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## Integral Calculus

### Unit 3: Lesson 1

Average value of a function

# Average value over a closed interval

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.B (LO)
,
CHA‑4.B.1 (EK)
Average value of a function over a closed interval.

## Want to join the conversation?

• Conceptually how can I get the average height by dividing the integral,which is the area,and its width?
• Suppose you had 16 people in a class and you want to find their average height. You would add up their heights and divide by the number of people. Similarly, you could line the people up against a wall, measure and sum the rectangular area each person covered (giving each person a 1 foot space to stand in), which would be the sum of their heights, and then divide by how long the line of people is against the wall (which would be the number of people). This is similar to finding the area under a function and dividing by its width. Remember than an integral is the sum of an infinite number of rectangles, whose area is calculated by height times width. Each (typically) has the same width, so to find the average height of all these rectangles, we add up all their areas and divide by their total width.
• Can't we just take the average of f(a) and f(b)? I hear Sal say we can't do that because the function is not linear, but can someone give me some intuition? An extreme case that demonstrates why not?
• Consider the average value of sin(x) from 0 to pi. a = 0, b = pi. Taking the average conventionally:
f(a) = sin(0) = 0
f(b) = sin(pi) = 0
f_avg = (0+0) / (pi - 0) = 0
Taking the average the conventional way would give you an average of 0. From examining the graph of sin(x), it should be apparent that the average value is NOT 0. Using integrals you would get:
f_avg = 2/pi
• At , how can the area of the rectangle exactly match the area under the curve from a to b? Is it an approximation of the actual area?
• , Sal assumes that the actual area equals the area of some arbitrary rectangle.
Length refers to the interval a to b, height here refers to the average height of the function in the given interval.
The dimensions of the rectangle are so calculated, that the area of the rectangle WILL equal the actual area. Or in other words, we choose a height that will give the actual area when multiplied with the length of interval
• Dear Mr. Khan,

Firstly, thanks a lot for the interesting videos on your blog.

I kindly would like to attract your attention to the following: (and I am sure you know that very well)
The height of the rectangle in the illustrating figure of your video needs to be taken on the curve itself and not above (or below) the curve, as you well know.

Kindest Regards,
Issa Qaqish
• Actually it is perfectly alright to take height of rectangle above curve. You see, it depends on width of rectangle chosen. In Sal's case, for understanding, he drew the rectangles large enough that some of area may have overflowed but if width is infinitesimally small, then each rectangles height(which are practically line segments now) are equal to height of curve at any point.
• Could one also apply the MVT for derivatives to f(x) on (a,b) to find this average value?
• Does this relate in any way to the Mean Value Theorem? I don't know but I feel like this could link to something I've already learned..
• It is a similar idea. The mean value theorem says that there is a value for the slope on a line in the interval that equals the average slope. This is the same but for average height. However, they are not directly related in that the two values will be differnt.
(1 vote)
• So basically, the average value theorem represents the average height of f(X)?
• Yes, essentially the Average Value Theorem provides you with the average y-value(or height) of the function over a designated interval. By adding up all of the y-values within the interval via the integral, and then dividing by the width of the interval, you obtain the average y-value(or height).
(1 vote)
• What is the practical use of finding the average value over a closed interval? In other words, what are some real world applications of this technique?
(1 vote)
• Does the mean value of integration and mean value of differentiation should be same?
For integral, there is a point c between the interval a,b, whose height is the average height of the area.
For differentiation, there is a point c between the interval a,b, whose slope at f(c) is the slope of the function.
Does both c should be same for the same function and same interval?
Thanks,
Matrix
(1 vote)
• integration and differentiation are both operations. Hence it does not make sense to find the mean value of it.

Your wording seems a bit off.

whose slope at f(c) is the slope of the function.

A function with the domain of real numbers can can have infinite number of slopes i.e. gradients

However perhaps you could come up with a function and see if matches the criteria you have mentioned.
(1 vote)
• Why is the area under the curve equal to the average area or the rectangle?
(1 vote)
• Because that is essentially the meaning of an average. Basically, the mean height or average height is the specific value for which half the values are above it and half the values are below it. Or, you can think of it as all value above this mean height get complemented by a value below. Thus, when you multiply by the width of the function you're left with the same area.

Another way you can think about it is the algebraic way that Sal represented it:

f_avg * (b-a) = ∫a,b [f(x)] dx.

Let's examine this equation. `f_avg` is the average height, for simplicity lets call it `h`. Next, `(b-a)`, if you look at the graph, is just the interval or width we are looking at for the average value. So, because its the width we are looking at, let's call it `w`. Finally, the integral at the end is, as you mentioned, the area under the curve in our interval w. Rewriting our equation, we get:

h * w = ∫w [f(x)] dx.

I rewrote the integral as being evaluated at w, even though this isn't correct notation, I am just showing that we are evaluating the integral over the interval of length w. What is `h*w`? It is just the equation for area of a rectangle. Now we can see that the area of the rectangle with a height of `f_avg` and a width of `w`, which is our interval, is just the area under the curve over the interval.
(1 vote)