Question

\(\int _0^π|sin2x|\,\text{d}x =\)

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(-1\)

Hint:

Split at pi/2 where sin 2x changes sign and double the first piece by symmetry.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
On \([0, \pi/2]\), \(2x\in[0,\pi]\) so \(\sin2x \geq 0\). On \([\pi/2, \pi]\), \(2x \in [\pi, 2\pi]\) so \(\sin2x \leq 0\).

Step 2: Split and integrate
\[ \int_0^{\pi}|\sin2x|\,dx = \int_0^{\pi/2}\sin2x\,dx + \int_{\pi/2}^{\pi}(-\sin2x)\,dx \]
\[ \int_0^{\pi/2}\sin2x\,dx = \left[-\frac{\cos2x}{2}\right]_0^{\pi/2} = \frac12 + \frac12 = 1 \]
The second piece is also 1 because \(|\sin 2x|\) repeats with period \(\pi/2\).
\[ \text{Total} = 1 + 1 = 2 \]
Option (A), 0, is what you get for \(\int_0^{\pi}\sin2x\,dx\) without the absolute value.

Final Answer:
The value is 2, option (C). \[ \boxed{2} \]