Question

What is the arclength of `f(x)=xcos(x-2)` on `x in [1,2]`?

Answer 1

show below

Explanation:

the solution is

`L=int_a^bsqrt(1+[f'(x)]^2)*dx`

`f'(x)=cos(x-2)-xsin(x-2)`

`L=int_1^2sqrt(1+[cos(x-2)-xsin(x-2)]^2)*dx`

`L=int_1^2sqrt(1+cos^2(x-2)-2xcos(x-2)*sin(x-2)+x^2sin^2(x-2))*dx`

then complete the steps manually