# Thread: Arc length of cubic function

1. ## Arc length of cubic function

So I know the formula for arc length through integration, but I'm a bit stuck with how you integrate from line 3 to 4. The original function is x^3 - 5x^2 + 2x + 24

2. ## Re: Arc length of cubic function

You don't go from line 3 to line 4, that's incorrect. It looks like some one thinks that the anti-derivative of $\displaystyle u^{1/2}dx$ is $\displaystyle \frac{2}{3}u^{3/2}\frac{du}{dx}$ but that is clearly wrong. It appears they are confusing the "chain rule" for differentiation with integration. If you are differentiating f(u) with respect to x, that is f'(u) times du/dx. But here we are not differentiating we are integrating. "Theoretically" we could divide by du/dx rather than multiplying but if it is a function of x we cannot just put that derivative into the integral.

It is true that if we had $\displaystyle \int (9x^4- 60x^3+ 88x^2+ 40x+ 5)^{1/2} (36x^3- 180x^3+ 176x+ 40)dx$ then we could "substitute" $\displaystyle u= 9x^4- 60x^3+ 88x^2+ 40x+ 5$ so that $\displaystyle du= (36x^3- 180x^3+ 176x+ 40)dx$ and then the integral becomes $\displaystyle \int u^{1/2}du= \frac{3}{2}u^{3/2}+ C= \frac{3}{2}(9x^4- 60x^3+ 88x^2+ 40x+ 5)+ C$. "Substitution" is, in a sense, the analog of the "chain rule" but that derivative has to already be in the integral.

3. ## Re: Arc length of cubic function

Thanks, that's exactly what I thought, but I was a bit sceptical because these notes have never been wrong before :/
Also I entering the function into Wolfram's arc length calculator and it finds the same length as these notes. How would you continue from line 3?

4. ## Re: Arc length of cubic function

You would get out a calculator and allow it to determine an accurate approximation ... many arc length integrals (like this one) have no elementary, closed-form antiderivative.

5. ## Re: Arc length of cubic function

Ok, thanks for the help