a
a
a
Now suppose c is another real number with a < c < b. If the closed interval
[a, c] is divisible into M intervals of length dx, where M is a positive infinite
integer less than N , then
N
M
N
f (x∗)dx =
f (x∗)dx +
f (x∗)dx
(2.3.9)
i
i
i
i=1
i=1
i=M +1
implies that
b
c
b
f (x)dx =
f (x)dx +
f (x)dx.
(2.3.10)
a
a
c
This is a reflection of our intuition that, for an object moving along a straight
line, the change in position from time t = a to time t = b is equal to the change
in position from time t = a to time t = c plus the change in position from time
t = c to time t = b. Although we assumed that [a, c] was divisible into an integer
number of subintervals of length dx, the result holds in general.
The final properties which we will consider revolve around a basic inequality.
If f and g are both continuous on [a, b] with f (x) ≤ g(x) for all x in [a, b], then
N
N
f (x∗)dx ≤
g(x∗)dx,
(2.3.11)
i
i
i=1
i=1
from which it follows that
b
b
f (x)dx ≤
g(x)dx.
(2.3.12)
a
a
74
CHAPTER 2. INTEGRALS
For example, if m and M are constants with m ≤ f (x) ≤ M for all x in [a, b],
then
b
b
b
mdx ≤
f (x)dx ≤
M dx,
(2.3.13)
a
a
a
and so
b
m(b − a) ≤
f (x)dx ≤ M (b − a).
(2.3.14)
a
Note, in particular, that if f (x) ≥ 0 for all x in [a, b], then
b
f (x)dx ≥ 0.
(2.3.15)
a
Example 2.3.1. From the observation that
1
f (x) = 1 + x2
is increasing on (−∞, 0] and decreasing on [0, ∞), it is easy to see that
1
1
≤
≤ 1
2
1 + x2
for all x in [−1, 1]. Hence
1
1
1 ≤
dx ≤ 2.
−1 1 + x2
We will eventually see, in Example 2.6.20, that
1
1
π
dx =
≈ 1.5708.
−1 1 + x2
2
Since for any real number a, −|a| ≤ a ≤ |a| (indeed, either a = |a| or
a = −|a|), we have
− |f (x)| ≤ f (x) ≤ |f (x)|
(2.3.16)
for all x in [a, b]. Hence
b
b
b
−
|f (x)|dx ≤
f (x)dx ≤
|f (x)|dx,
(2.3.17)
a
a
a
or, equivalently,
b
b
f (x)dx ≤
|f (x)|dx.
(2.3.18)
a
a
Notice that, since the definite integral is just a generalized version of summation,
this result is a generalization of the triangle inequality: Given any real numbers
a and b,
|a + b| ≤ |a| + |b|.
(2.3.19)
The next theorem summarizes the properties of definite integrals that we
have discussed above.
2.4. THE FUNDAMENTAL THEOREM OF INTEGRALS
75
Theorem 2.3.1. Suppose f and g are continuous functions on [a, b], c is any
real number with a < c < b, and k is a fixed real number. Then
b
(a)
kdx = k(b − a),
a
b
b
(b)
kf (x)dx = k
f (x)dx,
a
a
b
b
b
(c)
(f (x) + g(x))dx =
f (x)dx +
g(x)dx,
a
a
a
b
c
b
(d)
f (x)dx =
f (x)dx +
f (x)dx,
a
a
c
b
b
(e) if f (x) ≤ g(x) for all x in [a, b], then
f (x)dx ≤
g(x)dx,
a
a
b
(f) if m ≤ f (x) ≤ M for all x in [a, b], then m(b − a) ≤
f (x)dx ≤ M (b − a),
a
b
b
(g)
f (x)dx ≤
|f (x)|dx.
a
a
Exercise 2.3.1.
Show that
1
2 1
≤
dx ≤ 1.
2
1
x
2.4
The fundamental theorem of integrals
The main theorem of this section is key to understanding the importance of
definite integrals. In particular, we will invoke it in developing new applications
for definite integrals. Moreover, we will use it to verify the fundamental theorem
of calculus.
We first need some new notation and terminology. Suppose
is a nonzero
infinitesimal. Intuitively,
is infinitely smaller than any nonzero real number.
