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Just 2 integral problems today!

  1. Mar 22, 2005 #1
    Ok, ill try to explain what i did:

    1.
    integral of [square root of( e^t-3)dt]

    Sorry i didn't know how to do the square roots symbol and the integral symbol. Anyways, I tried to set e^t-3 as u and then got dt=du/e^t. Then I plugged in dt and couldn't go on after that.

    2. integral of x^2[e^(x^3)]dx

    so uh.... u substitution? but then what?

    Ok basically, i think i don't know what your supposed do when u need to do a u-sub with e^something. Your help would be most appreciated.
     
  2. jcsd
  3. Mar 22, 2005 #2

    HallsofIvy

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    1. I don't have any problem with the square root sign but is that e^(t-3) or
    (e^t)-3?
    If it is e^(t-3) then sqrt(e^(t-3))= (e^(t-3))^(1/2)= e^((t-3)/2). Let u= (t-3)/2 so
    du= (1/2)dt or dt= 2du. The integral becomes 2 integral e^u du.

    2. Yes, a substitution- seeing that "complicated" x^3 in the exponent and x^2 multiplying, you should immediately think of u= x^3 (NOT u= e^something- the e is not the problem!). Then du= 3 x^2 dx or (1/3)du= x^2 dx so the integral becomes
    (1/3) integral e^u du.
     
  4. Mar 22, 2005 #3
    Well, thinking of [tex] u = e^{x^3}[/tex] wouldn't hurt in this case regardless :rofl:
     
  5. Mar 22, 2005 #4
    thnx, i see how the second one works now.

    For the first one, its actually square root of (e^t)-3. Sorry i didnt make that clear.
     
  6. Mar 23, 2005 #5

    dextercioby

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    Then make the substitution
    [tex] e^{t}=u [/tex]
    ,then the substitution
    [tex]u=v^{2}+3 [/tex]

    Daniel.
     
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