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Integral in an old math book

  1. Mar 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi.
    Can anyone here solve this integral for me:
    ##\int\frac{d^3y}{\operatorname dx^3}\;y\;dx##


    2. Relevant equations
    I haven't seen such integral before, it is included in an old math book
    and it should be solved using "Integration by Parts" technique because it is an exercise of it.
    3. The attempt at a solution
    I don't no from where to start the solution so I don't know what to write here.

    Thank you very much!
     
  2. jcsd
  3. Mar 14, 2017 #2
    :welcome:

    I too started integration, Here in PhysicsForums we don't give direct answers. I will try to walk you through though.

    First it says integrate it with respect to dx.
    Second, In most of integration by parts questions. The best way to solve it is to just do the integration by parts a lot of times see if you arrive at a pattern.

    Do you know the table method? this should make this easier for you. If not just do the normal way and see if you can reach something you can make use of.
     
  4. Mar 14, 2017 #3
    You say that you don't know where to start but since it is an integration by parts problem you should know the procedure of it's doing. From the integral you choose your ##u## and you ##v'## which are the parts you seek. Here is a hint that will help you, usually for ##u## we choose functions that we don't know or are hard to integrate and for ##v'## we choose the function we know or are easy to integrate.
    An example to clarify would be ##\int x \ln x \, dx## where you are bound to take ##u=\ln x## and ##v'=x## since the first you know to differentiate and the second to integrate. Try to find whats best to fit the given description for your problem :) .
     
  5. Mar 14, 2017 #4
    Thank you very much for your help.
    I've done a lot of integration by parts problems, and I know the table method but still I don't think I'm solving correctly in this one.

    I've this so far:
    I took ##y## as ##D## and I took ##\frac{d^3y}{\operatorname dx^3}## as ##I## and made it to:
    ##y\;\frac{d^2y}{\operatorname dx^2}\;-\;\int\frac{dy}{\operatorname dx}\;\frac{d^2y}{\operatorname dx^2}\;dx##

    How about it?
     
  6. Mar 14, 2017 #5
    I didn't mean I don't know the integration by parts thing, but I mean I don't know what to take as ##u## and what as ##dv## in this problem.

    I forgot to thank you for your reply :)
     
  7. Mar 14, 2017 #6
    Excellent, Now notice the second part.
    If I do to it integration by parts a couple of more times just as you did with the first one, What will I get?
    Note: Notice what we are trying to find ##\int\frac{d^3y}{\operatorname dx^3}\;y\;dx##

    you will find that on both side we have the same integration.
     
  8. Mar 14, 2017 #7
    This is correct. Now may i suggest a different method of continuation than what Byker suggested. Look at the second integral. Can you introduce a substitution that would clarify everything without the need of further by parts integration?
     
  9. Mar 14, 2017 #8
    ##\frac{d^2y}{\operatorname dx^2}\;## is the derivative of ##\frac{dy}{\operatorname dx}##?
    Are you hinting for ##udu## substitution?
     
  10. Mar 14, 2017 #9
  11. Mar 14, 2017 #10
    I just want to clarify something:
    ##\frac{d^2y}{\operatorname dx^2}\;## is the derivative of ##\frac{dy}{\operatorname dx}## but not ##\frac{d^2y}{\operatorname dx^2}\;dx## right?
    I thought about making ##\frac{dy}{\operatorname dx}\;\frac{d^2y}{\operatorname dx^2}\;## as ##dv## and ##dx## as ##u##
    which leads us to:
    ##\begin{array}{l}dv=\frac{dy}{\operatorname dx}\;\frac{d^2y}{\operatorname dx^2}\\v=\frac12\left(\frac{dy}{dx}\right)^2\\u=dx\\du=0\end{array}##
    so:
    ##y\;\frac{d^2y}{\operatorname dx^2}\;-\;\frac12\left(\frac{dy}{dx}\right)^2dx\;-\;\int0##
    What do you think guys?
     
  12. Mar 14, 2017 #11
    That is not how substitution work, neither integration by parts. I thought I should lead you the way through integration by parts only so you can practice.

    When you do integration by parts, You do it to the functions inside the integral not to dx. You can think of dx as just a notation to tell you what you are integrating with respect to.

    You have two choices, Either continue integration by parts as you did in the first step or make a substitution and integrate (Which is easier).

    Here: https://www.mathsisfun.com/calculus/integration-by-substitution.html
     
  13. Mar 14, 2017 #12
    Sorry for that mistake, I had the idea that ##dx## is something more than a notation.
    So the answer will be:

    ##y\;\frac{d^2y}{\operatorname dx^2}\;-\;\frac12\left(\frac{dy}{dx}\right)^2+\;c##
    I hope I didn't mess it up this time.
     
  14. Mar 14, 2017 #13
    Excellent, good job :D.

    Integration means infinite sum of f(x) dx. But you can't just treat dx as part of what you are integrating. you can substitute with du in a sense (The reverse of the Chain rule).
     
  15. Mar 14, 2017 #14
    Thank you very very much!
    I'm glad that I got the help from great man like you Biker, and nice to meet you :)
    and thank you doktorwho for your help, you gave me a really great hint. and nice to meet you too :)
    Sorry for the inconvenience and sorry for my language (English is not my native language)
    not to forget, I completed the solution in the two methods you gave me and I understand it very much!
    thank you again!
    see you later :biggrin:
     
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