# True unit of average integral result

• B
• ecoo
In summary, the (b-a) factor in the formula for average value does have a unit, and it is dependent on the units of the x-axis.

#### ecoo

For example, when you have a function that describes the amount of water (liters) in a cup relative to time (the amount fluctuates for whatever reason). If you wanted to find the average amount of water in the cup in the interval [a,b], you would take the integral of the function in that interval and divide by (b-a).

The result can be interpreted as the average amount of liters in the cup in the interval, so you would write the unit as liters, but would the true unit be liters*time? If I were to remove the 1/(b-a) factor, would liters*time would be the true unit?

Thanks!

I am not sure what you mean by "true unit" here. The formula for the average value of a certain function had better return units identical to that function, or it is just not the right formula! The formula you gave is correct, and it gives the correct units - those of the integrand.

##\frac{1}{\Delta x}\int_{a}^{b} f(x) dx##

I mean that when one integrates, for example, a velocity to time function, the answer's unit is displacement (velocity*time). So I'm thinking that when I integrate a function that models liters to time, the integral result's unit should be liters*time. Of course, in the context the result can be interpreted as the average liters in the interval, so the unit can be assigned as liters in this case.

ecoo said:
So I'm thinking that when I integrate a function that models liters to time, the integral result's unit should be liters*time. Of course, in the context the result can be interpreted as the average liters in the interval, so the unit can be assigned as liters in this case.

But you are not simply integrating a function; you are using a formula which was derived for a specific purpose. Look at the formula again. Δx (=b-a) has the same units as dx does, so the only units in the average value formula are those of the integrand...

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mfig said:
But you are not simply integrating a function; you are using a formula which was derived for a specific purpose. Look at the formula again. Δx (=b-a) has the same units as dx does, so the only units in the average value formula are those of the integrand...

Oh, that makes sense.

Quick question, if I were to integrate the above function in some interval, then the units would be liters*time, right? And then if I were to multiply my result by some factor that happens to be 1/(b-a) with no goal on solving for average, simply part some calculation, then would the units be in liters*time or would the multiplication by the factor always cancel out the time (and in this case, multiplying by any other factor would not cancel out the time)?

If the multiplying factor were dimensionless then yes, it would not alter the units. For example, if I want to find the value of an even function over an even interval I can find the value of the function over the half-interval and multiply by the dimensionless value 2. As in:

##\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx \space \space \space (f(x)\space even \space about \space origin)##

In this case, the units of the result are the units of f(x) times the units of x, both with and without the multiplying (and dimensionless) factor.

ecoo
mfig said:
If the multiplying factor were dimensionless then yes, it would not alter the units. For example, if I want to find the value of an even function over an even interval I can find the value of the function over the half-interval and multiply by the dimensionless value 2. As in:

##\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx \space \space \space (f(x)\space even \space about \space origin)##

In this case, the units of the result are the units of f(x) times the units of x, both with and without the multiplying (and dimensionless) factor.

So in the formula to find average, the (b-a) factor does have a unit, and it is whatever the x-axis is defined as, if it did not have a unit then the answer would be liters*time.