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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 -

    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 -

    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 -

    If it's any help, the substitution 6x-2 = sinh(u) will simplify things a bit.

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