One way to express this is to note that for any nonzero real number r,
0,
(2.4.1)
r
that is, the ratio of
to r is an infinitesimal. Now we also have
2
=
0,
(2.4.2)
that is, the ratio of 2 to
is an infinitesimal. Intuitively, this means that 2
is infinitely smaller than
itself. This is related to a fact about real numbers:
76
CHAPTER 2. INTEGRALS
For any real number r with 0 < r < 1, r2 is smaller than r. For example, if
r = 0.01, then r2 = 0.0001.
Definition 2.4.1. Given a nonzero hyperreal number , we say another hyper-
real number δ is of an order less than
if δ is an infinitesimal, in which case we
write δ ∼ o( ).
In other words, we have
δ
δ ∼ o( ) if and only if
0.
(2.4.3)
Example 2.4.1. If α is any infinitesimal, than α ∼ o(1) since α = α is an
1
infinitesimal.
Example 2.4.2. If
is any nonzero infinitesimal, then 2 ∼ o( ) since
2
=
0.
Now suppose N is a positive infinite integer,
= 1 , and δ
N
i ∼ o( ) for
i = 1, 2, . . . , N . Then, for any positive real number r,
|δi| < r,
(2.4.4)
and so
N
|δi| < rN.
(2.4.5)
i=1
Multiplying both sides by , we have
N
1
|δi| < rN = rN
= r.
(2.4.6)
N
i=1
Since this holds for all positive real numbers r, it follows that
N
|δ
i=1
i| is an
infinitesimal. Now
N
N
δi ≤
|δi|,
(2.4.7)
i=1
i=1
and so we may conclude that
n
δ
i=1
i is an infinitesimal. In words, the sum of
N infinitesimals of order less than 1 is still an infinitesimal.
N
Now suppose B is a function that for any real numbers a < b in an open
interval I assigns a value B(a, b). Moreover, suppose B has the following two
properties:
• for any a < c < b in I, B(a, b) = B(a, c) + B(c, b), and
2.4. THE FUNDAMENTAL THEOREM OF INTEGRALS
77
• for some continuous function h and any nonzero infinitesimal dx,
B(x, x + dx) − h(x)dx ∼ o(dx)
(2.4.8)
for any x in I.
For a positive infinite integer N , let dx = b−a and let
N
a = x0 < x1 < x2 < · · · < xN = b
(2.4.9)
be a partition of [a, b] into N equal intervals of length dx. Then
B(a, b) = B(x0, x1) + B(x1, x2) + B(x2, x3) + · · · + B(xN−1, xN )
N
=
B(xi−1, xi−1 + dx)
i=1
N
=
((B(xi−1, xi−1 + dx) − h(xi−1)dx) + h(xi−1)dx)
i=1
N
N
=
(B(xi−1, xi−1 + dx) − h(xi−1)dx) +
h(xi−1)dx
i=1
i=1
N
b
(B(xi−1, xi−1 + dx) − h(xi−1)dx) +
h(x)dx.
(2.4.10)
i=1
a
Since the final sum on the right is the sum of N infinitesimals of order less than
1 , it follows that
N
b
B(a, b) =
h(x)dx.
(2.4.11)
a
This result is basic to understanding both the computation of definite inte-
grals and their applications. We call it the fundamental theorem of integrals.
Theorem 2.4.1. Suppose B is a function that for any real numbers a < b in
an open interval I assigns a value B(a, b) and satisfies
• for any a < c < b in I, B(a, b) = B(a, c) + B(c, b), and
• for some continuous function h and any nonzero infinitesimal dx,
B(x, x + dx) − h(x)dx ∼ o(dx)
(2.4.12)
for any x in I.
Then
b
B(a, b) =
h(x)dx
(2.4.13)
a
for any real numbers a and b in I.
78
CHAPTER 2. INTEGRALS
We will look at several applications of definite integrals in the next section.
For now, we note how this theorem provides a method for evaluating integrals.
Namely, given a function f which is differentiable on an open interval I, define,
for every a < b in I,
B(a, b) = f (b) − f (a).
(2.4.14)
Then, for any a, b, and c in I with a < c < b,
B(a, b) = f (b) − f (a)
= (f (b) − f (c)) + (f (c) − f (a))
= B(a, c) + B(c, b).
