If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Mean value theorem for integrals

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.B (LO)
,
CHA‑4.B.1 (EK)

## Video transcript

we have many videos on the mean value theorem but I'm going to do 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 continuous on the closed interval so it's including the endpoints from A to B and it is differentiable it is differentiable so the derivative is defined on the open interval from A to B so not necessarily have to be defined that differentiable at the boundaries as long as it's differentiable between the boundaries then we know then we know that there exists there exists some value or some number I guess is some number some number C such that C is between the two endpoints of our interval so I mean such that a is less than C is less than B so C is in this interval and and and this is kind of the the meet of it the meet of it is that the derivative of the derivative of our function at that point the derivative of the function at that point you could view it as the slope of the tangent line at that point is equal to essentially the average the average rate of change over the interval or you could even think about it as the slope between the two endpoints so the the slope between the two endpoints it's going to be your change in Y which is going to be a change in your function value so f of B minus F of a over over B minus a and once again we do this one 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 because I think it's always handy the mean value theorem that we learned in differential calculus just tells us a hey look you know 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 divided by so this right over here is f of B 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 sort of change in Y over change in X that gives us the slope this right over here gives us the slope of this line this the slope of the line that connects that connects these two points that's this quantity and this the mean value theorem tells us that there's some C in between a and B where you're going to 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 exist to see where the slope of the tangent line at that point is going to be the same so this would be AC right over there and we actually might have a couple of C's that's another candidate see 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 something might it might evoke some similarities with what we saw when we saw the the how we defined I guess you could say or the formula for the average value of a function remember of what we saw for the average value of function we said the average the average value of a function is going to be equal to one over B minus a notice 1 over B minus a fo b minus a in the denominator here times the definite integral from A to B of f of X DX now this is interesting because 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 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 retry to do that so once again this is 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 going take the antiderivative of F prime of X which is going to be f of X and you're going to evaluate it at be f of B and then from that you're going to subtract it evaluated a minus F of a these two things are identical and then you can of course divided by divided by B minus a now this is starting to get interesting one way to think about it is there must be a there must be a C if we there must be a C that takes on the average value of there must be AC that when you when you evaluate the derivative at C it takes on the average value of the derivative or another way to think about it another way to think about it if we were to just write if we were to just write let me say 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 one over is equal to one over B minus a one over B minus a so there exists a C where G of C is equal to one over B minus a times the definite integral from A to B of G of X G of X DX F prime of X is the same thing as G of X so another way of saying thinking about and this is actually another form of the mean value theorem it's called the mean value theorem for integrals mean value theorem for integrals let me so this is the mean I'll just write the acronym mean value theorem for for integrals or integration which essentially give it 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 continuous on this closed interval on going from A to B then there exists 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 is exist to see 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 it just to show 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 when in differential calculus but now there's a different notation and I guess you can have a slightly different interpretation we were thinking about it in in differential calculus we're thinking about having a point where the slope of the tangent line of the function at that point is the same as the average rate so that's when we had our kind of differential mode and we're kind of thinking 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 there's some C where the function evaluate at that point is equal to the average value so another way of thinking about it if I were to draw if I were to draw G of if I were to draw G of X if I were to draw G of X so that's X that is my Y axis this is the graph of y is equal to G of X which of course was the same thing as f prime of X but we've card I guess 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 we've already seen how to calculate the average value so maybe the average value is is that right over there so that is G average so our average value is this the mean value theorem for integrals there's some see where our function must take on where our function must take on that value at C where's that C is in the side where the C is in that interval
AP® is a registered trademark of the College Board, which has not reviewed this resource.