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Evaluating definite integral

  1. Jun 30, 2014 #1
    I know the value of the following definite integral


    I also have a realtion


    i.e. x is an explicit function of y but I do not have y as an explicit
    function of x. The relation between x and y is generally non linear.

    Now I want to get the following definite integral

    [itex]\int_{a}^{b}\left[\int ydx\right]xdx[/itex]

    i.e. [itex]\int ydx[/itex] multiplied by x evaluated over the interval [a,b].

    Is there an analytic (not numeric) way to evaluate this integral using
    for example mean value or similar averaging technique?
  2. jcsd
  3. Jul 3, 2014 #2
    Well, I think all you need to do is to find [itex]dx[/itex] in terms of [itex]dy[/itex] and then it should be simple since you already have [itex]x[/itex] in terms of [itex]y[/itex], just sub in the integral.

    [itex]\displaystyle\large\therefore \int_a^b \left[\int ydx\right]xdx=\int_a^b \left[\int yf'(y)dy\right]f(y)f'(y)dy[/itex]

    You already have the value of the integral [itex]\displaystyle\large\int_a^b ydx=\int_a^b yf'(y)dy[/itex]

    I believe (but do not recall) that there is a way to use sort of the "opposite" of the mean value theorem.

    Also, if possible, it would simply be easy enough to take [itex]f^{-1}(y)[/itex]
  4. Jul 3, 2014 #3

    I like Serena

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    Homework Helper

    Your integral is actually:
    \begin{aligned}\int_{a}^{b}\left[\int ydx\right]xdx
    &=\int_{a}^{b}\left[\int y(\xi) d\xi\right]xdx \\
    &= \int_{a}^{b} xdx \cdot \int y(\xi) d\xi \\
    &= \frac 1 2(b^2-a^2)\int y(\xi) d\xi \\
    &= \frac 1 2(b^2-a^2) \int y dx \\
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