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Integral Calculus (2017 edition)
Course: Integral Calculus (2017 edition) > Unit 5
Lesson 1: Definite integral evaluation- The fundamental theorem of calculus and definite integrals
- Intuition for second part of fundamental theorem of calculus
- Area between a curve and the x-axis
- Area between a curve and the x-axis: negative area
- Definite integrals: reverse power rule
- Definite integral of rational function
- Definite integral of radical function
- Definite integral of trig function
- Definite integral involving natural log
- Definite integrals: common functions
- Area using definite integrals
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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!(43 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 aboutaandb. Why does it needcanddfrom the start? Are they some kind of the limit off(x)?(15 votes)
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.(18 votes)
- I am a bit confused with the two parameters x and t. Can someone clarify their relationship for me?(11 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.(7 votes)
Why was
instead ofCapital 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⌠u
F(u) = | f(t) dt
⌡a
By the second fundamental theorem, we know that taking the derivative of this function with respect to u gives usf(u).
Now, what ifu = g(x)where g(x) is any function of x? This means that⌠u ⌠g(x)
| f(t) dt = | f(t) dt = F(g(x))
⌡a ⌡a
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 usd/dx F(g(x)) = f(g(x)) * g'(x)(given that F'(x) = f(x) ).
So we end up withthe derivative of the upper boundmultiplied bythe inner function (integrand) evaluated at the upper bound.
(Credit to MIT on edX for the explanation.)(7 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(5 votes)
It is only claimed that the continuity ofƒimplies the differentiability ofF.(5 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)
- For anyone who is confused, try thinking of it this way.
F(x) is the area under the curve of f(t) from some value c to any input value of your choice, x.
So, you start with the area under the curve of the function f(t) from c to b which equals F(b). That area is going to be larger than the area from c to a which equals F(a).
The first part of the fundamental theorem of calculus tells us that the derivative of F(x) (which is just the rate of change of the area under f[t] ) is equal to the function f(x) (which is exactly the same function as f(t) just with a different variable). In other words, if you take the anti-derivative of f(x), you get F(x), which shows us that if you have f(x), you can find F(x).
Ok so now the fun part.
Let me ask a question, what happens when you subtract a smaller number from a larger number? You get the distance between the numbers!
So the same applies here, when you subtract the smaller quantity F(a)=[area under curve from c to a] from the larger quantity F(b)[area under curve from c to b], you get the quantity of the area between a AND b.
Simple question, how do we denote the area under the curve of f(t) from a to b? Why, its just the definite integral of f(t) from a to b!
And since we know the relationship between f(t), f(x), and F(x), we can easily evaluate the definite integral of f(t) from a to b!
Tada(5 votes)
4:31Video transcript
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.