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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?
• 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!
• 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?
• 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!
• 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.
• 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?
• 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...
• 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!
• 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 )?
• 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.
• 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.
• 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?
• 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?
• 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.