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Stumped on Definite Integral Question

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Definite integral, from 0 to 3, of Square root of 1+t^3.


    2. Relevant equations
    Tried substitution

    3. The attempt at a solution
    There should be a simple way to do this but I can't seem to figure it out. Tried the substitution and whatnot but couldn't reach an answer.
     
  2. jcsd
  3. Jun 19, 2012 #2
    u-sub won't work because the derivative of the inside is not outside.

    trig-sub won't work because it should be t^2.

    You might try u^2=1+t^3, though I haven't tried it so no idea.

    i would say maybe some extra special clever application of integration by parts, but i wouldn't bet on it.

    How did the problem arise, are you sure this is the problem you are supposed to do?

    Is it possible that using a computer is expected?
     
  4. Jun 19, 2012 #3
    MAPLE output for both the definite and indefinite integral looks ridiculous with all kinds of special functions.
     
  5. Jun 19, 2012 #4
    Same with Wolfram. This integral is not expressable in elementary terms (verified by Risch's algorithm.) It involves the elliptic integral function. If that is acceptable, you can do the problem; but it will still be ridiculously long.
     
  6. Jun 19, 2012 #5
    As much as I have a warm place in my heart for MAPLE (created at UW?), I would just plug it in online at wolfram. It mentions something about the hypergeometric series, and about 7.3.
     
  7. Jun 19, 2012 #6
    Haha yea, I use both to be honest, certain things are easier for me to find on MAPLE and in this instance I had it open already.
     
  8. Jun 19, 2012 #7
    Well, the actual question is different from the question I asked. I didn't want to post the actual one because I partially solved it. Here it is though if there is in fact a different approach. I did find the derivative and was trying to get f^-1(0) by saying f^-1(0) = x so f(x) = 0, but couldn't solved the integral to be able to equal it to 0 to get x.

    If f(x) = integral, from 3 to x , of square root of (1+t^3) dt, find (f^-1)'(0).

    By the way, I would like to have the simplest and cleanest approach to this to be efficient.

    Thanks
     
  9. Jun 19, 2012 #8

    Curious3141

    User Avatar
    Homework Helper

    Let [itex]y = f(x) = \int_3^x \sqrt{1+t^3}dt[/itex]

    Use the Fundamental Theorem of Calculus to figure out [itex]\frac{dy}{dx} = f'(x)[/itex]

    Now, if [itex]y = f(x)[/itex], then [itex]f^{-1}(y) = x[/itex] and [itex]{(f^{-1})}'(y) = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(x)}[/itex].

    You're asked to determine [itex]{(f^{-1})}'(0)[/itex], so first figure out what (obvious) value of [itex]x[/itex] would make [itex]y[/itex] zero. The rest is trivial.
     
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