1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration - u substitution problem (Integration by parts?)

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the integral of 3x* (2x-5)^6*dx, let u= 2x -5.

    2. Relevant equations
    Im not sure if i am meant to use integration by parts or not?? I was able to do previous questions of the topic just using u sub to get rid of the first x variable.


    3. The attempt at a solution
    So i started by
    u= 2x-5
    du/dx = 2
    dx= du/2

    integral of 3x * (u)^6* du/2
    3x/2 *1/7 (u)^7 +c

    = 3x/14 * (2x-5)^7 +c
    However i know this is wrong because i took the integral in the form of a product (with two x variables).

    Please help i have been teaching this topic to myself via online videos as I do not go to school, and want to catch up before i start uni (majoring in physics ;))
     
  2. jcsd
  3. Feb 18, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You're fine up to here, but you need to get the integrand in terms of just u, so solve u=2x-5 for x and substitute the result into the integrand.
     
  4. Feb 18, 2012 #3
    Thanks for the quick reply, although

    doing this I get,
    x=(5+u)/2
    Hence, 3*(5+u/2)*1/2*1/7*(u)^7
    = 15/28 *(2x-5)^7 + c

    Butthe apparent answer is not this
    it is... 3/32 * (2x-5)^8 + 15/28 (2x-5) +c

    So i got the last part,, but where does the first part come into the equation??

    Thanks in advance.
     
  5. Feb 18, 2012 #4
    sorry, error in the answer it should be,

    3/32 * (2x-5)^8 + 15/28 (2x-5)^7 +c
     
  6. Feb 18, 2012 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You have to substitute for x first, multiply everything out, and then integrate.
     
  7. Feb 18, 2012 #6
    still dont understand because it looks as if he has integrated once to get u^7 and then again to get u^8? im lost.
     
  8. Feb 18, 2012 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You need to take it one step at a time. You got to
    $$\frac{3}{2} \int x u^6\,du$$ which you can't integrate yet. So you substitute for x and get
    $$\frac{3}{4} \int (u+5) u^6\,du$$ Now how do you integrate that?
     
  9. Feb 18, 2012 #8
    just got that ooooohhh moment.
    thanks so much for the help greatly appreciated.
    live long and prosper ;)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration - u substitution problem (Integration by parts?)
Loading...