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Integration by parts

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi, I've been having trouble solving the following problem, please help me.

    Question:
    (integration from 1 to 4) e^(x^(1/2))dx


    2. Relevant equations



    3. The attempt at a solution

    So far, i have done the following:

    u = e^(x^(1/2))
    du = (1/2)(x^(-1/2))e^(x^(1/2))dx

    dv = dx
    v = x

    so after applying the integration by parts formula, I got...

    xe^(x^(1/2)) - (integration sign)x(1/2)(x^(-1/2))e^(x^(1/2))
    dx

    The integration of the second part looks wrong because it seems that I missed a step somewhere and made it more complex. Thank you for helping me.
     
  2. jcsd
  3. Jan 27, 2007 #2

    Curious3141

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    Just to clarify, you're supposed to find [tex]\int_1^4{e^{\sqrt{x}}}dx[/tex] right?

    Start by evaluating the indefinite integral. The first step is to make an obvious substitution like [tex]x = u^2[/tex]. After you simplify that and put everything in terms of u, you'll find an expression that can more easily be integrated by parts. :smile:
     
  4. Jan 27, 2007 #3
    this isn't a by-parts questions.

    try letting z = sqrt(x)
     
  5. Jan 27, 2007 #4
    oh i see now. so after doing the subsitution for sqrt of x = y. I end up with integration of (e^y)(2ydy). which then do integration by parts through u=y and dv = e^y. thus i get x^(1/2)e^(x^(1/2))-e^(x^(1/2)). so after doing the limits it is 2e^2. is that how it is?
     
    Last edited: Jan 27, 2007
  6. Jan 27, 2007 #5

    Curious3141

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    You forgot a factor of 2. EDITED : You changed your post now, the final answer is correct, but the symbolic expression is still missing a factor of two. I'd suggest factoring it and tidying it up to look nicer when you present your answer.
     
    Last edited: Jan 27, 2007
  7. Jan 27, 2007 #6

    Gib Z

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    Hahaha lol, just remember integration by parts questions are for products :)
     
  8. Jan 28, 2007 #7
    thanks for helping me!!
     
  9. Jan 28, 2007 #8

    Gib Z

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    Here at physicsforums, We have no lives :P We have nothing else to do :D
     
  10. Jan 28, 2007 #9

    HallsofIvy

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    Not always. The standard way to integrate log(x) is by parts, letting u= 1and dv= log(x)dx.
     
  11. Jan 28, 2007 #10
    1*log(x) is a product isn't it? :p
     
  12. Jan 28, 2007 #11

    Gib Z

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    Whats wrong with f(x)=1 :(

    lol
     
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