Question

Evaluate:

`int_(-π/2)^(π/2) |sinx|dx`

Answer 1

I = `int_(-π/2)^(π/2) |sinx|dx`

When x < 0, sin x < 0

∴ |sin x| = – sin x

and when x > 0, sin x > 0

∴ |sin x| = sin x

∴ I = `int_(-π/2)^0 -sinx  dx + int_0^(π/2) sinx  dx`

= `[cosx]_(-π/2)^0 + [-cosx]_0^(π/2)`

= (1 – 0) + (0 + 1)

= 2