I have been trying to find the archlength of $\displaystyle y^2=4x$, where 0 ≤ x ≤ 2.

I first solved $\displaystyle y=\sqrt{4x}$.

Then I solved for y'.

$\displaystyle y'=\frac{4x^\frac{-1}{2}}{2}*4=2(4x)^\frac{-1}{2}$.

The formula I have to find arclength is $\displaystyle \int_a^b \sqrt{1+(y')^2} dx$.

So I plugged in and got this:

$\displaystyle \int_0^2 \sqrt{1+(2(4x)^\frac{-1}{2})^2}=\int_0^2 \sqrt{1+(4(4x)^{-1})}=\int_0^2 \sqrt{1+\frac{4}{4x}}=\int_0^2 \sqrt{1+\frac{1}{x}}$.

I evaluated $\displaystyle \frac{1}{x}=(\frac{1}{\sqrt{x}})^2$, because my book offers a formula for the form $\displaystyle \sqrt{a^2+u^2}$.

The formula was

My $\displaystyle a=1$ and my $\displaystyle u=\frac{1}{\sqrt{x}}$, so I plugged those into the formula.

What I got out was this:

$\displaystyle \int_0^2 \sqrt{1^2+\frac{1}{\sqrt{x}}^2} du=\frac{\frac{1}{\sqrt{x}}}{2}\sqrt{1^2+\frac{1}{ \sqrt{x}}^2}+\frac{1^2}{2}\ln(\frac{1}{\sqrt{x}}+\ sqrt{1^2+\frac{1}{\sqrt{x}}^2})=$

$\displaystyle \frac{1}{2\sqrt{x}}\sqrt{1+\frac{1}{\sqrt{x}}^2}+\ frac{1}{2}\ln(\frac{1}{\sqrt{x}}+\sqrt{1+(\frac{1} {\sqrt{x}})^2})$

This is what I got to. Now, I need to evaluate it from 0 to 2. But when I plug in 2 for the value of x, my answer is undefined. Did I do something wrong? Is there a different way to do this? Please...PLEASE help!