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Question

Question: \(\int _ { 0 } ^ { 2 \pi } | \sin x | d x =\)...

∫02π∣sin⁡x∣dx=\int _ { 0 } ^ { 2 \pi } | \sin x | d x =

A

0

B

1

C

2

D

4

Answer

4

Explanation

Solution

∫02π∣sin⁡x∣dx=∫0πsin⁡xdx+∫π2π−sin⁡xdx\int _ { 0 } ^ { 2 \pi } | \sin x | d x = \int _ { 0 } ^ { \pi } \sin x d x + \int _ { \pi } ^ { 2 \pi } - \sin x d x

=[−cos⁡x]0π+[cos⁡x]π2π=1+1+1+1=4= [ - \cos x ] _ { 0 } ^ { \pi } + [ \cos x ] _ { \pi } ^ { 2 \pi } = 1 + 1 + 1 + 1 = 4.