Results 1 to 12 of 12

Math Help - Help with Integration.

  1. #1
    Junior Member
    Joined
    Jun 2010
    Posts
    27

    Help with Integration.

    Hi all. It may be simple, but I do not seem to recall it. How can I solve the following
    Help with Integration.-integral.gif
    As an example u may like to set n to any integer, say, 5.
    Thanx.
    Nvd.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by beezee99 View Post
    Hi all. It may be simple, but I do not seem to recall it. How can I solve the following
    Click image for larger version. 

Name:	Integral.gif 
Views:	56 
Size:	1.2 KB 
ID:	19695
    As an example u may like to set n to any integer, say, 5.
    Thanx.
    Nvd.
    (it doesn't make sense to have x in the limits when it is the variable you're integrating with respect to...EDIT, if it is pi, then no worries )

    the way to handle an integral like this, is to simply expand the numerator, using the binomial theorem if n is large enough to make it useful, and then dividing each term by x. for the first term, the 1/x, the integral would be ln|x|, for the others, use the power rule for integrals: \displaystyle \int x^n ~dx = \frac {x^{n + 1}}{n + 1} + C

    then apply the fundamental theorem of calculus to deal with the limits on the integral.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by beezee99 View Post
    Hi all. It may be simple, but I do not seem to recall it. How can I solve the following
    Click image for larger version. 

Name:	Integral.gif 
Views:	56 
Size:	1.2 KB 
ID:	19695
    As an example u may like to set n to any integer, say, 5.
    Thanx.
    Nvd.

    It's hard to read what the upper limit is, but let's supose it is pi, so:

    \displaystyle{\int\limits^\pi_0\frac{(1-x)^n}{x}\,dx=\int\limits^\pi_0\frac{\sum\limits^n_  {k=0}\binom{n}{k}(-x)^k}{x}\,dx=\int\limits^\pi_0\left(\frac{1}{x}-n+\frac{n(n-1)}{2}x-\ldots\pm x^{n-1}\right)\,dx ,

    and now use linearity of the integral to solve each of the above...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
     \int \frac{(1-x)^n}{x} ~dx where  n is natural number .


    Sub.  x = 1 -t

     dx = -dt

    We have  - \int \frac{t^n}{1-t}~dt

     = - \int \frac{ 1 - (1-t^n) }{1-t}~dt

     = - \int \left[ \frac{1}{1-t} - (1+ t + t^2 + ... + t^{n-1} ) \right] ~dt

     = \frac{t}{1} + \frac{t^2}{2} + \frac{t^3}{3} + .... + \frac{t^n}{n}  + \ln|1-t| + C

    or  = \frac{(1-x)}{1} + \frac{(1-x)^2}{2} + \frac{(1-x)^3}{3} + ... + \frac{(1-x)^n}{n} + \ln|x| + C
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2010
    Posts
    27
    Hi guys. Sorry, the upper limit is 1.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,444
    Thanks
    1863
    It's the lower limit, 0, that is the problem!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2010
    Posts
    27
    So then how to resolve the issue for lower limit of 0 ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by beezee99 View Post
    So then how to resolve the issue for lower limit of 0 ?
    it's an improper integral then. you would have to compute:

    \displaystyle \lim_{N \to 0^+} \int_N^1 \frac {(1 - x)^n}x~dx

    (the integral doesn't converge...)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jun 2010
    Posts
    27
    How about if I have the following where m and n are different integers?
    Help with Integration.-integral.gif
    Will the ln(x) terms cancel out?
    Thanx.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by beezee99 View Post
    How about if I have the following where m and n are different integers?
    Click image for larger version. 

Name:	Integral.gif 
Views:	19 
Size:	1.8 KB 
ID:	19707
    Will the ln(x) terms cancel out?
    Thanx.
    No, both integrals would be considered separately. you apply the fundamental theorem one at a time since the limits are different
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jun 2010
    Posts
    27
    Hi. The limits are the same i.e. 0<=x<=1 for both the integrals. The only difference between the two is the powers of (1-x). So?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by beezee99 View Post
    Hi. The limits are the same i.e. 0<=x<=1 for both the integrals. The only difference between the two is the powers of (1-x). So?
    well in that case, you can put both under the jurisdiction of a single integral, and then yes, the 1/x 's will cancel. so you will have a convergent integral.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 05:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum