Question:
The value of the integral \(\int_0^{\pi} |\sin 2x|\,dx\) is ___________.
The value of the integral \(\int_0^{\pi} |\sin 2x|\,dx\) is ___________.
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\(|\sin 2x|\) is symmetric about \(x=\pi/2\) on \([0,\pi]\), so integrate over \([0,\pi/2]\) and double.
Updated On: Nov 4, 2026
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Correct Answer: 2
Solution and Explanation
The function \(|\sin 2x|\) has period \(\frac{\pi}{2}\) and is symmetric about \(x=\frac{\pi}{2}\) on \([0,\pi]\).
So \(\int_0^{\pi}|\sin 2x|\,dx=2\int_0^{\pi/2}\sin 2x\,dx\) (since \(\sin 2x\ge0\) on \([0,\frac{\pi}{2}]\)).
\(\int_0^{\pi/2}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2}=\frac12+\frac12=1\).
So the integral \(=2\times1=2\).
Answer: 2
So \(\int_0^{\pi}|\sin 2x|\,dx=2\int_0^{\pi/2}\sin 2x\,dx\) (since \(\sin 2x\ge0\) on \([0,\frac{\pi}{2}]\)).
\(\int_0^{\pi/2}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2}=\frac12+\frac12=1\).
So the integral \(=2\times1=2\).
Answer: 2
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