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### Course: Integral Calculus>Unit 3

Lesson 1: Average value of a function

# Mean value theorem for integrals

Here Sal goes through the connection between the mean value theorem and integration.

## Want to join the conversation?

• So, basically, the mean value theorem for integrals is just saying that there is a c equal to the average value of a function over [a,b], correct? And the mean value theorem is finding the points which have the same slope as the line between a and b?
(14 votes)
• If you mean in the case of MVT for integrals that there is a c in in the interval [a, b] such that f(c) is equal to the average value of the function over the interval, then you are correct, Sierra!
Well done!
(14 votes)
• How exactly is this useful? How does determining one or more point(s) within the interval [a,b] such that the gradient of that point is equal to the gradient between (a, f(a)) and (b, f(b))help us? Are we usually interested in obtaining the mean value of the derivative?
(5 votes)
• So, let me see if I get this right...

· MVTd ("for derivatives"):There's a point C where f'(c) = slope of secant that goes through a and b.· MVTi ("for integrals"):There's a point C where f(c) * (b-a) = area of f(x) between a and b.So, if you calculate MVTi of f'(x) you get MVTd of f(x)?

Thanks!
(7 votes)
• I'm not sure what you mean by "MVTi of f'(x)" and "MVTd of f(x)". The MVT is not a function; it is just a theorem that states the existence of a certain number c.
(3 votes)
• We actually prove fundamental theorem of calculus using mean value theorem of integration. Then how can we use the same result to verify the theorem?
(7 votes)
• Here g(x) is the derivative of the function f(x). Does that mean the theorem aplies only to functions which can be proven to be derivatives of other functions? I don't even know if there are functions that can be shown NOT to be the derivative of some function, but say there are...
(4 votes)
• The mean Value Theorem is about finding the average value of f over [a, b].
The issue you seem to be having is with the Fundamental Theorem of Calculus, and it is not called fundamental for nothing. You really need to understand the FToC. If you really get it, you would understand the reason for the initial conditions that f is continuous on [a,b] and that f is differentiable on (a,b) and the implications those two conditions hold.

You are now at the stage of math where it isn't enough to be able to manipulate and solve expressions based on a pattern you have learned, you need to understand the theory in order to draw conclusions so you can apply other theory correctly to arrive at a solution. It can take time. I suggest you do a review of the FToC here and elsewhere. Try this to start:
http://www.sosmath.com/calculus/integ/integ04/integ04.html
Keep Studying!
Keep Asking Questions!
(1 vote)
• so this means that f'(c) actually is the average slope for the function f(x) and it is also when looked from a different vantage point is the average height which gives me the area under the curve f'(x )?
(2 votes)
• No no no no, the "c" value where the slope of the function is equal to the slope of the secant line between 2 points is not the same as the "c" (or m or z or k) value where the function is equal to its average value between those 2 points.

BUT, that IS the point where the antiderivative of f has a slope equal to the secant line on the ANTIDERIVATIVE of f between those 2 points.

Lets say F(x) is the antiderivative of f(x), and f'(x) is the derivative of f(x).
If f(x) has its average SLOPE at x = c, then f'(x) has its average VALUE at x = c.
Take a step down the derivative chain.
If f(x) has its average VALUE at x = c, then F(x) has its average SLOPE at x = c.
(2 votes)
• 1. at , how to understand ∫ f'(x) dx =f(x)?

2. at , is it possible to have more than 1 c on [a,b]?
(1 vote)
• 1. It's a direct result of the Fundamental Theorem of Calculus. If I have a function $f(x)$ and its derivative $f'(x)$, it means the antiderivative of $f'(x)$ is $f(x)$. So, $\int\limits_{}f(x)dx$ would be $f(x) + c$. Taking the bounds into consideration, we get $f(b) - f(a)$.

2. Yep. The MVT guarantees one value of c to exist, but multiple can exist as well.
(4 votes)
• So does this essentially mean that if I plot F(x) = def integral (a,x) of f(x) dx, I would see that there will be some c of f(x) that would be equal to the average slope of F(x) from a to b?
(2 votes)
• I wish there was a feature here to search the questions and answers for certain keywords, so I can find if this question was already asked/answered.
As presented, the MVT for derivatives and the MVT for integrals seem to be a kind of reciprocal of the other or have some one-to-one relation. E.g. the point c was shown as the point where the derivative of the function has the average value (slope between a and b). Then, that same point c was used to show that there exists an average of the a function value. However the average of the derivative can exist in multiple points while the average of the function is unique (i.e. there is one only point where the function value is the average). Where is the contradiction here and how can this be clarified?
(2 votes)
• Howdy Mircea,

From what I gather, you are saying that there is only one point c, such that f(c) is equal to the average of f(a) and f(b) over the interval from a to b, correct?

If so, then your reasoning is not correct. There can be multiple points such that the value of the function is equal to the average over the interval. Let's look at 3 different scenarios:
(1.) A horizontal line will have infinite many x-values such that f(x) is equal to the average of f(a) and f(b).
(2.) A non-horizontal line from a to b (a diagonal) will have exactly one average. I think this is where you got confused. Note however, that in the third example we are not dealing with linear functions.
(3.) In this final example, we have a polynomial f(c) of degree 3 (look up one to see what it looks like). A polynomial of degree 3 has a level of symmetry that it is quite likely for the graph to have two points that are equal to the average of the function.

Hope this helps.
(1 vote)
• Isn't this already included in the Intermediate Value Theorem? Since the average of f(x) must be between f(a) and f(b), then according to the IVT, of course there is a c in [a,b] where f(x) equals the average. f(x) must equal ALL values between f(a) and f(b) on that interval.
(2 votes)

## Video transcript

- [Voiceover] We have many videos on the mean value theorem, but I'm going to review it a little bit, so that we can see how this connects the mean value theorem that we learned in differential calculus, how that connects to what we learned about the average value of a function using definite integrals. So the mean value theorem tells us that if I have some function f that is continuous on the closed interval, so it's including the endpoints, from a to b, and it is differentiable, so the derivative is defined on the open interval, from a to b, so it doesn't necessarily have to be differentiable at the boundaries, as long as it's differentiable between the boundaries, then we know that there exists some value, or some number c, such that c is between the two endpoints of our interval, so such that a < c < b, so c is in this interval, AND, and this is kind of the meat of it, is that the derivative of our function at that point, you could use as the slope of the tangent line at that point, is equal to essentially the average rate of change over the interval, or you could even think about it as the slope between the two endpoints. So the slope between the two endpoints is gonna be your change in y, which is going to be your change in your function value, so f of b, minus f of a, over b minus a, and once again, we go into much more depth in this when we covered it the first time in differential calculus, but just to give you a visualization of it, 'cause I think it's always handy, the mean value theorem that we learned in differential calculus just tells us, hey look, if this is a, this is b, I've got my function doing something interesting, so this is f of a, this is f of b, so this quantity right over here, where you're taking the change in the value of our function, so this right over here is f of b, minus f of a, is this change in the value of our function, divided by the change in our x-axis, so it's a change in y over change in x, that gives us the slope, this right over here gives us the slope of this line, the slope of the line that connects these two points, that's this quantity, and the mean value theorem tell us that there's some c in between a and b where you're gonna have the same slope, so it might be at LEAST one place, so it might be right over there, where you have the exact same slope, there exists a c where the slope of the tangent line at that point is going to be the same, so this would be a c right over there, and we actually might have a couple of c's, that's another candidate c. There's at least one c where the slope of the tangent line is the same as the average slope across the interval, and once again, we have to assume that f is continuous, and f is differentiable. Now when you see this, it might evoke some similarities with what we saw when we saw how we defined, I guess you could say, or the formula for the average value of a function. Remember, what we saw for the average value of a function, we said the average value of a function is going to be equal to 1 over b minus a, notice, 1 over b minus a, you have a b minus a in the denominator here, times the definite integral from a to b, of f of x dx. Now this is interesting, 'cause here we have a derivative, here we have an integral, but maybe we could connect these. Maybe we could connect these two things. Well one thing that might jump out at you is maybe we could rewrite this numerator right over here in this form somehow. And I encourage you to pause the video and see if you can, and I'll give you actually quite a huge hint, instead of it being an f of x here, what happens if there's an f prime of x there, so I encourage you to try to do that. So once again, let me rewrite all of this, this is going to be equal to... This over here is the exact same thing as the definite integral from a to b, of f prime of x dx. Think about it. You're gonna take the anti-derivative of f prime of x, which is going to be f of x, and you're going to evaluate it at b, f of b, and then from that you're going to subtract it evaluated at a, minus f of a. These two things are identical. And then, you can of course divide by b minus a. Now this is starting to get interesting. One way to think about it is, there must be a c that takes on the average value of, there must be a c, that when you evaluate the derivative at c, it takes on the average value of the derivative. Or another way to think about it, if we were to just write g of x is equal to f prime of x, then we get very close to what we have over here, because this right over here is going to be g of c, remember, f prime of c is the same thing as g of c, is equal to 1 over b minus a, so there exists a c where g of c is equal to 1 over b minus a, times the definite integral from a to b, of g of x dx, f prime of x is the same thing as g of x. So another way of thinking about it, this is actually another form of the mean value theorem, it's called the mean value theorem for integrals. I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, to give it in a slightly more formal sense, is if you have some function g, so if g is, let me actually go down a little bit, which tells us that if g of x is continuous on this closed interval, going from a to b, then there exists a c in this interval where g of c is equal to, what is this? This is the average value of our function. There exists a c where g of c is equal to the average value of your function over the interval. This was our definition of the average value of a function. So anyway, this is just another way of saying you might see some of the mean value theorem of integrals, and just to show you that it's really closely tied, it's using different notation, but it's usually, it's essentially the same exact idea as the mean value theorem you learned in differential calculus, but now your different notation, and I guess you could have a slightly different interpretation. We were thinking about it in differential calculus, we're thinking about having a point where the slope of the tangent line of the function of that point is the same as the average rate, so that's when we had our kind of differential mode, and we were kind of thinking in terms of slopes, and slopes of tangent lines, and now, when we're in integral mode, we're thinking much more in terms of average value, average value of the function, so there's some c, where g of c, there's some c, where the function evaluated at that point, is equal to the average value, so another way of thinking about it, if I were to draw g of x, that's x, that is my y-axis, this is the graph of y is equal to g of x, which of course is the same thing as f prime of x, but we've just rewritten it now to be more consistent with our average value formula, and we're talking about the interval from a to b, we've already seen how to calculate the average value, so maybe the average value is that right over there, so that is g average, so our average value is this, the mean value theorem for integrals just tells us there's some c where our function must take on that value at c, whereas that c is inside, where the c is in that interval.