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Math Help - [SOLVED] Finding the arclength of a curve

  1. #1
    Cidious
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    Exclamation [SOLVED] Finding the arclength of a curve

    Hi I am having a problem finding the arclength of the function
    f(x) = (1/3)(x^2 +2)^(3/2) x e [0,1]

    I know the arclength formula but when I plug in the values I come down to:

    ds = [integral from 0 to 1] underoot(1 + x^4 + 2x^2)dx

    and can't figure out what to do from here, can you please show me steps on what to do next ?

    thanks, EJ
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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by Cidious View Post
    Hi I am having a problem finding the arclength of the function
    f(x) = (1/3)(x^2 +2)^(3/2) x e [0,1]

    I know the arclength formula but when I plug in the values I come down to:

    ds = [integral from 0 to 1] underoot(1 + x^4 + 2x^2)dx

    and can't figure out what to do from here, can you please show me steps on what to do next ?

    thanks, EJ
    isn't it the formula is something like this?

    L = \int_a^b \sqrt{1 + \left[ f'(x) \right]^2} \, dx

    just do your integration..
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Santiago, Chile
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    kalagota, he actually did that, the problem is the integral:

    \int_0^1 {\sqrt {x^4  + 2x^2  + 1} \,dx}  = \int_0^1 {\sqrt {\left( {x^2  + 1} \right)^2 } \,dx}  = \int_0^1 {\left| {x^2  + 1} \right|\,dx} .

    Since x^2+1>0,\,\forall\,x\in\mathbb R, we can remove the absolute value bars and just write \int_0^1 {\left( {x^2  + 1} \right)\,dx}, the rest is routine.

    (Cidious see my signature for LaTeX typesetting.)
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