(2.4.15)
Moreover, for any infinitesimal dx and any x in I,
B(x, x + dx)
f (x + dx) − f (x)
=
f (x),
(2.4.16)
dx
dx
from which it follows that
B(x, x + dx) − f (x)dx
(2.4.17)
dx
is an infinitesimal. Hence
B(x, dx) − f (x)dx ∼ o(dx),
(2.4.18)
and so it follows from Theorem 2.4.1 that
b
f (b) − f (a) = B(a, b) =
f (x)dx.
(2.4.19)
a
This is the fundamental theorem of calculus.
Theorem 2.4.2. If f is differentiable on an open interval I, then for every
a < b in I,
b
f (x)dx = f (b) − f (a).
(2.4.20)
a
Example 2.4.3. To evaluate
1
xdx,
0
we first note that g(x) = x is the derivative of f (x) = 1 x2. Hence, by Theorem
2
1
1
1
xdx = f (1) − f (0) =
− 0 =
0
2
2
We will write
f (x)|b = f (b) − f (a)
(2.4.21)
a
to simplify the notation for evaluating an integral using Theorem 2.4.2. With this notation, the previous example becomes
1
1
1
1
1
xdx =
x2
=
− 0 =
.
0
2
2
2
0
2.5. APPLICATIONS OF DEFINITE INTEGRALS
79
Example 2.4.4. Since
1
x2dx =
x3 + c,
3
we have
2
2
1
8
1
7
x2dx =
x3
=
−
=
.
1
3
3
3
3
1
Example 2.4.5. Since
−20 sin(5x)dx = 4 cos(5x) + c,
we have
2π
−20 sin(5t)dt = 4 cos(5t)|2π = 4 − 4 = 0.
0
0
Note that if we consider an object moving along a straight line with velocity
v(t) = −20 sin(5t), then this definite integral computes the change in position
of the object from time t = 0 to time t = 2π. In this case, the object, although
always in motion, is in the same position at time t = 2π as it was at time t = 2π.
1
Exercise 2.4.1.
Evaluate
x4dx.
0
π
Exercise 2.4.2.
Evaluate
sin(x)dx.
0
Exercise 2.4.3.
Suppose the velocity of an object moving along a straight line
is v(t) = 10 sin(t) centimeters per second. Find the change in position of the
object from time t = 0 to time t = π.
2.5
Applications of definite integrals
In this section we will look at several examples of applications for definite inte-
grals.
2.5.1
Area between curves
Consider two continuous functions f and g on an open interval I with f (x) ≤
g(x) for all x in I. For any a < b in I, let R(a, b) be the region in the plane
consisting of the points (x, y) for which a ≤ x ≤ b and f (x) ≤ y ≤ g(x). That
is, R(a, b) is bounded above by the curve y = g(x), below by the curve y = f (x),
on the left by the vertical line x = a, and on the right by the vertical line x = b,
as in Figure 2.5.1. Let
A(a, b) = area of R(a, b).
(2.5.1)
80
CHAPTER 2. INTEGRALS
y
y = g(x)
x = a
R(a, b)
x = b
y = f (x)
a
x
b
Figure 2.5.1: Region R(a, b) between the graphs of y = g(x) and y = f (x)
Clearly, for any a ≤ c ≤ b,
A(a, b) = A(a, c) + A(c, b).
(2.5.2)
Now for an x in I and a positive infinitesimal dx, let c be the point at which
g(u) − f (u) attains its minimum value for x ≤ u ≤ x + dx and let d be the point
at which g(u) − f (u) attains its maximum value for x ≤ u ≤ x + dx. Then
(g(c) − f (c))dx ≤ A(x, x + dx) ≤ (g(d) − f (d))dx.
(2.5.3)
Moreover, since
g(c) − f (c) ≤ g(x) − f (x) ≤ g(d) − f (d),
(2.5.4)
we also have
(g(c) − f (c))dx ≤ (g(x) − f (x))dx ≤ (g(d) − f (d))dx.
(2.5.5)
Putting (2.5.3) and (2.5.5) together, we have
|A(x, dx) − (g(x) − f (x))dx| ≤ ((g(d) − f (d)) − (f (c) − g(c)))dx
(2.5.6)
or
|A(x, dx) − (g(x) − f (x))dx| ≤ (g(d) − f(d)) − (f(c) − g(c))
(2.5.7)
dx
Now since c
x and d
x,
(g(d) − f (d)) − (g(c) − f (c)) = (g(d) − g(c)) + (f (c) − f (d))
0.
