Length of a Parabolic Curve

Your train of thought is exactly right; you've single-handedly rederived the formula for the length of a curve given by $y=f(x)$ :-) This can be written as

$$ L=\int_a^b\sqrt{1+f'(x)^2}\,\mathrm dx $$

in general. In your case, as you rightly determined, $f'(x)=2x$, and we want the length from $a=0$ to $b=1$, so we have

$$ L=\int_0^1\sqrt{1+4x^2}\,\mathrm dx\;. $$

To solve this integral, you can use the substitution $\sqrt{1+4x^2}=\cosh u$, so $x=\frac12\sinh u$ and $\mathrm dx=\frac12\cosh u\,\mathrm du$, to get

$$ L=\frac12\int_0^{\operatorname{arcosh}\sqrt5}\cosh^2 u\,\mathrm du=\frac14\left[x+\sinh x\cosh x\right]_0^{\operatorname{arcosh}\sqrt5}=\frac14\left(\operatorname{arcosh}\sqrt5+2\sqrt5\right)\approx1.479$$

If you don't know how to solve such an integral, you can always ask Wolfram|Alpha; it will usually know the answer and can often tell you the steps to get there if you click on "step-by-step solution"; though the solution will sometimes, as in this case, not be the most simple one.

You might also be interested in the question Intuition behind arc length formula.