Results 1 to 5 of 5

Thread: Definite Integral

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    61

    Definite Integral

    $\displaystyle \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$
    break it up$\displaystyle 6\int_{x=1}^{x=2} (e^{3x}dx)-\int_{x=1}^{x=2} (\frac{1}{x}dx)$
    $\displaystyle 2\int (e^udu-\ln{x}+c$

    is this right so far, not sure what to do after.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    Quote Originally Posted by Jim Marnell View Post
    $\displaystyle \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$
    break it up$\displaystyle 6\int_{x=1}^{x=2} (e^{3x}dx)-\int_{x=1}^{x=2} (\frac{1}{x}dx)$
    $\displaystyle 2\int (e^udu-\ln{x}+c$

    is this right so far, not sure what to do after.
    Well, yes, but this is a definite integral. So after you find the antiderivative $\displaystyle F(x),$ you need to apply the fundamental theorem of calculus by taking $\displaystyle F(2)-F(1).$

    As for finding the antiderivative, what is $\displaystyle \int e^x\,dx?$ This is a basic integration rule, and replacing the $\displaystyle x$ with a $\displaystyle u$ won't make any difference.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    61
    yea i dont know why i put u in for the x, but, so now i would just insert the 2
    (x) values in and subtract them
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    Quote Originally Posted by Jim Marnell View Post
    yea i dont know why i put u in for the x, but, so now i would just insert the 2
    (x) values in and subtract them
    No, no, your substitution was fine. I was just pointing out the integration rule:

    $\displaystyle \int e^x\,dx=e^x+C.$

    So,

    $\displaystyle \int_1^2\left(6e^{3x}-\frac1x\right)dx$

    $\displaystyle =6\int_1^2e^{3x}\,dx-\int_1^2\frac1x\,dx$

    $\displaystyle =2\int_{x=1}^{x=2}e^u\,du-\bigg[\ln\lvert x\rvert\bigg]_1^2$

    $\displaystyle =2\bigg[e^u\bigg]_{x=1}^{x=2}-\ln 2$

    $\displaystyle =2\bigg[e^{3x}\bigg]_1^2-\ln 2$

    $\displaystyle =2(e^6-e^3)-\ln 2$

    $\displaystyle =2e^3(e^3-1)-\ln 2$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    61
    Thanks!, thats what i ended up getting so hopefully i understand this concept now
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Dec 5th 2011, 05:21 PM
  2. Replies: 4
    Last Post: Apr 13th 2011, 02:08 AM
  3. definite integral/ limit of integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 22nd 2010, 04:00 AM
  4. definite integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 1st 2008, 12:04 PM
  5. Definite integral ln(1+t)dt
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 26th 2008, 06:02 AM

Search Tags


/mathhelpforum @mathhelpforum