Question

Evaluate `int_0^(2π) |sin x| dx`

Answer 1

Let I = `int_0^(2π) |sin x| dx`

= `int_0^π |sin x| dx + int_π^(2π) |sin x| dx`

= `int_0^π sin x dx - int_π^(2π) sin x dx`

= `[- cos x]_0^π - [- cos x]_π^(2π)`

= [– cos π + cos 0] – [– cos 2π + cos π]

= [1 + 1] – [–1 – 1]

= 2 + 2

= 4