AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 6Lesson 7: The fundamental theorem of calculus and definite integrals
The fundamental theorem of calculus and definite integrals
There are really two versions of the fundamental theorem of calculus, and we go through the connection here. Created by Sal Khan.
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- I don't know if I'm just missing it in the videos, but the videos haven't answered this and my textbook doesn't answer this. What exactly does F(x) mean, versus f(x)? It's been confusing me the whole time Sal has been using it in his videos, and I've been waiting for him to address it specifically. It's keeping me from understanding what in the world is going on with these 2 theorems. Thanks!(42 votes)
- Actually that is a very good question.
F(x) is defined as the antiderivative of f(x).
To be a little more rigorous, F(x) is formally defined as:
F(x) = ∫ f(t) dt (lower bound a, upper bound x)
And F(x) is specified as being defined for all x over the closed interval [a,b].
Because of this definition:
F'(x) = f(x) for all x in the open interval (a,b)
(There are some more parts to this, but hopefully that should help you understand a little better.)(61 votes)
- This stuff is all about
b. Why does it need
dfrom the start? Are they some kind of the limit of
- you need C and D to calculate the area of some form(3 votes)
- 4:31I'm confused about the use of the word "anti-derivative", isn't it the same as saying "integral"?(8 votes)
- An antiderivative is an indefinite integral.
There are other kinds of integrals. The main one you'll be dealing with is the definite integral. In more advanced mathematics there are some subtleties of definitions of various kinds of integrals. I will leave that to a professional mathematician because I am not fully-versed in the nuances.(17 votes)
- I am a bit confused with the two parameters x and t. Can someone clarify their relationship for me?(10 votes)
- lower case f(t) is a derivative....so it is a rate.....some unit per second or y per second or y per minute or whatever your t values are would then be a rate...so it is f(t) , the derivative...then integrate to get the antiderivative.......which then means that F(t) is the anti derivative, or as I like to think of it as the original function that you are being given the derivative of. Y and x coordinates would give you the function and then you would get the derivative but we are integrating which is the reverse order....hope that helps and did not confuse you more.(5 votes)
- If F(x) = integral of f(t) with boundaries a to x,
then F(b) = integral of f(t) with boundaries a to b, right?
So, by the 2nd theorem, F(b) = F(b) - F(a) ? and F(a) = 0 ?!(7 votes)
- Given the way F(x) is defined, it should not be surprising that F(a) = 0, as it would be the integral of f(t) with boundaries a to a.(6 votes)
- Why was
Capital F (x) = ∫ f( t )dt
Capital F (x) = ∫ f( x ) dx(6 votes)
- Because the bounds of ∫ f(t) are [0,x]. That makes F(x) a function of x, not t.(8 votes)
- I really don't understand why the second fund thr can still be used if the upper limit is not x by finding the upperbound derivative. Can someone please explain?
-Test tomorrow!(4 votes)
- Consider this: instead of thinking of the second fundamental theorem in terms of x, let's think in terms of u. A function for the definite integral of a function f could be written as
F(u) = | f(t) dt
By the second fundamental theorem, we know that taking the derivative of this function with respect to u gives us
Now, what if
u = g(x)where g(x) is any function of x? This means that
| f(t) dt = | f(t) dt = F(g(x))
Then if take the derivative with respect to x of F(g(x)), which is the derivative of an integral with an upper bound other than x, we can just use the chain rule, which gives us
d/dx F(g(x)) = f(g(x)) * g'(x)(given that F'(x) = f(x) ).
So we end up with
the derivative of the upper boundmultiplied by
the inner function (integrand) evaluated at the upper bound.
(Credit to MIT on edX for the explanation.)(7 votes)
- what does the sign that looks like an elongated S signify?(4 votes)
- The "elongated S" is the integral symbol which signifies that we are integrating the function that comes after the symbol.(5 votes)
- At0:54, it is stated that if f(t) is continuous at a given interval, then it differentiable at every point in it's domain.......
But in the differential calculus course, we learned that a continuous function is not necessarily differentiable...eg:- f(x) = |x| is not differentiable at x=0, even though it is continuous.....
Please help(4 votes)
- It is only claimed that the continuity of
ƒimplies the differentiability of
- Why only this theorem is considered the Fundamental Theorem of Calculus not any other theorem or why is a certain theorem called the fundamental theorem?(3 votes)
- The fact that this theorem is called fundamental means that it has great significance. This theorem of calculus is considered fundamental because it shows that definite integration and differentiation are essentially inverses of each other.(3 votes)
Let's say we've got some function f that is continuous over the interval between c and d. And the reason why I'm using c and d instead of a and b is so I can use a and b for later. And let's say we set up some function capital F of x which is defined as the area under the curve between c and some value x, where x is in this interval where f is continuous, under the curve-- so it's the area under the curve between c and x-- so if this is x right over here-- under the curve f of t dt. So this right over here, F of x, is that area. That right over there is what F of x is. Now, the fundamental theorem of calculus tells us that if f is continuous over this interval, then F of x is differentiable at every x in the interval, and the derivative of capital F of x-- and let me be clear. Capital F of x is differentiable at every possible x between c and d, and the derivative of capital F of x is going to be equal to lowercase f of x. Fair enough. Now, what I want to do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. So let's think about what F of b minus F of a is, what this is, where both b and a are also in this interval. So F of b-- and we're going to assume that b is larger than a. So let's say that b is this right over here. And we'll do that in the same color. So let's say that b is right over here. F of b is going to be equal to-- we just literally replace the b where you see the x-- it's going to be equal to the definite integral between c and b of f of t dt, which is just another way of saying the area under the curve between c and b. So this F of b, capital F of b, is all of this business right over here. And from that, we are going to want to subtract capital F of a, which is just the integral between c and lowercase a of lowercase f of t dt. So let's say that this is a right over here. Capital F of a is just literally the area between c and a under the curve lowercase f of t. So it's this right over here. It's all of this business right over here. So if you have this blue area, which is all of this, and you subtract out to this magenta area, what are you left with? Well, you're left with this green area right over here. And how would we represent that? How would we denote that? Well, we could denote that as the definite integral between a and b of f of t dt. And there you have it. This right over here is the second fundamental theorem of calculus. It tells us that if f is continuous on the interval, that this is going to be equal to the antiderivative, or an antiderivative, of f. And we see right over here that capital F is the antiderivative of f. So we could view this as capital F antiderivative-- this is how we defined capital F-- the antiderivative-- or we didn't define it that way, but the fundamental theorem of calculus tells us that capital F is an antiderivative of lowercase f. So right over here, this tells you, if you have a definite integral like this, it's completely equivalent to an antiderivative of it evaluated at b, and from that, you subtract it evaluated at a. So normally it looks like this. I've just switched the order. The definite integral from a to b of f of t dt is equal to an antiderivative of f, so capital F, evaluated at b, and from that, subtract the antiderivative evaluated at a. And this is the second part of the fundamental theorem of calculus, or the second fundamental theorem of calculus. And it's really the core of an integral calculus class, because it's how you actually evaluate definite integrals.