Question

What is the arclength of `f(x)=cos^2x-x^2` in the interval `[0,pi/3]`?

Answer 1

arc length `=intsqrt[1+(dy/dx)^2]`

Explanation:

`dy/dx=-2cosx(sinx)-2x`

`int_0^(pi/3)sqrt[1+(-2cosxsinx-2x)^2dx]~~2.20`

(Note: Used TI-84 calculator to integrate)

Hope that helped