Question

integrate mod(sin x -cos x) in limit 0 to pi/2

Answer 1

|cosx -sinx|.dx limit x =0, to π/2

we know,
sinx and cosx has one common value in 0 to π/2 . e.g π/4

in interval [ 0 , π/4 ] cosx ≥ sinx

so, modulus break in [0, π/4 ] with positive sign
e.g ( cosx -sinx).dx limit [ 0 , π/4]

in interval [ π/4 , π/2 ] , sinx≥ cosx

so, modulus beak with negative sign in [π/4 , π/2]

now ,

integration without modulus is -----

I =(cosx-sinx)dx limit [0 , π/4] - (cosx -sinx)dx limit [ π/4, π/2]

={ sinx +cosx } limit [0, π/4] + {-sinx-cosx} limit [ π/4 , π/2]

={1/√2 +1/√2 -1 } + { -1 -(-1/√2 -1/√2)}

=√2 -1 - 1 +√2

=2√2 -2