Results 1 to 6 of 6

Math Help - Integration

  1. #1
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83

    Integration

    integral(16/(xln(4x)dx [e, e^7]

    Can anyone help?

    I tried putting it in my calculator and it is unable to work it out.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    16\int_e^{e^7} \frac{{\color{blue}1}}{{\color{blue}x} {\color{red}\ln (4x)}} \ {\color{blue}dx}

    Let: {\color{red}u = \ln (4x)} \ \Rightarrow \ du = \frac{1}{4x} \cdot 4 \ dx \ \Leftrightarrow \ {\color{blue}du = \frac{1}{x} \ dx}

    Don't forget to change your upper and lower limits.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    Would a final answer of 20 be correct? If not how do I get the bounds because Im not sure if I did that right.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by CalcGeek31 View Post
    Would a final answer of 20 be correct? If not how do I get the bounds because Im not sure if I did that right.
    No. Show your work so that your mistakes can be explained.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member CalcGeek31's Avatar
    Joined
    Aug 2008
    Posts
    83
    Okay. So to get the new bounds you plug into u right?

    so u = ln(4(e)) for the bottom

    and u = ln(4(e^7)) for the top?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by CalcGeek31 View Post
    Okay. So to get the new bounds you plug into u right?

    so u = ln(4(e)) for the bottom

    and u = ln(4(e^7)) for the top?
    Correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum