# Average Value Theorem

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• 0:06 Average Value Theorem
• 2:03 Example
• 4:20 Lesson Summary
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Lesson Transcript
Instructor: Jeff Calareso

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

If you know you've gone 120 miles in 2 hours, you're averaging 60 mph. But what if you know your velocity at every point in time and not how far you've gone? In this lesson, learn how to calculate average values using integrals.

## Average Value Theorem

Let's consider the graph of your velocity as a function of time. Here, I've got your velocity between some time a and time b. I know that the area underneath this curve is going to be equal to the distance that I have traveled from time a to time b. I also know that the area under this curve is going to be the integral of v(t)dt from t=a to t=b.

Let's say that I found that area to be 120 miles. And my total time from a to b is 2 hours. Say a=0 and b=2. If I've gone 120 miles in 2 hours, then you know that the average speed was 60 mph. That's 120 / 2. One way that we could write this mathematically is by saying that the average equals 1 / delta t (which is b - a, my total time) times the integral from a to b v(t)dt, which is the area under the curve. So my average is the area under the curve divided by my width here. That's going to give me an average height, if you will. Further, because my velocity is continuous, I know that at some point on this graph, I was going exactly 60 mph. This is because I had to average 60 mph. Sometimes I was going more than 60, sometimes I was going less than 60. But at at least one point on this graph, I was going exactly 60 mph. I had to have been going the average.

## Example

Let's do an example. Let's say that y = 9 - x^2 between 0 and 3. I'm going to take the integral from 0 to 3 of 9 - x^2. Let's say that I can find the area under this curve exactly. That area is 18.

I know that my average value of y is going to be 1 / delta x - that's x on one side minus x on the other - times the integral from a to b, so the integral over the region, of f(x)dx. All this is is the area divided by the width. Here, my width is 3 - because it's 3 - 0, that's my width in x - and my area I've told you I calculated to be 18. So the average value of y along this entire region here is going to be 6.

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