Is there a way to measure the arc-length of a curve created by a 4-degrees polynomial, between a given range?
For example, I want to measure the length of the polynomial $-\frac{1}{200}x^3+x+1$ between [0,10] (the blue line in the image below):
This is just an example, the only thing I know is that the range always starts at $x=0$, and the polynomial is a 4-degree polynomial. Is there a formula to measure the arc-length of the section?
On
This may be a dumb question, but isn't $\int_{a}^{b}\sqrt{1+(1- 3t^2/200)^2 } dt$ = $\sqrt{\int_{a}^{b}1+(1- 3t^2/200)^2 } dt$ which is easily expanded to $\sqrt{\int_{a}^{b}(2-3t^2/100 + 9t^4/40,000)} dt $ and integrated as a polynomial to $\sqrt{2t - t^3/100 + 9t^5/(5*40,000) +C }$ between a and b?
This is my first use of Latex, so bear with me!
The way to calculate arc-length of a curve given parametrically is to integrate speed, where speed is the magnitude of the velocity vector.
For a function $f(x)$ on an interval $[a, b]$, the parametric curve is $$t \to (t, f(t)),$$ the velocity is $$t \to (1, f'(t)),$$ the speed is $$t \to \sqrt{1+\left(f'(t)\right)^2},$$ and the arc-length is therefore $$\int_a^b \sqrt{1+\left(f'(t)\right)^2}\mathrm{dt}.$$ So for your example, $f(x) = -\frac{1}{200}x^3+x+1$, $f'(x) = -\frac{3}{100}x^2+1$, and the arclength between $0$ and $10$ is $$\int_0^{10} \sqrt{1+\left(1-\frac{3t^2}{200}\right)^2}\mathrm{dt}.$$ I don't know whether that integral can be solved analytically, but it is worth noting that if you just need the number, Wolfram Alpha does the algebra of deriving the integrand for you.