Find the actual art length of the given curve. y^2=4x, 0<_x<_2?

It might be best to do this with the parametric equations for this curve.

x = t^2 , y = 2t

The end points x = 0 and x = 2 correspond to t = 0 and t = sqrt2

Then you can use the formula for arc length in parametric form which is

INT sqrt [(dy/dt)^2 + (dx/dt)^2] dt

In this case it gives

INT [0, sqrt2] sqrt(4 + 4t^2) dt

Can you take it from there? I think that the integral needs the substitution t = 2sinh y, dt = 2(cosh y) dy. At different points you will need to use

(cosh y)^2 = 1 + (sinh y)^2 and (cosh y)^2 = (1/2) + (1/2)cosh 2y

For such a simple curve, finding an arc length is surprisingly difficult.