Current time:0:00Total duration:12:24
0 energy points
Studying for a test? Prepare with these 9 lessons on Antiderivatives and the fundamental theorem of calculus.
See 9 lessons

Intuition for second part of fundamental theorem of calculus

Video transcript
Let's say that I have some function, s of t, which is positioned as a function of time. And let me graph a potential s of t right over here. We have a horizontal axis as the time axis. Let me just graph something. I'll draw it kind of parabola-looking. Although I could have done it general, but just to make things a little bit simpler for me. So I'll draw it kind of parabola-looking. We call this the y-axis. We could even call this y equals s of t as a reasonable way to graph our position as a function of time function. And now let's think about what happens if we want to think about the change in position between two times, let's say between time a-- let's say that's time a right over there-- and then this right over here is time b. So time b is right over here. So what would be the change in position between time a and between time b? Well, at time b, we are at s of b position. And at time a we were at s of a position. So the change in position between time a and time b-- let me write this down-- the change in position between-- and this might be obvious to you, but I'll write it down-- between times a and b is going to be equal to s of b, this position, minus this position, minus s of a. So nothing earth-shattering so far. But now let's think about what happens if we take the derivative of this function right over here. So what happens when we take the derivative of a position as a function of time? So remember, the derivative gives us the slope of the tangent line at any point. So let's say we're looking at a point right over there, the slope of the tangent line. It tells us for a very small change in t-- I'm exaggerating it visually-- for a very, very small change in t, how much are we changing in position? So we write that as ds dt is the derivative of our position function at any given time. So when we're talking about the rate at which position changes with respect to time, what is that? Well, that is equal to velocity. So this is equal to velocity. But let me write this in different notations. So this itself is going to be a function of time. So we could write this is equal to s prime of t. These are just two different ways of writing the derivative of s with respect to t. This makes it a little bit clearer that this itself is a function of time. And we know that this is the exact same thing as velocity as function of time, which we will write as v of t. So let's graph what v of t might look like down here. Let's graph it. So let me put another axis down here that looks pretty close to the original. I'll give myself some real estate, so that looks pretty good. And then let me try to graph v of t. So once again, if this is my y-axis, this is my t-axis, and I'm going to graph y is equal to v of t. And if this really is a parabola, then the slope over here is 0, the rate of change is 0, and then it keeps increasing. The slope gets steeper and steeper and steeper. And so v of t might look something like this. So this is the graph of y is equal to v of t. Now, using this graph, let's think if we can conceptualize the distance, or the change in position, between time a and between time b. Well, let's go back to our Riemann sums. Let's think about what an area of a very small rectangle would represent. So let's divide this into a bunch of rectangles. So I'll do it fairly large rectangles just so we have some space to work with. You can imagine much smaller ones. And I'm going to do a left Riemann sum here, just because we've done those a bunch. But we could do it right Riemann sum. We could do a trapezoidal sum. We could do anything we want. And then we could keep going all the way-- actually, let me just do three right now. Let me just do three right over here. And so this is actually a very rough approximation, but you can imagine it might get closer. But what is the area of each of these rectangles trying-- what is it an approximation for? Well, this one right over here, you have f of a, or actually I should say v of a. So your velocity at time a is the height right over here. And then this distance right over here is a change in time, times delta t. So the area for that rectangle is your velocity at that moment times your change in time. What is the velocity at that moment times your change in time? Well, that's going to be your change in position. So this will tell you-- this is an approximation of your change in position over this time. Then the area of this rectangle is another approximation for your change in position over the next delta t. And then, you can imagine, this right over here is an approximation for your change in position for the next delta t. So if you really wanted to figure out your change in position between a and b, you might want to just do a Riemann sum if you wanted to approximate it. You would want to take the sum from i equals 1 to i equals n of v of-- and I'll do a left Riemann sum, but once again, we could use a midpoint. We could do trapezoids. We could do the right Riemann sum. But I'll just do a left one, because that's what I depicted right here-- v of t of i minus 1. So this would be t0, would be a. So this is the first rectangle. So the first rectangle, you use the function evaluated at t0. For the second rectangle, you use the function evaluated at t1. We've done this in multiple videos already. And then we multiply it times each of the changes in time. This will be an approximation for our total-- and let me make it clear-- where delta t is equal to b minus a over the number of intervals we have. We already know, from many, many videos when we looked at Riemann sums, that this will be an approximation for two things. We just talked about it'll be an approximation for our change in position, but it's also an approximation for our area. So this right over here. So we're trying to approximate change in position. And this is also approximate of the area under the curve. So hopefully this satisfies you that if you are able to calculate the area under the curve-- and actually, this one's pretty easy, because it's a trapezoid. But even if this was a function, if it was a wacky function, it would still apply that when you're calculating the area under the curve of the velocity function, you are actually figuring out the change in position. These are the two things. Well, we already know, what could we do to get the exact area under the curve, or to get the exact change in position? Well, we just have a ton of rectangles. We take the limit as the number of rectangles we have approaches infinity. We take the limit as n approaches infinity. And as n approaches infinity, because delta t is b minus a divided by n, delta t is going to become infinitely small. It's going to turn into dt, is one way to think about it. And we already have notation for this. This is one way to think about a Riemann integral. We just use the left Riemann sum. Once again, we could use the right Riemann sum, et cetera, et cetera. We could have used a more general Riemann sum, but this one will work. So this will be equal to the definite integral from a to b of v of t dt. So this right over here is one way of saying, look, if we want the exact area under the curve, of the velocity curve, which is going to be the exact change in position between a and b, we can denote it this way. It's the limit of this Riemann sum as n approaches infinity, or the definite integral from a to b of v of t dt. But what did we just figure out? So remember, this is the-- we could call this the exact change in position between times a and b. But we already figured out what the exact change in position between times a and b are. It's this thing right over here. And so this gets interesting. We now have a way of evaluating this definite integral. Conceptually, we knew that this was the exact change in position between a and b. But we already figured out a way to figure out the exact change of position between a and b. So let me write all this down. We have that the definite integral between a and b of v of t dt is equal to s of b minus s of a where-- let me write this in a new color-- where s of t is the-- we know v of t is the derivative of s of t, so we can say where s of t is the antiderivative of v of t. And this notion, although I've written it in a very nontraditional-- I've used position velocity-- this is the second fundamental theorem of calculus. And you're probably wondering about the first. We'll talk about that in another video. But this is a super useful way of evaluating definite integrals and finding the area under a curve, second fundamental theorem of calculus, very closely tied to the first fundamental theorem, which we won't talk about now. So why is this such a big deal? Well, let me write it in the more general notation, the way that you might be used to seeing it in your calculus book. It's telling us that if we want the area under the curve between two x points a and b of f of x-- and so this is how we would denote the area under the curve between those two intervals. So let me draw that just to make it clear what I'm talking about in general terms. So this right over here could be f of x. And we care about the area under the curve between a and b. If we want to find the exact area under the curve, we can figure it out by taking the antiderivative of f. And let's just say that capital F of x is the antiderivative-- or is an antiderivative, because you can have multiple that are shifted by constants-- is an antiderivative of f. Then you just have to take-- evaluate-- the antiderivative at the endpoints and take the difference. So you take the endpoint first. I guess you subtract the antiderivative evaluated at the starting point from the antiderivative evaluated at the end point. So you get capital F of b minus capital F of a. So if you want to figure out the exact area under the curve, you take the antiderivative of it and evaluate that at the endpoint, and from that, you subtract the starting point. So hopefully, that makes sense. In the next few videos, we'll actually apply it.