Results 1 to 11 of 11

Thread: [SOLVED] Help needed with a definite integral

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    68

    [SOLVED] Help needed with a definite integral

    <br />
\int_0^1 x^2e^{x^{1/5}}~dx<br />

    I assume that this is solved using integration by parts, but what should I choose for u and dv?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    I'm not sure on this but I think you can do integration by parts if you make a substituion.

    w = x^{\frac{1}{5}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 11rdc11 View Post
    I'm not sure on this but I think you can do integration by parts if you make a substituion.

    w = x^{\frac{1}{5}}
    yeah, but it would be crazy! you would have to do it like 14 times. the resulting integral is:

    5 \int u^{14}e^u~du
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    How would that work? Would I just solve the problem like normal, but totally ignore the x^1/5 part until the very end, and simply swap w and x^1/5?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Quote Originally Posted by Jhevon View Post
    yeah, but it would be crazy! you would have to do it like 14 times. the resulting integral is:

    5 \int u^{14}e^u~du
    Lol yea I just noticed. Umm can't we use a reduction formula after?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Quote Originally Posted by redman223 View Post
    How would that work? Would I just solve the problem like normal, but totally ignore the x^1/5 part until the very end, and simply swap w and x^1/5?
    Yep you would just solve the problem like normal and you don't have to worry bout swapping x^{\frac{1}{5}} for w because all you have to do is change your lower and upper bound to correlate with w.

    For example

    Your lower bound

    0^{\frac{1}{5}} = 0

    and upper bound

    1^{\frac{1}{5}} = 1

    5 \int_0^1 w^{14}e^w~dw
    Last edited by 11rdc11; Sep 7th 2008 at 08:23 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    So what would be u in this case? Would that be the same thing as the w mentioned before?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Oops made corrections. I used variable w so I wouldn't get mixed up when doing integration by parts, since integration by parts normally uses u and v.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Just put this integral in the TI89 and the solution is very long
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 11rdc11 View Post
    Just put this integral in the TI89 and the solution is very long
    yup. when i saw the integral, i thought it would be messy so i put it into the integrator, the solution there was messy as well.

    i hope there is a mistake in this problem

    maybe there is a shortcut. we can wait on Kriz or one of the other integral people to see
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    \int_0^1 x^2 e^{x^{1/5}}~dx ~=~ 5\int_0^1 u^{14} e^u~du

    \int u^{14}e^u~du = u^{14} e^u - 14 u^{13}e^u + 14.13 u^{12}e^u ..... - 14! u^1 e^u+14! e^u~+  C = \sum_{i=0}^{14}u^i e^u (-1)^i \frac{14!}{i!}~+C

    Evaluating this from 0 to 1 and multiplying by 5 gives us 5\left [ \left ( \sum_{i=0}^{14} e (-1)^i \frac{14!}{i!} \right )-14! \right ] \approx 0.8526185
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Apr 7th 2010, 03:15 PM
  2. [SOLVED] Definite Integral Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 4th 2009, 11:27 AM
  3. Need this problem solved definite integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 26th 2009, 12:50 PM
  4. Definite integral to be solved
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Jun 8th 2008, 12:31 AM
  5. [SOLVED] definite integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 9th 2008, 10:58 PM

Search Tags


/mathhelpforum @mathhelpforum