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Homework Help: How to find the integral of e^(sqrt(x)) as x goes from 0 to 1

  1. Dec 20, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_0^1 e ^(sqrt. x) dx [/tex]

    2. Relevant equations

    3. The attempt at a solution

    i don´t know what to let u equal to... maybe sqrt.x? but what good would that do when u is differentiated?
  2. jcsd
  3. Dec 20, 2007 #2
    hmm, your latek messed up

    [tex]\int_{0}^{1} e^{\sqrt{x}}dx[/tex]

  4. Dec 20, 2007 #3
    yes, thanks... i´m still figuring out how to use it :smile:
  5. Dec 20, 2007 #4
    Let u=[tex]\sqrt{x}[/tex]

    then, use another "simple" substitution so that you only have 1 variable.

    then, you will need to use integration by parts.
  6. Dec 20, 2007 #5
    i still haven´t figured the problem itself though...
  7. Dec 20, 2007 #6
    shoot, at least you've already almost got it down. there are ppl with like 500+ posts and still don't use latek, hurts my eyes :p
  8. Dec 20, 2007 #7
    that was before learning integration by parts so i guess i´m not supposed to use it
  9. Dec 20, 2007 #8
    what technique are you not allowed to use?
  10. Dec 20, 2007 #9
    integration by parts... thats why i´m stuck
  11. Dec 20, 2007 #10
    there's no way around this problem without the use of integration by parts, want to work this problem together by use of parts?
  12. Dec 20, 2007 #11
    do i let u = [tex]e^\sqrt{x}[/tex] ?
  13. Dec 20, 2007 #12
    the derivative of your substitution would then become



    messy. then you will need to take the ln of your u-substitution so you can get your problem in terms of 1 variable.
    Last edited: Dec 20, 2007
  14. Dec 20, 2007 #13
    so then u is just [tex]\sqrt{x}[/tex] ?
  15. Dec 20, 2007 #14
    yes and i can help you through integration by parts, b/c we can't integrate this problem by simple substitutions.
  16. Dec 20, 2007 #15
    ok... u = [tex]\sqrt{x}[/tex]
    and du = [tex]du=\frac{1}{2\sqrt{x}}dx[/tex]
  17. Dec 20, 2007 #16
    what would dv be?
  18. Dec 20, 2007 #17
    well first, let's make it a "t-substitution" b/c we'll need to use U and V for parts.



    now we need our integration in terms of 1 variable, so



    [tex]\int e^{\sqrt{x}}dx \rightarrow 2\int te^{t}dt[/tex]

    now we do integration by parts.
    Last edited: Dec 20, 2007
  19. Dec 20, 2007 #18
    no, the last step was a simple substitution ... this next step we will use parts.
  20. Dec 20, 2007 #19
    wait.. i´m confused... now i have [tex]\sqrt{x}[/tex] as u
    and then i have to square it?
    Last edited: Dec 20, 2007
  21. Dec 20, 2007 #20
    [tex]2\int te^{t}dt[/tex]

    [tex]\begin{array}{cc}u=t & dV=e^{t}dt \\ du=dt & V=e^{t}\end{array}[/tex]

    i'm sure you can take it from here! (judging from your other post)
  22. Dec 20, 2007 #21
    omg i´m sorry hehe
    where did u get the t from?
  23. Dec 20, 2007 #22
    oh my god.. nevermind... i got it
  24. Dec 20, 2007 #23
    i didn´t read your other post!! thankss hehehe :smile:
  25. Dec 20, 2007 #24
    i just updated it again, idk why i squared it. lol sorry, it's not supposed to be squared.

    at times you will need to square it, but not in this situation. (you could, but you'll get the same result which is a waste of time)
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