The Fundamental Theorem of Calculus
This theorem bridges the antiderivative concept with the area
problem. Indeed, let f (x) be a function defined and continuous
on [a, b]. Consider the function
F(
x) =
f (
t)
dt
defined on [a, b]. Then we have
Since
dt = h, we get

f (
x) =
f (
t) 
f (
x)
dt .
Since F(t) is continuous, we can easily prove that

f (
x) = 0 .
This means that the function F(x) is differentiable and
F '(x) = f (x). In other words, the function F(x) is an
antiderivative of f (x). From this and what we learned about
antiderivatives, we obtain the following fundamental result:
The Fundamental Theorem of Calculus Let f (x) be
continuous on [a, b]. If F(x) is any antiderivative of
f (x), then
f (
x)
dx =
F(
b) 
F(
a) .
Remark. The number
F(b)  F(a) is also denoted by
[F(x)]^{b}_{a} (or
F(x) ). So
f (
x)
dx = [
F(
x)]
^{b}_{a}
Example. We have
cos(
t)
dt = [sin(
t)]
_{0}^{x} = sin(
x)  sin(0) = sin(
x) .
So
Example. We have
dt = [arctan(
t)]
_{0}^{x} = arctan(
x) .
In particular, we have
Combining the Chain Rule with the Fundamental Theorem of Calculus,
we can generate some nice results. Indeed, let f (x) be
continuous on [a, b] and u(x) be differentiable on [a, b].
Define the function
F(
x) =
f (
t)
dt .
Then the Chain Rule implies that F(x) is differentiable and
F '(
x) =
fu(
x)
u '(
x) .
We can generalize this a little bit more to find the derivative
of a function of the form
H(
x) =
f (
t)
dt
where u(x) and v(x) are both differentiable on [a, b]. We
have
H '(
x) =
fv(
x)
v '(
x) 
fu(
x)
u '(
x) .
Example. Find the derivative of
cos(
t^{2})
dt .
Set
F(
x) =
cos(
t^{2})
dt .
From the Fundamental Theorem of Calculus, we know that F(x) is
an antiderivative of cos(x^{2}). We have
cos(
t^{2})
dt =
F(
x^{2}) .
The Chain Rule then implies that
cos(
t^{2})
dt =
F '(
x^{2})2
x = 2
x cos
(
x^{2})
^{2} = 2
x cos(
x^{4}) .
Note that F(x) does not have an explicit form. So it is quite
amazing that even if F(x) is defined via some theoretical
result, we are still able to find the derivative of the given
function.
Example. Find the derivative of
e^{t2}dt .
Set
F(
x) =
e^{t2}dt .
From the Fundamental Theorem of Calculus, we know that F(x) is
an antiderivative of e^{x2}. We have
e^{t2}dt =
F(
x^{2}) 
F(
x) .
The Chain Rule implies
e^{t2}dt =
F '(
x^{2})2
x 
F '(
x) = 2
xe^{x4} 
e^{x2} .
Exercise 1. Let f (x) be a function defined and continuous
on [a, b]. Compare
Answer.
Exercise 2. Find a function F(x) such that
F '(x) = sin(x^{2}) and F(1) = 2.
Answer.
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