5.3The Fundamental
Theorem of Calculus
INTEGRALS
In this section, we will learn about:
The Fundamental Theorem of Calculus
and its significance.
The Fundamental Theorem of Calculus (FTC) is
appropriately named.
It establishes a connection between the two branches of calculus—differential calculus and integral calculus.
FUNDAMENTAL THEOREM OF CALCULUS
FTC
Differential calculus arose from the tangent
problem.
Integral calculus arose from a seemingly unrelated
problem—the area problem.
Newton’s mentor at Cambridge, Isaac Barrow
(1630–1677), discovered that these two problems
are actually closely related.
In fact, he realized that differentiation and integration are inverse processes.
FTC
The FTC gives the precise inverse relationship
between the derivative and the integral.
FTC
It was Newton and Leibniz who exploited this
relationship and used it to develop calculus into a
systematic mathematical method.
In particular, they saw that the FTC enabled them to compute areas and integrals very easily without having to compute them as limits of sums—as we did
in Sections 5.1 and 5.2
FTC
The first part of the FTC deals with functions
defined by an equation of the form
where f is a continuous function on [a, b] and x
varies between a and b.
( ) ( )x
ag x f t dt
Equation 1FTC
Observe that g depends only on x, which appears as the variable upper limit in the integral.
If x is a fixed number, then the integral is a definite number.
If we then let x vary, the number also varies and defines a function of x denoted by g(x).
FTC
( ) ( )x
ag x f t dt
If f happens to be a positive function, then g(x) can
be interpreted as the area under the graph of f from
a to x, where x can vary from a to b.
Think of g as the ‘area so far’ function, as seen here.
FTC
If f is the function
whose graph is shown
and ,
find the values of:
g(0), g(1), g(2), g(3),
g(4), and g(5).
Then, sketch a rough graph of g.
Example 1
0( ) ( )
xg x f t dt
FTC
First, we notice that:
0
0(0) ( ) 0g f t dt
FTC Example 1
From the figure, we see that g(1)
is the area of a triangle:
1
0
12
(1) ( )
(1 2)
1
g f t dt
Example 1FTC
To find g(2), we add to g(1)
the area of a rectangle:
2
0
1 2
0 1
(2) ( )
( ) ( )
1 (1 2)
3
g f t dt
f t dt f t dt
Example 1FTC
We estimate that the area under f from 2 to 3 is
about 1.3.
So, 3
2(3) (2) ( )
3 1.3
4.3
g g f t dt
Example 1FTC
For t > 3, f(t) is negative.
So, we start subtracting areas, as follows.
Example 1FTC
Thus, 4
3(4) (3) ( ) 4.3 ( 1.3) 3.0g g f t dt
FTC Example 1
5
4(5) (4) ( ) 3 ( 1.3) 1.7g g f t dt
We use these values to sketch the graph of g. Notice that, because f(t)
is positive for t < 3, we keep adding area for t < 3.
So, g is increasing up to x = 3, where it attains a maximum value.
For x > 3, g decreases because f(t) is negative.
Example 1FTC
If we take f(t) = t and a = 0, then, using Exercise
27 in Section 5.2, we have:
2
0( )
2
x xg x tdt
FTC
Notice that g’(x) = x, that is, g’ = f.
In other words, if g is defined as the integral of f by Equation 1, g turns out to be an antiderivative of f—at least in this case.
FTC
If we sketch the derivative of
the function g, as in the first
figure, by estimating slopes
of tangents, we get a graph
like that of f in the second
figure.
So, we suspect that g’ = f in Example 1 too.
FTC
To see why this might be generally true, we
consider a continuous function f with f (x) ≥ 0.
Then, can be interpreted as the
area under the graph of f from a to x.
( ) ( )x
ag x f t dt
FTC
To compute g’(x) from the definition of derivative,
we first observe that, for h > 0, g(x + h) – g(x) is
obtained by subtracting areas.
It is the area under the graph of f from x to x + h (the gold area).
FTC
For small h, we see that this area is approximately
equal to the area of the rectangle with height f(x)
and width h:
So,
FTC
( ) ( ) ( )g x h g x hf x
( ) ( )
( )
g x h g x
hf x
Intuitively, we therefore expect that:
The fact that this is true, even when f is not necessarily positive, is the first part of the FTC (FTC1).
0
( ) ( )'( ) lim ( )
h
g x h g xg x f x
h
FTC
Fundamental Theorem of Calculus Version 1
If f is continuous on [a, b], then the function g
defined by
is continuous on [a, b] and differentiable on
(a, b), and g’(x) = f(x).
