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Average Value of a Function

  1. Feb 17, 2016 #1
    Hey everyone,

    I am preparing for a Calc 2 lecture tomorrow with KhanAcademy videos. I was watching a video on average value of a function in which he said, basically, you'll have some definite integral on a to b divided by (b-a). My question is, is this essentially calculating average speed?

    I was thinking of the graph of an integrand expressed in terms of a velocity-vs-time axis. By taking the integral we get a position function, but then we are dividing it by time. Since the integral is definite, the result would be that we have change in position/change in time. This would give distance/time, or average speed? Perhaps more specifically, average velocity. I'm just trying to understand this in a physical context.

    Thanks.
     
  2. jcsd
  3. Feb 17, 2016 #2

    RUber

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    If you have velocity v(t) vs. time t in two dimensions, you are assuming a constant direction of travel. When you take the integral of v(t), you will get total distance.
    Total distance divided by total time is average velocity.
    Your question about velocity vs. speed is not entirely clear. For example, velocity could be negative, indicating a reversal of direction, but speed is defined as the magnitude of velocity. So, for a velocity function that is all positive, the two are interchangeable. However, if you have a velocity function that is positive then negative, like throwing a ball into the air and having it return to your hand, your average velocity and average speed are two different things.
     
  4. Feb 17, 2016 #3
    I follow you, but the first part of your response is kind of what I was getting at. If you're divided an integral by its upper and lower bounds (in this case the x-axis is time), when you take the integral, you'd get a distance over time, so average velocity, right?

    However, I think this spawns another question based on what you said. Isn't total distance divided by total time average speed, not average velocity considering the speed doesn't include direction of motion? Yet, taking a definite integral is solely our net area, which would include direction, would it not (provided the graph falls below the x-axis we would be subtracting distance). I guess to simplify it by taking the integral and dividing by its bound we wind up getting Δy/Δx, velocity.

    Perhaps I'm reading too far into your response and confusing myself in the process. Haha.

    Thanks.
     
  5. Feb 17, 2016 #4

    RUber

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    By plotting velocity vs. time on a 2D graph, you have already nearly taken the directional component out of velocity. But, you still have +y and -y directionality.
    If you had a plot of a ball going up and coming down, your average velocity would be zero, since your total distance travelled would be zero. This can be though of as average velocity in the +y direction.
    However, your average speed would be the integral of the absolute value of velocity, so it would not be zero.
     
  6. Feb 17, 2016 #5

    HallsofIvy

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    Given a function f(x), its "average value" over interval [a, b], is given by
    [tex]\frac{\int_a^b f(x)dx}{b- a}[/tex]
    Is that what you are asking?

    RUber's last sentence-
    "However, your average speed would be the integral of the absolute value of velocity, so it would not be zero." is referring to the distinction between "velocity", which is a vector quantity, and "speed" which is the norm of the velocity vector.
     
  7. Feb 22, 2016 #6

    Svein

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    What if you are driving around in circles? After a while you would come back to where you started, so your "distance" is zero (but your average speed is not zero). You need a slightly more complex integral in order to find the true average speed:

    When driving along a flat road, assume that your position is given as [itex] (x(t), y(t))[/itex]. The length element corresponding to dt is then [itex]\sqrt{(\frac{dx(t)}{dt})^{2}+(\frac{dy(t)}{dt})^{2}} [/itex]. So, in order to calculate the length of the road you have travelled in the time interval [a, b] you must calculate the integral [itex] \int_{a}^{b}\sqrt{(\frac{dx(t)}{dt})^{2}+(\frac{dy(t)}{dt})^{2}} dt[/itex].
     
  8. Feb 22, 2016 #7

    SteamKing

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    Instead of looking at the average value of a function in terms of a speed-time curve, which drags in a whole host of other issues, look at the integral of a function as an area under the curve, such that

    $$A =\int_a^b f(x)\,dx$$

    The length of the base of this area is then (b - a). The value ##\frac{A}{(b-a)}## is the average length of all the ordinates under the curve of f(x) from a to b, and A = favg ⋅ (b - a).


    trap6.gif
     
  9. Feb 22, 2016 #8

    HallsofIvy

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    '
    If you drive around in a circle, you displacement is 0 but the distance driven is the circumference of the circle.
     
    Last edited: Feb 23, 2016
  10. Feb 22, 2016 #9
    You're on a road trip and you want to find your average speed over the trip. So you start out by writing down the starting odometer reading and the time you left. During your trip, you don't travel to your destination in a straight line. The roads don't run that way. And, you take some side trips. And, you stop for lunch and for getting gas. There are stop signs and traffic lights. The speed limit changes along the way. When you arrive at your destination, you write down your odometer reading and the time that you arrived. How do you calculate your average speed for the trip?
     
  11. Feb 23, 2016 #10
    This is my attempt to repeat what others have said, but without some restrictions that I believe are unnecessary:

    Let C be a smooth curve located in any space you like (such as Rn for any n ≥ 1) with a specified direction (orientation). Assume the points of C are assigned position numbers in such a way that the difference between two position numbers is the distance along C between the corresponding points in the specified direction (or the negative of the distance if in the opposite direction.)

    Let s(t) denote the position of a point moving smoothly on C at time t, for a ≤ t ≤ b.

    Then v(t), the velocity at time t, is the derivative ds(t)/dt, defined for a <= t <= b. (Positive velocity means you are moving in the specified direction; negative velocity means you are moving in the opposite direction. The function v(t) might have both positive and negative values.)

    Thus if you draw the graph of v(t) as the y-coordinate vs. t as the x-coordinate, you can picture integrating v(t) between t = a and t = b, to get s(b) - s(a), or the net distance traveled along C. (Which is clear, since v(t) = ds(t)/dt.) The total time is b-a, and thus the average velocity, being the net distance divided by the total time, is

    vavg = (1/(b-a)) ∫ v(t) dt​

    where the integral is from t = a to t = b.

    Of course the same formalism works for any function f(x) for a ≤ x ≤ b in place of v(t) for a ≤ t ≤ b, so that the average value of f(x) becomes

    favg = (1/(b-a)) ∫ f(x) dx​

    where likewise the integral is from x = a to x = b.
     
  12. Feb 25, 2016 #11

    phion

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    Isn't this calculus one?
     
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