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!

Trapezium rule, what does dx represent

  1. Jun 14, 2014 #1
    1. The problem statement, all variables and given/known data
    What does the "dx", associated with the definite integral represent for the trapezium rule? Could dx=h? (the heights of the trapeziums)
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 14, 2014 #2
    I am not entirely sure what you're asking here. If you are referring to the trapezium rule for approximating definite integrals, then dx should not appear in the approximation at all.

    If you are asking what dx is an "analog" to in the physical trapezoid, well, nothing really. It is a differential element of the line of the trapezoid on the x-axis, I suppose..
     
  4. Jun 14, 2014 #3
    It's seems to me that dx represents the heights and the rest of the definite integral represents
    {2(y0+yn)+(y1+...+yn-1)}
     
  5. Jun 14, 2014 #4
    If it was (dx)^ 2 what change would it make to the rule?
     
  6. Jun 14, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The point of the "trapezoid rule", explained in any Calculus textbook, is that instead of approximating f(x) by a constant (as is done in the "Euler" rule), we approximate it by the straight line from [itex]x_0, f(x_0))[/itex] to the "next point", [itex](x_0+ h, f(x_0+ h))[/itex]. In that case, rather than rectangles we have trapezoid where the two "bases" are [itex]f(x_0)[/itex] and the other is [itex]f(x_0+ h)[/itex]. Yes, the distance along the x-axis, from [itex]x_0[/itex] to [itex]x_0+ h[/itex], which is, of course, h, is the "height" of the trapezoid. ("h" in the trapezoid rule is NOT "dx"- you should be thinking "[itex]\Delta x[/itex]" instead.)

    By the way, if you use an "Euler" rule, always choosing the value of the function at the left hand end of the interval as the height of the rectangle, then do another "Euler" rule, always choosing the value of the function at the right end of the interval as the height of the rectangle, and then average the two answers, you get exactly the result of the "trapezoid" rule. Draw a few examples to convince yourself of that.
     
  7. Jun 14, 2014 #6
    I'm trying to manipulate
    /lambda
    | sqrt(1+cos(x)) (dx)^2
    / 0
    Into the form described by the trapezium rule, how would I do that?
     
  8. Jun 14, 2014 #7
    Unless this is just something I know nothing about, I would say having dx squared is completely nonsensical.
     
  9. Jun 14, 2014 #8

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's not clear what the (dx)^2 represents in your integral. Where did this integral come from.

    As a point of information, for the trapezoidal rule, Δx represents the spacing of the abscissas, while y represents the ordinates, or heights, of the function being integrated.
     
  10. Jun 14, 2014 #9
    /lambda
    | Sqrt(1+cos(x)) dx= (u)
    /0

    u(dx)=m This is because "u" represents mass per unit length

    Therefore I said

    m= /lambda
    |Sqrt(1+cos(x))(dx)^2
    /0
     
  11. Jun 14, 2014 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do you know Calculus?

    Is you original problem written something like the following?

    [itex]\displaystyle \int_{\!0}^{\!\!1}\sqrt{1+\cos(x)}\,dx = (u)[/itex]
     
  12. Jun 14, 2014 #11
    It is not clear to me why some of these parentheses are present (in (u) and u(dx)), but it seems to me that he says

    \begin{align*}
    m &= udx, \\
    u &= \int_0^{\lambda} \sqrt{1 + \cos x} \, dx,
    \end{align*}

    whence

    \begin{equation*}
    m = \int_0^{\lambda} \sqrt{1 + \cos x} \, (dx)^2.
    \end{equation*}

    This is, of course, nonsensical. Taking a stab in the dark, perhaps it started as

    \begin{align*}
    dm &= udx, \\
    u &= \sqrt{1 + \cos x},
    \end{align*}

    but I could be completely wrong.
     
  13. Jun 14, 2014 #12
    You are right!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trapezium rule, what does dx represent
  1. Trapezium rule (Replies: 2)

  2. Trapezium Rule (Replies: 1)

Loading...