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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.

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  • $\begingroup$ Your application of the fundamental theorem merely erased the integral; does this seem reasonable? $\endgroup$ Commented Oct 18, 2017 at 0:45
  • $\begingroup$ Your second line should be $y'$ which makes a world of difference. $\endgroup$ Commented Oct 18, 2017 at 0:45
  • $\begingroup$ So the question is: What is the function here? $\endgroup$ Commented Oct 18, 2017 at 0:58

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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.

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  • $\begingroup$ Ah - FTC means that it is y' not y. Thanks! $\endgroup$ Commented Oct 18, 2017 at 4:18

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