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Math Help - maclaurin series with radical integral

  1. #1
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    maclaurin series with radical integral

    I am trying to get the first four terms of the Maclaurin Series for the following functiong:

    integral from 0 to x of sqrt(1+t^3) dt

    Sadly, I'm stuck trying to find the integral so that I can get the first f(0) term. Using u-substitution I get:

    u = 1 + t^3
    du = 3t^2 dt
    dt= du / 3t^2

    I think that I need to put 3t^2 in terms of u to be able to continue with the integration, but I cannot see a connection. If I make u = t^3 then would 3t^2 = 3u^2/3 ??

    thanks!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    hello again

    remember any integral from 0 to 0 or in fact a to a is 0.

    No one can do the integral
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  3. #3
    MHF Contributor
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    Hi

    The Maclaurin series of f is \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!}\:x^n

    f(0) = 0

    f'(x) = \sqrt{1+x^3} \implies f'(0) = 1

    You can show that f^{(2)}(0) = 0 and f^{(3)}(0) = 0

    Spoiler:
    Therefore the first four terms of the Maclaurin Series is ... x
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