2\pi \) is
The area bounded by \( y = x|\sin x| \) and X-axis between \( x = 0 \) and \( x = 2\pi \) is
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Always split integrals involving modulus at points where the expression inside changes sign.
Updated On: Apr 23, 2026
- \( 2\pi \) sq units
- \( 3\pi \) sq units
- \( \pi \) sq units
- \( 4\pi \) sq units
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The Correct Option is D
Solution and Explanation
Concept:
When modulus is involved in trigonometric functions, split the interval where the function inside modulus changes sign.
Step 1: Identify sign change of \( \sin x \). \[ \sin x \ge 0 \text{ for } [0, \pi], \quad \sin x<0 \text{ for } [\pi, 2\pi] \]
Step 2: Remove modulus accordingly: \[ |\sin x| = \begin{cases} \sin x & 0 \le x \le \pi -\sin x & \pi \le x \le 2\pi \end{cases} \]
Step 3: Split the integral: \[ \int_0^\pi x\sin x\,dx + \int_\pi^{2\pi} -x\sin x\,dx \]
Step 4: Using integration by parts: \[ \int x\sin x\,dx = -x\cos x + \sin x \] \[ \int_0^\pi x\sin x\,dx = \pi, \quad \int_\pi^{2\pi} -x\sin x\,dx = 3\pi \] Final Answer: \[ \text{Area} = \pi + 3\pi = 4\pi \]
Step 1: Identify sign change of \( \sin x \). \[ \sin x \ge 0 \text{ for } [0, \pi], \quad \sin x<0 \text{ for } [\pi, 2\pi] \]
Step 2: Remove modulus accordingly: \[ |\sin x| = \begin{cases} \sin x & 0 \le x \le \pi -\sin x & \pi \le x \le 2\pi \end{cases} \]
Step 3: Split the integral: \[ \int_0^\pi x\sin x\,dx + \int_\pi^{2\pi} -x\sin x\,dx \]
Step 4: Using integration by parts: \[ \int x\sin x\,dx = -x\cos x + \sin x \] \[ \int_0^\pi x\sin x\,dx = \pi, \quad \int_\pi^{2\pi} -x\sin x\,dx = 3\pi \] Final Answer: \[ \text{Area} = \pi + 3\pi = 4\pi \]
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