( ) ( )x
ag x f t dt a x b
In words, the FTC1 says that the derivative of a
definite integral with respect to its upper limit is
the integrand evaluated at the upper limit.
FTC1
Using Leibniz notation for derivatives, we can
write the FTC1 as
when f is continuous. Roughly speaking, Equation 5 says that, if we first integrate f
and then differentiate the result, we get back to the original function f.
( ) ( )x
a
df t dt f x
dx
FTC1
Find the derivative of the function
As is continuous, the FTC1 gives:
Example 2
2
0( ) 1
xg x t dt
2( ) 1f t t
2'( ) 1g x x
FTC1
A formula of the form
may seem like a strange way of defining a
function.
However, books on physics, chemistry, and statistics are full of such functions.
( ) ( )x
ag x f t dt
FTC1
FRESNEL FUNCTION
For instance, consider the Fresnel function
It is named after the French physicist Augustin Fresnel (1788–1827), famous for his works in optics.
It first appeared in Fresnel’s theory of the diffraction of light waves.
More recently, it has been applied to the design of highways.
2
0( ) sin( / 2)
xS x t dt
Example 3
FRESNEL FUNCTION
The FTC1 tells us how to differentiate the Fresnel
function:
This means that we can apply all the methods of differential calculus to analyze S.
Example 3
2( ) sin( / 2)S x t
The figure shows the graphs of f (x) = sin(πx2/2)
and the Fresnel function
A computer was used to graph S by computing the value of this integral for many values of x.
0( ) ( )
xS x f t dt
Example 3FRESNEL FUNCTION
It does indeed look as if S(x) is the area under the
graph of f from 0 to x (until x ≈ 1.4, when S(x)
becomes a difference of areas).
Example 3FRESNEL FUNCTION
The other figure shows a larger part of the graph
of S.
Example 3FRESNEL FUNCTION
If we now start with the graph of S here and think
about what its derivative should look like, it seems
reasonable that S’(x) = f(x).
For instance, S is increasing when f(x) > 0 and decreasing when f(x) < 0.
Example 3FRESNEL FUNCTION
FRESNEL FUNCTION
So, this gives a visual confirmation of the FTC1.
Example 3
Find
Here, we have to be careful to use the Chain Rule in conjunction with the FTC1.
4
1sec
xdt dt
dx
Example 4FTC1
Let u = x4. Then,
4
1 1
1
4 3
sec sec
(Chain Rule)
sec (FTC1)
sec( ) 4
x u
u
d dt dt t dt
dx dxd du
sec t dtdu dx
duu
dx
x x
Example 4FTC1
In Section 5.2, we computed integrals from the
definition as a limit of Riemann sums and saw that
this procedure is sometimes long and difficult.
The second part of the FTC (FTC2), which follows easily from the first part, provides us with a much simpler method for the evaluation of integrals.
FTC1
If f is continuous on [a, b], then
where F is any antiderivative of f, that is, a
function such that F ’ = f.
( ) ( ) ( )b
af x dx F b F a
Fundamental Theorem of Calculus Version 2
FTC2
Let
We know from the FTC1 that g’(x) = f(x),
that is, g is an antiderivative of f.
( ) ( )x
ag x f t dt
Proof
FTC2
If F is any other antiderivative of f on [a, b], then
we know from Corollary 7 in Section 4.2 that F
and g differ by a constant
F(x) = g(x) + C
for a < x < b.
Proof—Equation 6
FTC2
However, both F and g are continuous on [a, b].
Thus, by taking limits of both sides of Equation 6
(as x → a+ and x → b- ), we see it also holds when
x = a and x = b.
Proof
FTC2
If we put x = a in the formula for g(x), we get:
Proof
( ) ( ) 0a
ag a f t dt
FTC2
So, using Equation 6 with x = b and x = a, we
have:
( ) ( ) [ ( ) ] [ ( ) ]
( ) ( )
( )
( )b
a
F b F a g b C g a C
g b g a
g b
f t dt
Proof
FTC2
The FTC2 states that, if we know an antiderivative
F of f, then we can evaluate
simply by subtracting the values of F at the
endpoints of the interval [a, b].
( )b
af x dx
FTC2
It is very surprising that , which
was defined by a complicated procedure
involving all the values of f(x) for a ≤ x ≤ b,
can be found by knowing the values of F(x) at
only two points, a and b.
( )b
af x dx
FTC2
At first glance, the theorem may be surprising.
However, it becomes plausible if we interpret it in physical terms.
FTC2
If v(t) is the velocity of an object and s(t) is its
position at time t, then v(t) = s’(t).
So, s is an antiderivative of v.
FTC2
In Section 5.1, we considered an object that
always moves in the positive direction.
