Question

π/2
∫ | sinx-cosx | dx ,Evaluate it.
0

Answer 1

Refer to attachment.......

Answer 2

Answer:

2($\sqrt{2}-1$)

Step-by-step explanation:

We have,  | sinx-cosx | =sinx-cosx, when sinx≥cosx i.e. π/4 <x<π/2 .... (1)

                                     = cosx-sinx when sinx<cosx i.e. 0 <x < π/4 .....(2)

Hence, $\int\limits^\frac{\pi }{2} _0 { | sinx-cosx | } \, dx$

=$\int\limits^\frac{\pi }{4} _0 { | sinx-cosx | } \, dx +\int\limits^\frac{\pi }{2} _\frac{\pi }{4}  { | sinx-cosx | } \, dx$

=$\int\limits^\frac{\pi }{4} _0 {(cosx-sinx) } \, dx +\int\limits^\frac{\pi }{2} _\frac{\pi }{4}  { (sinx-cosx ) } \, dx$

{Getting from the definition (1) and (2)}

=$[sinx+cosx]^{\frac{\pi }{4} } _{0} +[-sinx-cosx]^{\frac{\pi }{2} } _{\frac{\pi }{4} }$

=($\sqrt{2} -1$)-($1-\sqrt{2}$)

=2($\sqrt{2}-1$) (Answer)