The second part of the fundamental theorem of calculus tells us that to find the definite integral of a function ƒ from 𝘢 to 𝘣, we need to take an antiderivative of ƒ, call it 𝘍, and calculate 𝘍(𝘣)-𝘍(𝘢). Get some intuition into why this is true. Created by Sal Khan.
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- i dont entirely understand how finding the area under the line of v(t) would give u the change in position if change in position is equal to change in position/t(35 votes)
- v(t) is just a function of velocity with respect to time. that means that velocity will be the dependent variable (y) and time will be the independent variable (x). since velocity is the same thing as distance/time we can rewrite the y variable as distance/time. Therefore, when we find the area under the v(t) function, we are multiplying the combination of a bunch of rectangles together with the base being time (x) and the height being velocity (y or distance/time). Therefore, when we do the following calculation: time*(distance/time), the 'time's cancel out and we are left with just distance travelled which is the same thing as change in position.(64 votes)
- At10:25, I thought that this was First fundamental theorem of calculus (basically, how to take an integral).
I learned that the second fundamental theorem of calculus was: that if you take the derivate of an integral from 0 to x, you get f(x). Is this incorrect?(11 votes)
- Yes, that is what my calculus book says too. Farther down on the playlist for "indefinite and definite integrals" is a set of videos for "fundamental theorem of calculus". In these videos it becomes clear what Sal's designation of "first" and "second" is just switched of what our book says. Another comment in those section of videos says that in other references the first and second theorems of calculus are just parts of one theorem of calculus. I don't think it matters which you view as first and which you view as second. They are really just different viewpoints of the same idea.(2 votes)
- Please point to the best calculator online , for these equations .. Thanks(3 votes)
- At11:38, Sal says F(x) is 'an' anti-derivative of f(x). How can a single curve have multiple anti-derivatives?(3 votes)
- When you take the derivative, if there is a constant term, it disappears. so you're losing information. When you find the antiderivative, you don't know what the constant term was, which is why a single curve has (infinitely) many antiderivatives.(11 votes)
- Is the area under the upper graph useful for anything?
It looks like it would be a Reimann sum of delta-t * s(t-1), which is a sum of times * distances, or in other words the sum of velocities at specific moments. And that doesn't make much sense.(5 votes)
- Thinking of it in measurements, the unit of the area under the first curve would be m*s (position times time) and that certainly doesn't make sense. Abstractly speaking though, area under the curve of a function (for example f(x) ) can describe an aspect of of it's anti-derivative (F(x) ) so long as it has one.(4 votes)
- Why the capital F notation?(4 votes)
- In the beginning I always thought of the antiderivative as the Original Function, so if I saw f(x) I would think derivative and F the anti or original function.(2 votes)
- Is this the first fundamental theorem? My book has the second fundamental theorem as being d/dx integral on [x,a] of f(t)dt=f(x). Where can you find the video on the second fundamental theorem?(2 votes)
- Different authors may present these theorems in a different order depending on how they developed the background material leading to the definition of the FToC. Some call both the FToC, but one is part 1 and the other part 2 and some even call one of them (research!) a definition, not a theorem.
Anyhoo, try this: https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/definite_integrals/v/connecting-the-first-and-second-fundamental-theorems-of-calculus(3 votes)
- I don't understand how to solve problems by this theory explained(2 votes)
- I can relate to your position. As you keep getting higher and higher in math, the amount of "solution by application of rules and formulas" starts to decrease, and the "solution by application of implications and consequences of theorems" starts to increase. In a first year calculus program, that usually happens first with the formal definition of limits, and then again here at the fundamental theorem of calculus. Here is a brief explanation of the relation of the F.T.o.C. and the type of problem you may encounter testing your insight/understanding of the theorem. I hope it helps. If not, do not hesitate to ask for clarification. I like to take breaks from my regular work and come here to help students, so I will answer as fast as I can. Here is the link: http://bajasound.com/khan/khan0002.jpg.
Keep studying!(3 votes)
- Would it have been better to say "distance travelled" rather than "change in position", as we are not looking at displacement, or have I completely misunderstood the video and we ARE in fact talking about displacement of an object from a specific point? Does it even matter?(2 votes)
- In fact, for a strictly increasing position function, these two concepts will be exactly the same. However, Sal's description describes a change in position, and if you look at a function with a change in direction ( in which velocity changes signs), his description will give displacement, not total distance traveled. It might be useful to think of this displacement as a 1-dimensional vector, with direction being given by the sign.(2 votes)
- My teacher taught this as the first fundamental theorem of calculus, not second.(1 vote)
- I believe most authors use the terminology your teacher uses. Not a big deal, though: it's basically an arbitrary choice as to which one you call first and which you call second. You just have to be aware of which one Sal (or your teacher) is talking about.(3 votes)
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.