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Video transcript

let's say that we've got some function f that is continuous continuous on the interval A to B so let's try to see if we can visualize that so this is my y-axis that's my y-axis this right over here I'm going to make it my t axis well we'll use X a little bit later so I'll call this my T axis and then let's say that this right over here is the graph of y is equal to f of T y is equal to f of T and we're saying it's continuous on the interval from A to B so this is T is equal to a this is T is equal to B so we're saying that it is continuous it is continuous over this whole interval now for fun let's define a function capital f of X and I will do it in blue let's define capital f of X as equal to the definite integral from a from a as a lower bound to X to X of F of T let me do that of F of T of F of T DT where X is in this interval where where a is less than or equal to X is less than or equal to B or that's just another way of saying that X is in this interval right over here now when you see this you might say oh you know the definite integral this has to do with differentiation and anti derivatives and all that but we don't know that yet all we know right now is that this is the area under the curve F between a and X so between a and let's say this right over here this right over here is X so f of X is just is just this area this area right over here that's all we know about it we don't know it has anything to do with anti derivatives just yet that's what we're going to try to prove in this video so just for fun let's take the derivative of F and we're going to do it just using the definition of derivatives and see what that what we get when we take the derivative using the definition derivatives so we would get so the derivative F prime of X well this definition of derivatives it's the limit as Delta X approaches zero of capital f of X plus Delta X minus f of X all of that all of that over Delta X this is just the definition of the derivative now what is this equal to well let me rewrite it using these integrals right up here this is going to be equal to this is going to be equal to the limit the limit as Delta X approaches zero of what's f of X plus Delta X well put X in right over here you're going to get the definite integral from a to X plus Delta X of F of T DT and then from that you are going to subtract you are going to subtract this business f of X which we've already written as the integral from a the definite integral from a to X of F of T DT and then all of that is over Delta X all of this is over all of this is over Delta X now what is this represent remember we don't know anything about definite integrals to somehow dealing with something with an antiderivative and all that we just know this this is another way of saying the area under the curve F between a and X plus Delta X so it's the area under the curve F between a and X plus Delta X X plus Delta X so it's this entire area it's this entire area right over here so that's this part we already know what this blue stuff is this blue stuff in that same shade of blue so this blue stuff right over here this is equal to all of this business we've already shaded this in it's equal to all of this business right over here so if you were to take all of this green area which is from a to X plus Delta X and subtract out this blue area which is exact what we're doing in the numerator what are you left with well you're going to be left with you're going to be left with what color have I not used yet maybe I will use this pink color well now I already use that I'll use this purple color you're going to be left with this area you're going to be left with this area right over here so what's another way of writing that well another way of writing this area right over here is the definite integral between X and X plus Delta X of F of T of F of T DT so we can rewrite this entire expression the derivative of capital f of X this is capital F prime of X we can rewrite it now as being equal to the limit as Delta X approaches 0 this I can write as 1 over Delta X times the numerator the numerator we already figured out the numerator the green area minus the blue area is just the purple area which is and another way of sewing is denoting that area is this expression right over here so 1 over Delta X times the definite integral from X to X plus Delta X of F of T F of T DT now this expression is interesting this might look familiar from the mean value theorem of definite integrals the mean value theorem of definite integrals tells us so the mean mean value theorem of definite definite integrals definite integrals tells us there exists there exists a C in in the interval so I could see where I'll write this way where a is less than or equal to C which is less than or actually let me make it clear the interval that we now care about is between X and X plus Delta X where at where X is less than or equal to C which is less than or equal to X plus Delta X such that such that the function evaluated at C the function of value C so let me draw this see so there's a see someplace over here so if I were to take the function evaluated at the C so that's F of C right over here so if I were to take the function evaluate the C which would essentially be the height of this line and I multiply it times the base this interval if I multiply it times the interval which and this interval is just Delta X X plus Delta X minus X is just Delta X so if we just multiply the height times the base times the base that this is going to be equal to the area under the curve is going to be equal to the area under the curve which is the definite integral from X to X plus Delta X X to X plus Delta X of F of T f of T DT this is what the mean value theorem of integrals tells us if f is a continuous function there exists two C in this interval between our two endpoints between our two end points where