1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Just 2 integral problems today!

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

    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


    User Avatar
    Science Advisor

    1. I don't have any problem with the square root sign but is that e^(t-3) or
    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


    User Avatar
    Science Advisor
    Homework Helper

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook