The Fundamental Theorem of Calculus
Section 4.3
Difference Equations
to
Differential Equations
The Fundamental Theorem
of Calculus
We are now ready to make the long-promised connection between differentiation and integration, between areas and tangent lines. We will look at two closely related theorems,
both of which are known as the Fundamental Theorem of Calculus. We will call the first
of these the Fundamental Theorem of Integral Calculus.
Suppose f is integrable on [a, b] and F is an antiderivative of f on (a, b) which is continuous on [a, b]. In particular, F 0 (x) = f (x) for all x in (a, b). Let P = {x0 , x1 , x2 , . . . , xn }
be a partition of [a, b] and, as usual, let ∆xi = xi − xi−1 , i = 1, 2, 3, . . . , n. Now
F (b) − F (a) = F (xn ) − F (x0 )
= F (xn ) + (F (xn−1 ) − F (xn−1 )) + (F (xn−2 ) − F (xn−2 )) + · · ·
+ (F (x1 ) − F (x1 )) − F (x0 )
= (F (xn ) − F (xn−1 )) + (F (xn−1 ) − F (xn−2 )) + · · ·
+ (F (x1 ) − F (x0 ))
n
X
=
(F (xi ) − F (xi−1 )).
(4.3.1)
i=1
By the Mean Value Theorem, for every i = 1, 2, 3, . . . , n, there exists a point ci in the
interval [xi−1 , xi ] such that
F 0 (ci ) =
F (xi ) − F (xi−1 )
.
xi − xi−1
(4.3.2)
Since F 0 (ci ) = f (ci ) and xi − xi−1 = ∆xi , it follows that
F (xi ) − F (xi−1 ) = f (ci )∆xi .
(4.3.3)
Hence, putting (4.3.3) into (4.3.1),
F (b) − F (a) =
n
X
f (ci )∆xi .
(4.3.4)
i=1
Thus F (b) − F (a) is equal to the value of a Riemann sum using the partition P , and so
must lie between the upper and lower sums for P . That is, we have shown that for any
partition P ,
L(f, P ) ≤ F (b) − F (a) ≤ U (f, P ).
(4.3.5)
1
Copyright c by Dan Sloughter 2000
2
The Fundamental Theorem of Calculus
Section 4.3
1
0.8
0.6
0.4
0.2
1
0.5
-0.5
1.5
Figure 4.3.1 Region beneath the graph of f (x) = x2 over the interval [0, 1]
But, since f is integrable, there is only one number that has this property, namely,
Rb
f (x)dx. In other words, we have shown that
a
Z b
f (x)dx = F (b) − F (a).
(4.3.6)
a
Fundamental Theorem of Integral Calculus If f is integrable on [a, b] and F is an
antiderivative of f on (a, b) which is continuous on [a, b], then
Z b
f (x)dx = F (b) − F (a).
(4.3.7)
a
This result reveals a sense in which integration is the inverse of differentiation: The
definite integral of a function f may be evaluated easily, using (4.3.7), provided we can
find a function F whose derivative is f .
It is common to write
b
F (x)
a
for F (b) − F (a). With this notation, (4.3.7) becomes
Z b
b
f (x)dx = F (x) .
Example
(4.3.8)
a
a
Since
F (x) =
1 3
x
3
is an antiderivative of f (x) = x2 , we have
Z 1
1 1
1
1
x2 dx = x3 = − 0 = .
3 0
3
3
0
Thus the area under the parabola y = x2 and above the interval [0, 1] on the x-axis is
exactly 31 . See Figure 4.3.1.
Section 4.3
The Fundamental Theorem of Calculus
3
1
0.8
0.6
0.4
0.2
0.5
1
1.5
2
2.5
3
Figure 4.3.2 Region beneath the graph of y = sin(x) over the interval [0, π]
Note that F in the previous example is but one of an infinite number of antiderivatives
of f . We can in fact use any antiderivative of f we want in applying (4.3.7), although we
typically choose the simplest one we can find.
Example
Since
G(x) =
1 3
x +x
3
is an antiderivative of g(x) = x2 + 1 (you may check by differentiating G), we have
Z 2
2
(x + 1)dx =
−1
2
1 3
8
1
x +x
=
+ 2 − − − 1 = 6,
3
3
3
−1
as we claimed in Section 4.2.
Example
If A is the area under one arch of the curve y = sin(x), then
Z π
sin(x)dx.
A=
0
Since F (x) = − cos(x) is an antiderivative of f (x) = sin(x), we have
Z π
π
sin(x)dx = − cos(x)
A=
= − cos(π) − (− cos(0)) = 1 + 1 = 2.
0
0
See Figure 4.3.2.
Example
Since
F (x) =
4 3 1 2
x − x + 2x
3
2
4
The Fundamental Theorem of Calculus
Section 4.3
is an antiderivative of f (x) = 4x2 − x + 2 (again, you may check this by differentiating F ),
we have
3
Z 3
4 3 1 2
2
(4x − x + 2)dx =
x − x + 2x
3
2
−2
−2
32
108 9
=
− +6 − − −2−4
3
2
3
325
=
.
6
Example
Since
F (t) =
is an antiderivative of f (t) =
√
t, we have
Z 4√
0
2 3
t2
3
4
t dt =
16
16
2 3
−0=
.
t2 =
3
3
3 0
As can be seen from these examples, the Fundamental Theorem of Integral Calculus
provides us with a powerful tool for evaluating definite integrals exactly. However, to utilize
the theorem we must first find an antiderivative for the function we are integrating. This
turns out to be a difficult problem in general, and we will devote the next two sections, as
well as parts of Chapter 6, to developing techniques to aid in finding antiderivatives. For
example,
1
3
3
3
F (x) = − x3 cos(2x) + x2 sin(2x) + x cos(2x) − sin(2x)
2
4
4
8
is an antiderivative of f (x) = x3 sin(2x), as may be checked by differentiation, but at this
point it is not clear how to find such an antiderivative in the first place. Moreover, there
are integrable functions, even relatively simple ones such as
f (x) =
sin(x)
,
x
which do not have antiderivatives expressible in terms of the elementary functions studied
in calculus.
The Fundamental Theorem of Integral Calculus tells us that if a function f has an
antiderivative, then we may use that antiderivative to evaluate a definite integral of f , but
it does not tell us which functions have antiderivatives. The Fundamental Theorem of Differential Calculus will tell us, in part, that every continuous function has an antiderivative.
Before beginning that discussion, we need to extend the definition of the definite integral
slightly.
Rb
The definition of a f (x)dx in Section 4.1 implicitly assumes that a < b. For the work
we are about to do, we need to extend the definition to include a ≥ b, as we did in Problem
9 of Section 4.1. First of all, if a = b, it would seem reasonable for the value of the definite
Section 4.3
The Fundamental Theorem of Calculus
5
integral to be 0 since the region between the graph of the function and the x-axis has been
reduced to a line segment. Hence we make the following definition.
Definition
For any function f defined at a point a, we define
Z a
f (x)dx = 0.
(4.3.9)
a
Note that with this definition, the statement
Z b
Z c
f (x)dx =
a
Z b
f (x)dx +
a
f (x)dx,
(4.3.10)
c
which we discussed in Section 4.1 in the case a < c < b, holds true even if a = c, b = c, or
a = b = c. Now suppose we have a < b < c. Then
Z c
Z b
f (x)dx =
Z c
f (x)dx +
a
a
b
Z b
Z c
Z c
f (x)dx,
(4.3.11)
f (x)dx.
(4.3.12)
from which it follows that
f (x)dx −
f (x)dx =
a
a
b
If we define
Z b
Z c
f (x)dx = −
c
f (x)dx,
(4.3.13)
b
then we may rewrite (4.3.12) in the form of (4.3.10). For this reason, we make the following
definition.
Definition
If b < a and f is integrable on [b, a], we define
Z b
Z a
f (x)dx = −
a
f (x)dx.
(4.3.14)
b
You may check that with these two extensions to the definition of the definite integral,
we may now state the following proposition.
Proposition
If f is integrable on a closed interval containing the points a, b, and c, then
Z b
Z c
f (x)dx =
a
Z b
f (x)dx +
a
f (x)dx.
c
(4.3.15)
6
The Fundamental Theorem of Calculus
Section 4.3
y = f(t)
a
x
Figure 4.3.3 F (x) =
Rx
a
b
f (t)dt is the area from a to x
We may now return to our discussion of antiderivatives and the Fundamental Theorem
of Differential Calculus. Suppose f is continuous on the interval [a, b]. We want to construct
an antiderivative for f on (a, b). From the Fundamental Theorem of Integral Calculus, we
know that if F is an antiderivative of f on (a, b) which is continuous on [a, b], then for any
x in (a, b) we would have
Z
x
f (t)dt = F (x) − F (a),
(4.3.16)
a
that is,
Z x
F (x) = F (a) +
f (t)dt.
(4.3.17)
a
Hence, if we are seeking an antiderivative for f , it makes sense to define
Z x
F (x) =
f (t)dt
(4.3.18)
a
and verify that F 0 (x) = f (x) for all x in (a, b). Note that F (x), geometrically, is the
cumulative area between the graph of f and the x-axis from a to x, as shown in Figure
4.3.3. We need to compute
!
Z x+h
Z x
1
F
(x
+
h)
−
F
(x)
F 0 (x) = lim
= lim
f (t)dt −
f (t)dt
(4.3.19)
h→0
h→0 h
h
a
a
for x in (a, b). Now
Z x+h
Z x
f (t)dt =
a
so
Z x+h
f (t)dt +
a
x
Z x
Z x+h
f (t)dt −
a
Z x+h
f (t)dt =
a
x
f (t)dt,
(4.3.20)
f (t)dt.
(4.3.21)
Section 4.3
The Fundamental Theorem of Calculus
7
y = f(t)
x
a
Figure 4.3.4
R x+h
a
x+h
b
R x+h
Ra
f (t)dt − x f (t)dt = x f (t)dt
See Figure 4.3.4. Thus
1
F (x) = lim
h→0 h
0
Z x+h
f (t)dt.
(4.3.22)
x
Suppose h > 0. Since f is continuous, f has a minimum value m(h) and a maximum value
M (h) on the interval [x, x + h]. Hence m(h) ≤ f (x) ≤ M (h) for all x in [x, x + h], from
which it follows that
Z x+h
Z x+h
m(h)dt ≤
Z x+h
f (t)dt ≤
x
x
M (h)dt.
(4.3.23)
x
Since m(h) and M (h) are constants, (4.3.23) implies
Z x+h
m(h)h ≤
f (t)dt ≤ M (h)h.
(4.3.24)
x
Thus
1
m(h) ≤
h
Z x+h
f (t)dt ≤ M (h).
(4.3.25)
x
Now m(h) = f (c) for some c in [x, x + h]. Moreover, as h approaches 0, x + h approaches
x, and so c must also approach x. Hence, since f is continuous,
lim m(h) = lim+ f (c) = f (x).
(4.3.26)
lim M (h) = f (x).
(4.3.27)
h→0+
h→0
Similarly,
h→0+
8
The Fundamental Theorem of Calculus
Section 4.3
It now follows from (4.3.25) that
1
lim
h→0+ h
Z x+h
f (t)dt = f (x).
(4.3.28)
f (t)dt = f (x),
(4.3.29)
x
A similar argument shows that
1
lim−
h→0 h
Z x+h
x
and so we may conclude that
1
F (x) = lim
h→0 h
0
Z x+h
f (t)dt = f (x).
(4.3.30)
x
That is,
Z x
F (x) =
f (t)dt
a
is an antiderivative of f on (a, b).
Fundamental Theorem of Differential Calculus
interval [a, b] and F is the function on (a, b) defined by
If f is continuous on the closed
Z x
F (x) =
f (t)dt,
(4.3.31)
a
then F is differentiable on (a, b) with F 0 (x) = f (x) for all x in (a, b). In other words,
d
dx
Z x
f (t)dt = f (x)
(4.3.32)
a
for all x in (a, b).
It is worth noting that (4.3.32) holds for x < a as well, as long as f is continuous on
a closed interval which contains both x and a.
Example
Let
sin(x) , if x 6= 0,
x
f (x) =
1,
if x = 0.
Then f is continuous on (−∞, ∞), so
Z x
F (x) =
f (t)dt
0
Section 4.3
The Fundamental Theorem of Calculus
9
1
1.5
0.8
1
0.6
0.5
0.4
-10
5
-5
10
-0.5
0.2
-1
-10
5
-5
10
-1.5
-0.2
sin(x)
Figure 4.3.5 Graphs of f (x) =
and F (x) =
x
Z x
0
sin(t)
dt
t
is an antiderivative of f on (−∞, ∞). In particular,
F 0 (x) =
sin(x)
x
for all x 6= 0. The graphs of F and f are shown in Figure 4.3.5. Geometrically, F (x) is the
cumulative area between the graph of f and the x-axis from 0 to x and F 0 (x) is the rate
at which area is accumulating as x increases. Since the rate at which area is accumulating
depends on the height of the curve, it is natural to expect, and the Fundamental Theorem
of Differential Calculus confirms, that F 0 (x) = f (x). The function F is known as the sine
integral function. It may be shown that it is not representable in closed form in terms of
the elementary functions of calculus.
Example
Using Leibniz notation,
Z x
d
sin(t2 )dt = sin(x2 ).
dx 0
Example
Suppose
Z 3x
G(x) =
sin(t2 )dt.
0
Then G(x) = F (h(x)), where h(x) = 3x and
Z x
F (x) =
sin(t2 )dt.
0
Hence, using the chain rule,
G0 (x) = F 0 (h(x))h0 (x) = sin((3x)2 )(3) = 3 sin(9x2 ).
Example
Suppose
Z 0
H(x) =
x
1
dt.
1 + t4
10
The Fundamental Theorem of Calculus
Then, using (4.3.14),
Z x
H(x) = −
0
so
d
H (x) = −
dx
0
Example
Z x
0
Section 4.3
1
dt,
1 + t4
1
1
dt
=
−
.
1 + t4
1 + x4
Suppose
F (x) =
Z x2 p
1 + t4 dt.
2x
Then, using (4.3.15) and (4.3.14),
F (x) =
Z 0p
1 + t4 dt +
2x
Z x2 p
Z 2x p
Z x2 p
4
4
1 + t dt = −
1 + t dt +
1 + t4 dt.
0
0
0
Note that there is nothing special about
√ using 0 in this decomposition, other than the
requirement that the function f (t) = 1 + t4 be integrable on all of the relevant intervals.
Now we have
Z 2x p
Z x2 p
d
d
0
4
F (x) = −
1 + t dt +
1 + t4 dt
dx 0
dx 0
p
p
= − 1 + (2x)4 (2) + 1 + (x2 )4 (2x)
p
p
= 2x 1 + x8 − 2 1 + 16x4 .
To summarize this section, the Fundamental Theorem of Integral Calculus provides
us with an elegant method for evaluating definite integrals, but is useful only when we
can find an antiderivative for the function being integrated. The Fundamental Theorem
of Differential Calculus tells us that every continuous function has an antiderivative and
shows how to construct one using the definite integral. Unfortunately, this brings us in
circle and does not provide us with an effective means for finding antiderivatives to use in
applying the Fundamental Theorem of Integral Calculus. For example, we know that
Z x
sin(t)
F (x) =
dt
t
1
is an antiderivative of
f (x) =
but this is of no help in evaluating, say,
Z 4
1
sin(x)
,
x
sin(x)
dx.
x
Hence in order to fully utilize the Fundamental Theorem of Integral Calculus in the evaluation of definite integrals, we must develop some procedures to aid in finding antiderivatives.
We will turn to this problem in the next section.
Section 4.3
The Fundamental Theorem of Calculus
11
Problems
1. Evaluate the following definite integrals using the Fundamental Theorem of Integral
Calculus.
Z 1
Z 3
(a)
xdx
(b)
(x2 + 2x)dx
0
0
Z 2
Z 1
x3 dx
(c)
(d)
0
−1
Z 2
Z 4
(e)
1
dx
2
1 x
Z 8
√
(h)
t + 1 dt
(2x3 + 3x2 + x − 4)dx
(f)
0
Z 4
(g)
1
1
√ dt
t
0
Z π2
(i)
x3 dx
Z π
sin(x)dx
(j)
cos(z)dz
−π
0
2. Evaluate the following definite integrals using the Fundamental Theorem of Integral
Calculus.
Z 2
Z 2
2
(a)
(x + 1) dx
(b)
(2x + 1)2 dx
0
0
Z 4
Z π4
(c)
√
1 + 2t dt
(d)
0
0
Z π
Z π
(e)
sin(2x)dx
(f)
5 cos(3x)dx
0
0
Z π3
Z π
(g)
4 sin(3x)dx
(h)
(i)
Z 2
2
2x sin(x )dx
(j)
(k)
2x(1 + x2 )5 dx
−1
0
Z √π
8 cos(5θ)dθ
−π
0
Z √π
sec2 (x)dx
Z 2
x sin(x2 )dx
(l)
x(1 + x2 )5 dx
−1
0
3. For each of the following functions, graph both f and
Z x
F (x) =
f (t)dt
0
together over the given interval.
(a) f (x) = sin(x) on [−2π, 2π]
1
(c) f (x) =
on [−3, 3]
1 + x4
(b) f (x) = sin(x2 ) on [0, 10]
1
(d) f (x) =
on [0, 10]
x+1
12
The Fundamental Theorem of Calculus
Section 4.3
4. Find the derivatives of each of the following functions.
Z x
Z x
2
(a) F (x) =
sin (4t)dt
(b) g(x) =
3
dt
2 t+2
Z 0p
(d) G(t) =
4 − z 2 dz
0
Z π
(c) F (x) =
cos3 (t)dt
x
Z x2
(e) f (x) =
0
t
1
ds
1 + s2
(f) h(z) =
Z 3z p
1 + t2 dt
z
5. Evaluate the following derivatives.
Z x
1
d
(a)
dt
dx 1 1 + t2
Z
d 5 2
(c)
sin (3x)dx
dt t2
d
(b)
dx
Z 3x
d
dx
Z x2
(d)
1
3x
sin(3t)
dt
t
√
1
dt
1 + t2
6. Find the area of the region beneath one arch of the curve y = 3 sin(2x).
7. Let R be the region bounded by the curves y = x2 and y = (x − 2)2 and the x-axis.
Find the area of R.
8. Explain why the integral
Z 1
(x − x2 )dx
0
is the area of the region bounded by the curves y = x2 and y = x. Find this area.
9. Explain why the integral
Z 1
(2 − 2x2 )dx
−1
is the area of the region bounded by the curves y = 1 − x2 and y = x2 − 1. Find this
area.