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Question

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

∫0π/2∣sin⁡x−cos⁡x∣dx=\int _ { 0 } ^ { \pi / 2 } | \sin x - \cos x | d x =

A

0

B

2(2−1)2 ( \sqrt { 2 } - 1 )

C

2−1\sqrt { 2 } - 1

D

2(2+1)2 ( \sqrt { 2 } + 1 )

Answer

2(2−1)2 ( \sqrt { 2 } - 1 )

Explanation

Solution

∫0π/2∣sin⁡x−cos⁡x∣dx\int _ { 0 } ^ { \pi / 2 } | \sin x - \cos x | d x

.