Time average in a continuum probability distribution

[ricard.py]

Homework Statement

I have a question that looks so stupid that I have never dared to ask.
If I want to measure the time average from t=0s to t=1s of a given f(t), the solution is compute the following integral:

TA = 1/T*∫F(t)dt

However, I have some doubts about this calculus.

Homework Equations

1) An integral computes an area under the curve. I am not really interested in computing an area, I just want the summation of the f(T) values along the interval. So, what does an area do here?

2) The integral sum up a lot of values which depend on the step size h. Assuming that we selected h=0.0001 (very close to 0), the total number of summations is 1/0.0001=10000. However, you only normalize the integral by 1 (1/T). Why is that? In a discrete probability distribution, I would have normalized by the number of added values.

 

[BiGyElLoWhAt]

Area/length = height

If you compute the total area (variable height) and divide by the total length, you get the average height

You're not summing up values with a definitive step size, the step size is dt, which is infintesimal, you're thinking of a riemann sum, which is the precurser to the integral. The integral is exact.

 

[BiGyElLoWhAt]

I was actually done with calc 2 before I really understood why this is the case. This is the explanation that solidified my understanding and convinced me 100%

http://math.stackexchange.com/questions/15294/why-is-the-area-under-a-curve-the-integral

 

[BiGyElLoWhAt]

Maybe this illustration will help.

 

[STEMucator]

1) When you are dividing by $T$, which is the change in time, you are obtaining the "time average" of the given $f(t)$ over the interval.

2) $dt$ is the infinitesimally small time width. The integral is the summation of the $f(t)$ over a suitable partition as $n → ∞$.

 

[ricard.py]

Area/length = height

If you compute the total area (variable height) and divide by the total length, you get the average height

You're not summing up values with a definitive step size, the step size is dt, which is infintesimal, you're thinking of a riemann sum, which is the precurser to the integral. The integral is exact.

OK I think I am understanding it better, thanks! I have on more question. The surface calculated with an integral is usually not a rectangle such that area = length*height. However, I suppose that what you can do is "rearrange" the surface keeping the area constant and making it look like a rectangle, and then you apply the area/length=height. However, how can you be sure that the height is actually the AVERAGE time?

 

[HallsofIvy]

Homework Statement

I have a question that looks so stupid that I have never dared to ask.
If I want to measure the time average from t=0s to t=1s of a given f(t), the solution is compute the following integral:

TA = 1/T*∫F(t)dt

However, I have some doubts about this calculus.

Homework Equations

1) An integral computes an area under the curve.

This is NOT true. "Area under the curve" is one application of the integral. Essentially the integral is a "continuous sum". Just as the average of a finite number of data points is the sum divided by the number of points, so the average of a continuous function is its "sum" (integral) divided by the length of the interval.

I am not really interested in computing an area, I just want the summation of the f(T) values along the interval. So, what does an area do here?

2) The integral sum up a lot of values which depend on the step size h. Assuming that we selected h=0.0001 (very close to 0), the total number of summations is 1/0.0001=10000. However, you only normalize the integral by 1 (1/T). Why is that? In a discrete probability distribution, I would have normalized by the number of added values.

This also is NOT true. The sum of "a lot of values which depend on step size h" is an approximation to the integral. The integral is the limit of that as h goes to 0- and so the number of steps, your "number of added values" goes to infinity.