Question

What is the arc length of `f(x)=-xsinx+xcos(x-pi/2)` on `x in [0,(pi)/4]`?

Answer 1

`pi/4`

Explanation:

The arc length of `f(x)`, `x in[a.b]` is given by:
`S_x=int_b^af(x)sqrt(1+f'(x)^2)dx`

`f(x)=-xsinx+xcos(x-pi/2)=-xsinx+xsinx=0`
`f'(x)=0`

Since we just have `y=0` we can just take the length of s straight line between `0to pi/4` which is `pi/4-0=pi/4`