Average value over a closed interval
Average value of a function over a closed interval.
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- Conceptually how can I get the average height by dividing the integral,which is the area,and its width?(15 votes)
- 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.(73 votes)
- 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?(4 votes)
- 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(16 votes)
- At1:58, 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?(7 votes)
- 1:58, 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(5 votes)
- 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.
Issa Qaqish(4 votes)
- 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.(2 votes)
- Could one also apply the MVT for derivatives to f(x) on (a,b) to find this average value?(4 votes)
- 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..(2 votes)
- 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)?(2 votes)
- 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?
- 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_avgis the average height, for simplicity lets call it
(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_avgand a width of
w, which is our interval, is just the area under the curve over the interval.(1 vote)
what I want to do in this video is think about the idea of an average value of a function over some closed interval so what do I mean by that and how could we think about what average value of a function even means so let's say that's my y-axis and let's say that this is my this right over here is my x-axis and let me draw a function here so let's say the function looks something like that that's the graph of y is equal to Y is equal to f of X and now let's think about a closed interval so we're going to think about the closed interval between a and B including a and B that's what makes it closed or including our endpoints we're going to think about this interval right over here so between x is equal to a and X is equal to B what is the average value of this function one way to think about it is what is the average height of this function so how could we what would that mean well one way to think about it it would be some height so that if we multiply it times the width of this interval we'll get the area under the curve so another way to think about it is the area under the curve right over here the area let me do this in a different color so the area under this curve right over here I'll shaded in yellow we already know that we can express this as the definite integral from A to B of f of X DX the average value of our function over this closed interval a B let me write that over over the closed interval between a and B including a and B we could think about it as some height some height let me do this in a new color some value of our function some height let me think about maybe some height right over here so that if we multiply this height times this width we're going to get the area of a rectangle the rectangle is would be the area of this rectangle right over here and that rectangle is going to have the same area as the area under the curve which is a reasonable way if you kind of remember how when you even think about finding the area or one way to find think about the area of a trapezoid you can multiply if you have a trapezoid if you have a trapezoid like this you have a trapezoid like this this is you can kind of you kind of turn 90 degrees but your multiply the height times the average width of the trapezoid and then that will give you its area so this would be the average width which in a which in a trapezoid like this would just be halfway between this function is not linear so it's not necessarily going to be halfway in between but it's that same idea so how could we use this idea where this is right over here this height this height right over here we could call this we could call this the functions average the functions average how could we use all of this to come back to with a formula for the the the average of a function over this closed interval well let's just Express in math what we've already said we already said that this function average should be some height so let's say the function average so if I that's a height and if I multiply it times the width of this interval so this width right over here this width right over here is just going to be the larger value minus the smaller value so that's going to be B minus a so the average value of the function times the width of the interval should give us an area that is equivalent to the area under the curve so it should be equal to the definite integral from A to B of f of X DX and so if we just if we knew all of this other stuff we could solve for the functions average the functions average we divide both sides by B minus a the functions average is going to be equal to just dividing both sides by B minus a you're going to get 1 over B minus a times the definite integral the definite integral from A to B of f of X DX or another way to think about it you're going to figure out what the area under the curve is over that interval you're going to divide that by the width and then you're going to have the functions average you want me to think about you're going to have the average height and once again I'd like to remind you that because this is you shouldn't just sit there and try to memorize this thing and just just get a conceptual understanding of what this is really just trying to say area under the curve divided by the width well that's just going to give you the average height that's going to give you the average of the function in the next video we'll actually apply this formula to see that it's actually straightforward to calculate if you if you can figure out the definite integral