Question

$\int _ { 0 } ^ { \pi / 2 } | \sin x - \cos x | d x$ is equal to

• A. 0
• B. $\sqrt { 2 }$– 1
• C. 2($\sqrt { 2 }$– 1)
• D. 2( $\sqrt { 2 }$ + 1)

Answer 1

Answer: (C)

2($\sqrt { 2 }$– 1)

Answer 2

I = $\int _ { 0 } ^ { \pi / 2 } \mathrm { f } ( \mathrm { x } ) \mathrm { dx }$ +

where f(x) = |sin x – cos x|

=$\int _ { 0 } ^ { \pi / 4 } ( \cos x - \sin x ) d x$+$\int _ { \pi / 4 } ^ { \pi / 2 } ( \sin x - \cos x ) d x$

= $[ \sin x + \cos x ] _ { 0 } ^ { \pi / 4 }$ + $[ - \cos x - \sin x ] _ { \pi / 4 } ^ { \pi / 2 }$

= 2$\left( \frac { 1 } { \sqrt { 2 } } \right)$ – 1 + (– 1) + 2 $\left( \frac { 1 } { \sqrt { 2 } } \right)$ = 2$( \sqrt { 2 } - 1 )$