the function evaluated the C is essentially you can view it as the mean height and if you take that mean value of the function and you multiply it times the base you're going to get the area of the curve or another way of rewriting this you could say that F of C there exists a C in that interval where f of C is equal to 1 over Delta X I'm just dividing both sides by Delta X times the definite integral from X to X plus Delta X of F of T DT and this is often viewed as the mean value of the function over the interval why is that well this part right over here this part right over here gives you the area and then you divide the area by the base and you get the mean height or another way you could say it is if you were to take the height right over here multiply it times the base you get a rectangle that has the exact same area as the area under the curve well this is useful because this is exactly what we got as the derivative of F prime of X so there must exist a C such that such that F of C is equal to this stuff or we could say that the limit and let me rewrite all of this down a new color so there exists there exists a C in the interval X to X plus Delta X where we're F prime of X which we know is equal to this if we can now say is now equal to the limit as Delta X approaches zero instead of writing this we know that there's some C that's equal to all of this business of f of C now we're in the homestretch we just have to figure out what the limit as Delta X approaches 0 of f of C is and the main realization is this part right over here we know that C is always sandwiched in between X and X plus Delta X and intuitively you could tell that look as Delta X approaches 0 is Delta X as this Green Line as this green line right over here moves more and more to the left as it approaches this as it approaches this blue line as it approaches this blue line the C has to be in between and so the C is going to approach X so we know intuitively we know intuitively that C approaches X as Delta X Delta X approaches 0 or another way of saying it is that F of C is going to approach f of X as Delta X as Delta X approaches 0 and so intuitively we could say that this is going to be equal to F of this is going to be equal to f of X now you might say ok that's intuitively but we're kind of working on a little bit of a proof here Sal tell them let me know for sure that X is going to approach you don't just do this little thing where you drew this diagram and it makes sense that C is going to have to get closer and closer to X and if you want that you could just resort to the squeeze theorem and to resort to the squeeze theorem you just have to view C as a function of Delta X and it really is depending on your Delta X C is going to be further to the left or to the right possibly and so I can just rewrite this expression as X is less than or equal to C as a function of Delta X which is less than or equal to X plus Delta X so now you see that C is always sandwiched between X and X plus Delta X well what's the limit of X as Delta X approaches 0 well X is independent on Delta X in any way so this is just going to be equal to X what's the limit of X plus Delta X as Delta X approaches zero well as Delta X approaches 0 this is just going to be equal to X so if this approaches X is X Delta X approaches zero and it's less than this function and if this approaches X is Delta X approaches zero and it's always greater than this then we know from the squeeze theorem or the sandwich theorem that the limit as Delta X approaches 0 of C as a function of Delta X is going to be equal to is going to be equal to X as well it has to approach the same thing that that and that is it's sandwiched in between and so that's a slight we resort to the sandwich theorem it's a little bit more rigorous to get to this exact result as Delta X approaches 0 C approaches X if C is approaching X then F of C is going to approach f of X and then we essentially have our proof F is a continuous function we defined f in this way capital f in this way and we were able to use just the definition of the derivative to figure out that the derivative of capital f of X is equal to is equal to is equal to f of X and once again why is this a big deal well it tells you that if you have any continuous function f and that's what we assume we assume that f is continuous over the interval there exists some function there exist a function you can just define the function this way is the area under the curve between between some endpoint or the the beginning of the interval and some X if you define a function in that way the derivative of this function is going to be equal to your continuous function or another way of saying it is that you always have an antiderivative that any continuous function has an antiderivative and so it's a couple of cool things any continuous function has an antiderivative is going to be that capital it's going to be that capital f of X and this is why it's called the fundamental theorem of calculus it ties together these two ideas and you have differential calculus you have the ideas you have the idea of a derivative and then an integral calculus you have the idea of an integral before this proof all we viewed an integral as is the area under the curve is it was just literally a notation to say the area under the curve but now we've been able to make a connection that there's a connection between the integral and the derivative or connection between the integral and the antiderivative in particular so it connects all of calculus together in a very very very powerful and and we were so used to it now and and and now we can say almost a somewhat obvious way but it wasn't obvious remember we always think of integrals as somehow doing an antiderivative but it wasn't clear if you just viewed an integral as only an area you would have to go through this process a Wow no it's connected it's connected to the process of taking a derivative