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Integral problem

  1. Oct 4, 2004 #1
    I have this integral: (The first is the original, the one I need to solve)
    http://www.absinthen.dk/math.jpg [Broken]

    Well, I have a program that can calculate it for me, but I need to do it in hand - but even though I keep trying, I just don't end up with the result my program says it is, which is:

    I've been trying everything, but I going crazy very soon :cry:

    I really hope you guys can give me a hint, of what may be wrong.

    - Ylle
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Oct 4, 2004 #2
    I don't understand exactly what it is you've done to the integral, but...

    [tex]\int e^{\sqrt{x}} x^{-1/2} dx = \int e^{\sqrt{x}} \cdot \frac{dx}{\sqrt{x}}[/tex]

    Let [tex]u = \sqrt{x}[/tex]. Then [tex]\frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{x}}[/tex], so 2du = 1/sqrt(x) dx. The integral turns in to:

    [tex]\int e^{u} \cdot 2 du[/tex]

    After finding an antiderivative, putting in the limits should be easy... ;)
    Last edited: Oct 4, 2004
  4. Oct 4, 2004 #3
    hehe, and i don't understand what you are doing :D
    I don't think they teach us to solve the integral the same way, as they do to you :(

    But another example:
    http://www.absinthen.dk/math2.jpg [Broken]

    This integral is solved correctly this time, and I've done the same thing as I would do in the one I gave you. But in the one I gave you, it just wont do as I want it to do :confused:
    Last edited by a moderator: May 1, 2017
  5. Oct 4, 2004 #4


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    I really doubt that anyone "taught" you to replace "x" with "t" without saying what in the world the relationship between x and t is!

    I also note that when you make the substitution, there is no "dt" in the integral. You are not being sufficiently careful- that may be where your problem is.

    State clearly what substitution you are making and how you are replacing dx.
    Last edited by a moderator: Oct 5, 2004
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