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Homework Help: Integration with respect to a higher power

  1. Jun 23, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \int {x^2 ,d(x^4)}
    [/tex]


    2. Relevant equations
    as a starter; the previous problem was:
    [tex]
    \int {x^2 ,d(x^2)}
    [/tex]

    and I managed to solve this by letting u=x^2 then integrating u:

    [tex]
    \int {u}
    [/tex]

    =[tex]
    \frac{x^4}{2}+C
    [/tex]


    3. The attempt at a solution
    can someone please explain the thoery behind this?
    my textbook gives no explanation and i dont really know what im looking for...
    I tried following the working from the previous problem through, but got (x^6)/3 when the answer was supposed to be 2(x^6)/3

    thanks in advance
    Steven
     
  2. jcsd
  3. Jun 23, 2007 #2
    In the previous problem, you made the differential to du. That is an interesting tactic for this problem too..
     
  4. Jun 23, 2007 #3
    so I am on the right track i take it?
    If so;
    [tex]\int {x^2 d(x^4)} = \int {u^2 d(u)}[/tex]
    am I correct?
     
  5. Jun 23, 2007 #4
    Not quite. If

    [tex]u = x^4[/tex]

    then

    [tex]x^2 = ?[/tex]
     
  6. Jun 23, 2007 #5

    MathematicalPhysicist

    User Avatar
    Gold Member

    dx^4/dx=4x^3
    dx^4=4x^3dx
     
  7. Jun 23, 2007 #6

    HallsofIvy

    User Avatar
    Science Advisor

    In general,
    [tex]\int f(x)dg(x)= \int f(x) \frac{dg}{dx} dx[/tex]
     
  8. Jun 24, 2007 #7
    thankyou all for your help, I understand it now :)
    [tex] dx^4 = 4x^3 dx [/tex]
    therefore
    [tex] \int {x^2 dx^4} [/tex]
    [tex]= \int {x^2 4x^3 dx}[/tex]
    [tex]= 4 \int {x^2 x^3 dx}[/tex]
    [tex]= 4 \int {x^5 dx}[/tex]
    [tex]= 4 \frac {x^6}{6} + C[/tex]
    [tex]= \frac {2x^6}{3} + C[/tex]

    cheers
    Steven
     
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