Results 1 to 6 of 6

Thread: Integration

  1. #1
    Len
    Len is offline
    Member
    Joined
    Mar 2008
    Posts
    93
    Thanks
    1

    Integration

    Alright so I'm a little confused with basic integration.

    $\displaystyle \int\frac{dx}{x}=ln|x|+c$
    But what if theres a constant infront of dx.
    For example
    $\displaystyle \int\frac{7dx}{x}=\int\frac{1}{x}*7dx=?$

    What changes then?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    As 7 is a constant, it is not affected by the integration, so you can do this:

    $\displaystyle
    \int\frac{7dx}{x}$

    $\displaystyle = 7\int\frac{dx}{x}$

    $\displaystyle = 7ln|x| + c$
    $\displaystyle = ln |x^7| + c
    $

    or if you like to think of it this way:

    $\displaystyle 7\int\frac{7dx}{x}(\frac{1}{7})$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Len
    Len is offline
    Member
    Joined
    Mar 2008
    Posts
    93
    Thanks
    1
    Quote Originally Posted by Gusbob View Post
    As 7 is a constant, it is not affected by the integration, so you can do this:

    $\displaystyle
    \int\frac{7dx}{x}$

    $\displaystyle = 7\int\frac{dx}{x}$

    $\displaystyle = 7ln|x| + c$
    $\displaystyle = ln |x^7| + c
    $

    or if you like to think of it this way:

    $\displaystyle 7\int\frac{7dx}{x}(\frac{1}{7})$
    Thanks, another quick question,
    $\displaystyle \int\frac{7dx}{4x+2}=?=7ln|4x+2|$
    Or do I have to multiply it by the derivative of 4x+2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    $\displaystyle
    \int\frac{7dx}{4x+2}
    $

    You multiply by the derivative, but don't forget to divide it.

    For this one, I'll take the 7 out again and use the substitution u = 4x + 2. du = 4 dx.

    You need a 4 on top, so multiply by $\displaystyle \frac{4}{4}$.

    $\displaystyle
    7\int\frac{4dx}{4x+2}(\frac{1}{4})
    $

    Since 1/4 is a constant, I took it out like I did with the 7. (Note: only constants that are multiplied or divided can be taken out, addition or subtraction of constants are integrated.)

    $\displaystyle \frac{7}{4} \int\frac{du}{u}$ (since du = 4dx; u = 4x+2)

    =$\displaystyle \frac{7}{4} ln|4x+2| +c$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Len
    Len is offline
    Member
    Joined
    Mar 2008
    Posts
    93
    Thanks
    1
    Quote Originally Posted by Gusbob View Post
    $\displaystyle
    \int\frac{7dx}{4x+2}
    $

    You multiply by the derivative, but don't forget to divide it.

    For this one, I'll take the 7 out again and use the substitution u = 4x + 2. du = 4 dx.

    You need a 4 on top, so multiply by $\displaystyle \frac{4}{4}$.

    $\displaystyle
    7\int\frac{4dx}{4x+2}(\frac{1}{4})
    $

    Since 1/4 is a constant, I took it out like I did with the 7.

    $\displaystyle \frac{7}{4} \int\frac{du}{u}$ (since du = 4dx; u = 4x+2)

    =$\displaystyle \frac{7}{4} ln|4x+2| +c$
    Thanks a ton, things are a lot clearer now!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    I dont know it this helps

    but. $\displaystyle \int\frac{1}{ax+b}dx$=$\displaystyle \frac{1}{a}\ln(ax+b)\$ using the same technique by isolating the derivative of the quantity
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

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

Search Tags


/mathhelpforum @mathhelpforum