_0^ 2 | x| d x= - Vidyadip

MCQ

$\int_0^{2 \pi}|\sin x| d x=$

A
$0$
B
1
C
2
✓
4

✓

Answer

Correct option: D.

4

(D)
Since $\sin x$ is positive in the interval $(0, \pi)$ and negative in the interval $(\pi, 2 \pi)$.
$\therefore \quad \int_0^{2 \pi}|\sin x| d x=\int_0^\pi \sin x d x+\int_\pi^{2 \pi}(-\sin x) d x$
$\begin{array}{l}=[-\cos x]_0^\pi+[\cos x]_\pi^{2 \pi} \\ =1+1+1+1=4\end{array}$

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