Results 1 to 6 of 6

Math Help - Different forms of same integral?

  1. #1
    Xei
    Xei is offline
    Newbie
    Joined
    Apr 2008
    Posts
    10

    Different forms of same integral?

    Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

    \int \frac{1}{3x}dx

    Normally I would do this as following

    \frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k

    However, is it not true that I can also make this of the form

    \int \frac{f'(x)}{f(x)}dx

    via

    \frac{1}{3}\int \frac{3}{3x}dx

    which via a simple u sub yields

    \frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3}  \ln{3}+j

    Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?

    Also, how would I go about expressing the answers

    y=\frac{1}{3}\ln{x}+k and \frac{1}{3}\ln{3x}+j

    as explicit functions of x?

    Ta muchly.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Xei View Post
    Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

    \int \frac{1}{3x}dx

    Normally I would do this as following

    \frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k

    However, is it not true that I can also make this of the form

    \int \frac{f'(x)}{f(x)}dx

    via

    \frac{1}{3}\int \frac{3}{3x}dx

    which via a simple u sub yields

    \frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3}  \ln{3}+j

    Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?

    Mr F says: They are equivalent. j + ln 3 is just as arbitrary as k .....

    Also, how would I go about expressing the answers

    y=\frac{1}{3}\ln{x}+k and \frac{1}{3}\ln{3x}+j

    as explicit functions of x?

    Mr F says: They already are explicit functions of x. Do you mean how to express them as explicit functions of y ....?

    Ta muchly.
    ..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,924
    Thanks
    332
    Awards
    1
    Quote Originally Posted by Xei View Post
    Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

    \int \frac{1}{3x}dx

    Normally I would do this as following

    \frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k

    However, is it not true that I can also make this of the form

    \int \frac{f'(x)}{f(x)}dx

    via

    \frac{1}{3}\int \frac{3}{3x}dx

    which via a simple u sub yields

    \frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3}  \ln{3}+j

    Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
    Since j and k are just arbitrary constants one can be set equal to the other.

    By the way, just so that you are aware...
    \int \frac{1}{x}~dx = ln|x| + C

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Xei
    Xei is offline
    Newbie
    Joined
    Apr 2008
    Posts
    10
    Yeah it's all right, I just wasn't bothered with the mod bit.

    I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.

    Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?

    Mr F: Yeah, y please. As simple as it'll go.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,735
    Thanks
    642
    Hello, Xei!

    You are basically correct . . .


    \int \frac{1}{3x}\,dx \;=\;\frac{1}{3}\int \frac{1}{x}\,dx\;=\;\frac{1}{3}\ln{x}+K .[1]

    \frac{1}{3}\int\frac{3}{3x}\,dx \;=\;\frac{1}{3}\ln{3x}+J .[2]


    Is it the case that both are valid, but the constants K and J are different? .
    . . . yes

    Is K\:=\:J+\ln3? .
    . . . um, not quite

    From [2], we have: . \frac{1}{3}\ln(3x) + J \;=\;\frac{1}{3}\left[\ln(3) + \ln(x)\right] + J \;=\;\frac{1}{3}\ln(3) + \frac{1}{3}\ln(x) + J

    . . = \;\frac{1}{3}\ln(x) + \underbrace{\frac{1}{3}\ln(3) + J}_{\text{a constant}} \;=\;\frac{1}{3}\ln(x) + K

    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 Xei View Post
    Yeah it's all right, I just wasn't bothered with the mod bit.

    I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.

    Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?

    Mr F: Yeah, y please. As simple as it'll go.
    y=\frac{1}{3}\ln{|x|}+k \Rightarrow 3y - 3k = \ln |x| \Rightarrow e^{3y - 3k} = x

    \Rightarrow x = e^{3y}\, e{-3k} = A e^{3y}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Definition about the sum of 1-forms
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 26th 2010, 04:49 AM
  2. Indeterminate Forms
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 8th 2010, 03:31 PM
  3. forms of groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 6th 2008, 12:18 PM
  4. 2-forms
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 10th 2007, 01:52 PM
  5. Equivalent Forms
    Posted in the Algebra Forum
    Replies: 13
    Last Post: July 25th 2006, 03:46 AM

Search Tags


/mathhelpforum @mathhelpforum