Forgot how to integrate fractions

  1. Pengwuino

    Pengwuino 7,118
    Gold Member

    Ok so I have to find the arc length from t=1 to t=2.

    [tex]
    \begin{array}{l}
    L = \int_a^b {|r'(t)|dt} \\
    |r'(t)| = \frac{{2(1 + 2t^4 )}}{{t^3 }} \\
    \end{array}
    [/tex]

    And I have completely forgotten how to integrate fractions.....

    Oh wait... i THINK i know what to do. Should i set u=1+2t^4?
     
    Last edited: Nov 3, 2005
  2. jcsd
  3. I hope I'm not missing something.

    [tex]\frac{{2\left( {1 + 2t^4 } \right)}}{{t^3 }} = 2\left( {\frac{1}{{t^3 }} + \frac{{2t^4 }}{{t^3 }}} \right)[/tex] then it should just be the normal power rule right?
     
  4. dextercioby

    dextercioby 12,317
    Science Advisor
    Homework Helper

    No need for any substitution. Benny's right.

    Daniel.
     
  5. Pengwuino

    Pengwuino 7,118
    Gold Member

    heh, i don't even remember the power rule.
     
  6. Power rule for integration is:
    [tex]\int x^n dx = \frac{x^{n +1}}{n + 1} + C[/tex]
     
  7. dextercioby

    dextercioby 12,317
    Science Advisor
    Homework Helper

    Don't forget the condition

    [tex] \int x^{n} \ dx=\frac{x^{n+1}}{n+1} +C , n\in{\mathbb{C}-\{-1\}} [/tex]

    Daniel.
     
  8. Also remember that [tex]\frac{1}{x^n} = x^{-n}[/tex]
     
  9. disregardthat

    disregardthat 1,840
    Science Advisor

    I'm sorry to bring this up, but how do you integrate a fraction?

    If you have 4x/3, how do you integrate it?
     
  10. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    Remember that a fraction is nothing else than an ordinary number. Also remember that you can write:
    [tex]\frac{4x}{3}=\frac{4}{3}*x[/tex]
    f
    Thus, an anti-derivative is:
    [tex]\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C[/tex]
     
  11. Also think of the following: If there is a constant inside an integral you can move it outside the integral (sorry, don't know this formula editor):

    INTEGRAL(c*f(x)dx)=c*INTEGRAL(f(x)dx)

    So in your case 4/3 is a constant (independant of x) and you can move it outside the integral, getting:

    INTEGRAL(4x/3dx)=4/3*INTEGRAL(xdx)
     
  12. disregardthat

    disregardthat 1,840
    Science Advisor

    Thanks,

    [tex]\int{C*f(x)dx} = C[/tex] [tex]\int{f(x)dx}[/tex]

    But Kruger, wouldn't the dx be over/beside the fraction, and not under: like this?

    [tex]\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}[/tex]
     
    Last edited: Mar 7, 2007
  13. HallsofIvy

    HallsofIvy 40,807
    Staff Emeritus
    Science Advisor


    That's what he meant. Most people interpret 4/3a to be (4/3)a, not 4/(3a).
    It's always best to use the parentheses, though!
     
  14. disregardthat

    disregardthat 1,840
    Science Advisor

    We all love parantheses!
     
  15. Hi, found this thread and hoping to get a reply:
    I want to integrate curve eqn to get volume of curve rotated around x-axis
    The curve eqn is 6/(5-2x) and the x limits are 0 and 1
     
  16. Mark44

    Staff: Mentor

    Welcome to Physics Forums!

    This thread is more than four years old. When you tack an unrelated question onto an existing thread, that's called "hijacking" the thread. Please use the New Thread button to start a new thread.
     
  17. jtbell

    Staff: Mentor

    Actually, the button is labeled "New Topic" not "New Thread", otherwise Mark's advice is good. Also, please show the work that you have already done on the problem, or at least tell us what is confusing you about how to solve it.
     
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