Results 1 to 6 of 6

Math Help - Compute the Integral?

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    14

    Compute the Integral?

    Compute integral x on top, -1 on bottom, sin(t^2) times 2t dt


    I have a calc final in three hours and I don't know how to solve this problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    \int_{-1}^x 2t\sin t^2 dt

    Note that \frac{d}{dt}\left(-\cos t^2 \right)=2t\sin t^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,674
    Thanks
    445
    Quote Originally Posted by masterofcheese View Post
    Compute integral x on top, -1 on bottom, sin(t^2) times 2t dt


    I have a calc final in three hours and I don't know how to solve this problem.
    \int_{-1}^x \sin(t^2) \cdot 2t \, dt

    substitution ... let u = t^2

    du = 2t \, dt

    substitute and reset the limits of integration ...

    \int_1^{x^2} \sin(u) \, du = \left[-\cos(u)\right]_{1}^{x^2}

    \cos(1) - \cos(x^2)
    Last edited by skeeter; June 6th 2009 at 01:38 PM. Reason: corrected the typo that Spec caught ... thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2009
    Posts
    53
    Skeeter, I understand your solution except for the new limits: 1 and x^2.

    Why did you introduce new limits to the integral and how did you get them?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    He made the substitution u=t^2, and since -1 \leq t \leq x we get 1 \leq u \leq x^2

    His final answer is correct, but he made a typo in the process by the way. \int \sin x dx = \textcolor{red}{-}\cos x +C
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,639
    Thanks
    1592
    Awards
    1
    I have one negative comment. I find it absolutely astounding that someone three hours away from a calculus final exam would not see at once that
    \int {2t\sin \left( {t^2 } \right)dt =  - \cos \left( {t^2 } \right)} .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute this integral.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 6th 2010, 02:25 PM
  2. compute integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 26th 2010, 12:12 AM
  3. Compute an integral
    Posted in the Math Challenge Problems Forum
    Replies: 5
    Last Post: September 22nd 2009, 03:32 AM
  4. Compute the line integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 25th 2009, 02:36 PM
  5. Work compute with integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 14th 2008, 01:13 PM

Search Tags


/mathhelpforum @mathhelpforum