Posted by **W** on Tuesday, November 15, 2011 at 9:29pm.

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!

- math -
**MathMate**, Tuesday, November 15, 2011 at 10:07pm
The length of a curve for the interval [x1,x2] is given by the integral:

x2

∫sqrt(1+(dy/dx)^2)dx

x1

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.

- math -
**Reiny**, Tuesday, November 15, 2011 at 10:10pm
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.

- math -
**Steve**, Wednesday, November 16, 2011 at 5:21am
If it's any help, the substitution 6x-2 = sinh(u) will simplify things a bit.

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