Then, we guessed that the area under the velocity
curve equals the distance traveled.
In symbols,
That is exactly what the FTC2 says in this context.
( ) ( ) ( )b
av t dt s b s a
FTC2
Evaluate the integral
The function f(x) = ex is continuous everywhere and we know that an antiderivative is F(x) = ex.
So, the FTC2 gives:
Example 5
3
1
xe dx
3
1
3
(3) (1)xe dx F F
e e
FTC2
Notice that the FTC2 says that we can use any antiderivative F of f.
So, we may as well use the simplest one, namely F(x) = ex, instead of ex + 7 or ex + C.
Example 5
FTC2
We often use the notation
So, the equation of the FTC2 can be written as:
( ) | ( ) ( )baF x F b F a
( ) ( ) | where 'b
baa
f x dx F x F f
FTC2
Find the area under the parabola y = x2 from 0 to 1.
An antiderivative of f(x) = x2 is F(x) = (1/3)x3. The required area is found using the FTC2:
Example 6
13 3 312
00
1 0 1
3 3 3 3
xA x dx
FTC2
If you compare the calculation in Example 6 with
the one in Example 2 in Section 5.1, you will see
the FTC gives a much shorter method.
FTC2
Evaluate
An antiderivative of f(x) = 1/x is F(x) = ln |x|.
As 3 ≤ x ≤ 6, we can write F(x) = ln x.
Example 7
6
3
1dx
x
FTC2
Therefore, 6
633
1ln ]
ln 6 ln 3
6ln
3ln 2
dx xx
Example 7
FTC2
Find the area under the cosine curve from 0 to b,
where 0 ≤ b ≤ π/2.
As an antiderivative of f(x) = cos x is F(x) = sin x, we have:
Example 8
00cos sin ] sin sin 0 sin
bbA x dx x b b
FTC2
In particular, taking b = π/2, we have proved that the area under the cosine curve from 0 to π/2 is:
sin(π/2) = 1
Example 8
FTC2
When the French mathematician Gilles de
Roberval first found the area under the sine and
cosine curves in 1635, this was a very challenging
problem that required a great deal of ingenuity.
FTC2
If we didn’t have the benefit of the FTC, we would
have to compute a difficult limit of sums using
either:
Obscure trigonometric identities
A computer algebra system (CAS), as in Section 5.1
FTC2
It was even more difficult for Roberval.
The apparatus of limits had not been invented in 1635.
FTC2
However, in the 1660s and 1670s, when the FTC
was discovered by Barrow and exploited by
Newton and Leibniz, such problems became very
easy.
You can see this from Example 8.
FTC2
What is wrong with this calculation?
313
211
1 1 41
1 3 3
xdx
x
Example 9
FTC2
To start, we notice that the calculation must
be wrong because the answer is negative but
f (x) = 1/x2 ≥ 0 and Property 6 of integrals says that
when f ≥ 0.( ) 0b
af x dx
Example 9
FTC2
The FTC applies to continuous functions.
It cannot be applied here because f(x) = 1/x2
is not continuous on [-1, 3].
In fact, f has an infinite discontinuity at x = 0.
So, does not exist.3
21
1dx
x
Example 9
INVERSE PROCESSES
We end this section by bringing together the two
parts of the FTC.
FTC
Suppose f is continuous on [a, b].
1.If , then g’(x) = f(x).
2. , where F is
any antiderivative of f, that is, F’ = f.
( ) ( )x
ag x f t dt
( ) ( ) ( )b
af x dx F b F a
INVERSE PROCESSES
We noted that the FTC1 can be rewritten as:
This says that, if f is integrated and then the result is differentiated, we arrive back at the original function f.
( ) ( )x
a
df t dt f x
dx
INVERSE PROCESSES
As F’(x) = f(x), the FTC2 can be rewritten as:
This version says that, if we take a function F, first differentiate it, and then integrate the result, we arrive back at the original function F.
However, it is in the form F(b) - F(a).
'( ) ( ) ( )b
aF x dx F b F a
INVERSE PROCESSES
Taken together, the two parts of the FTC say that
differentiation and integration are inverse
processes.
Each undoes what the other does.
SUMMARY
The FTC is unquestionably the most important
theorem in calculus.
Indeed, it ranks as one of the great accomplishments of the human mind.
SUMMARY
Before it was discovered—from the time of
Eudoxus and Archimedes to that of Galileo and
Fermat—problems of finding areas, volumes, and
lengths of curves were so difficult that only a
genius could meet the challenge.
SUMMARY
Now, armed with the systematic method that
Newton and Leibniz fashioned out of the
theorem, we will see in the chapters to come that
these challenging problems are accessible to all
of us.