The value of \(\int_{0}^{2\pi}\displaystyle |\sin x|\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
sin x is non-negative on \([0, \pi]\) and negative on \((\pi, 2\pi)\), so the integral is split at \(\pi\):
\( \displaystyle \int_{0}^{2\pi}|\sin x|\,dx = \int_{0}^{\pi}\sin x\,dx + \int_{\pi}^{2\pi}(-\sin x)\,dx \)
The two parts:
\( \displaystyle \int_{0}^{\pi}\sin x\,dx = \Big[-\cos x\Big]_{0}^{\pi} = 1 + 1 = 2 \)
\( \displaystyle \int_{\pi}^{2\pi}(-\sin x)\,dx = \Big[\cos x\Big]_{\pi}^{2\pi} = 1 + 1 = 2 \)
Adding:
\( 2 + 2 = 4 \)
Why the other options are incorrect
Option (1) 0 — Incorrect. This is \(\int_{0}^{2\pi}\sin x\,dx\) without the modulus; the positive and negative halves cancel. With the modulus nothing cancels.
Option (2) 2 — Incorrect. This is only the part from 0 to \(\pi\).
Option (4) \(2\pi\) — Incorrect. This would need \(|\sin x| = 1\) everywhere; the value is 4.