posted by W .
How do I find the length of curve f(x)=3x^2-2x+3 on the interval [1,4]? I have no idea on how to start this. Anyways, thanks in advance!
The length of a curve for the interval [x1,x2] is given by the integral:
and where dy/dx = 6x-2
For the present case, as an estimate,
y(1)=4, y(4)=43, so you'd expect the length to be a little over 39, or more precisely, 39.1414... approximately.
Post if you need more help.
length of curve
= ∫√(1 + (dy/dx)^2 ) dx from a to b
so for yours
length = ∫(1 + (6x-2)^2 )^(1/2) dx from 1 to 4
= ∫(36x^2 -24x + 5)^(1/2) dx
Now the mess begins.
At this point I will admit that my integration skills are so rusty, that I can't see a way out.
If it's any help, the substitution 6x-2 = sinh(u) will simplify things a bit.