For my calculus class we are learning how to find the arc length of a function. I understand the basic premise which is to find the derivative of the function and integrate between the bounds for the square root of one plus the derivative.
The problem I am stuck on though has the function as an integral.
$$y=\int_1^x\sqrt{t^3-1}\\1\le x\le4$$
I think by using the fundamental theorem of calculus I can say that:
$$y=\sqrt{x^3-1}$$
The derivative with respect to $x$ is:
$$y'=\frac{3x^2}{2\sqrt{x^3-1}}$$
When I try to calculate the arc length from here I get stuck. My equation looks like this:
$$\int_{1}^{4}\sqrt{\frac{9x^4+4x^3-4}{8x^3-8}}dx$$
That thing is ugly so I think I went wrong somewhere but I am not sure which step. Any help would be greatly appreciated.
By the FTC, $y' = \sqrt{x^3-1}$. Now, $\sqrt{1+(y')^2} = \sqrt{1+x^3-1} = x^{3/2}$. You need to integrate this now to get the total length, and this is really doable.