(2.5.8)
2.5. APPLICATIONS OF DEFINITE INTEGRALS
81
5
y
4
y = x + 2
3
2
R
1
y = x2
0
1
0
1
2
−
x
Figure 2.5.2: The region bounded by the curves y = x + 2 and y = x2
Hence
A(x, dx) − (g(x) − f (x))dx ∼ o(dx).
(2.5.9)
It now follows from Theorem 2.4.1 that
b
A(a, b) =
(g(x) − f (x))dx.
(2.5.10)
a
Example 2.5.1. Let A be the area of the region R bounded by the curves
with equations y = x2 and y = x + 2. Note that these curves intersect when
x2 = x + 2, that is when
0 = x2 − x − 2 = (x + 1)(x − 2).
Hence they intersect at the points (−1, 1) and (2, 4), and so R is the region in
the plane bounded above by the curve y = x + 2, below by the curve y = x2, on
the right by x = −1, and on the left by x = 2. See Figure 2.5.2. Thus we have 2
A =
(x + 2 − x2)dx
−1
2
1
1
=
x2 + 2x −
x3
2
3
−1
8
1
1
=
2 + 4 −
−
− 2 +
3
2
3
9
=
.
2
82
CHAPTER 2. INTEGRALS
y
y = f (x)
x = a
R
x = b
a
b
x
Figure 2.5.3: Area beneath the graph of a function f
Exercise 2.5.1.
Find the area of the region bounded by the curves y = x and
y = x2.
Exercise 2.5.2.
Find the area of the region bounded by the curves y = x2
and y = 2 − x2.
√
Exercise 2.5.3.
Find the area of the region bounded by the curves y =
x
and y = x.
Now consider a continuous function f on an interval [a, b] with f (x) ≥ 0 for
all x in [a, b]. If A is the area of the region R bounded above by the graph of
y = f (x), below by the graph of y = 0 (that is, the x-axis), on the right by the
vertical line x = a, and on the left by the graph of x = b (see Figure 2.5.3), then b
b
A =
(f (x) − 0)dx =
f (x)dx.
(2.5.11)
a
a
This gives us a geometric interpretation for a the definite integral of a nonneg-
ative function f over an interval [a, b] as the area beneath the graph of f and
above the x-axis.
Example 2.5.2. If A is the area of the region R beneath the graph of y = sin(x)
over the interval [0, π], as in Figure 2.5.4, then
π
A =
sin(x)dx = − cos(x)|π = 1 + 1 = 2.
0
0
On the other hand, if f is continuous on [a, b] with f (x) ≤ 0 for all x in
[a, b], and A is the area of the region R above the graph of y = f (x) and below
2.5. APPLICATIONS OF DEFINITE INTEGRALS
83
y 1.5
1
y = sin(x)
0.5
R
0
0
1
2
3 x
Figure 2.5.4: Area beneath the graph of y = sin(x)
the x axis, then
b
b
A =
(0 − f (x))dx = −
f (x)dx.
(2.5.12)
a
a
That is, the definite integral of a non-positive function f over an interval [a, b]
is the negative of the area above the graph of f and beneath the x-axis.
In general, given a continuous function f on an interval let R be the region
bounded by the x-axis and the graph of y = f (x). If A+ is the area of the part
of R which lies above the x-axis and A− is the area of the part of R which lies
below the x-axis, then
b
f (x)dx = A+ − A−.
(2.5.13)
a
Example 2.5.3. Note that
2π
sin(x)dx = − cos(x)|2π = −1 + 1 = 0.
0
0
Geometrically, we can see this result in Figure 2.5.5. If R+ is the the region beneath the graph of y = sin(x) over the interval [0, π] and R− is the region
above the graph of y = sin(x) over the interval [0, 2π], then these two regions
have the same area. Hence the integral, which is the area of R+ minus the area
of R−, is 0.
Exercise 2.5.4.
Evaluate
1
x3dx
−1
and explain the result geometrically.
84
CHAPTER 2. INTEGRALS
y
1.5
y = sin(x)
0.5
R+
0
1
2
3
4
5
6 x
−0.5
R−
−1.5
Figure 2.5.5: Area of R+ is the same as the area of R−
Exercise 2.5.5.
Evaluate
2
xdx
−1
and explain the result geometrically.
Exercise 2.5.6.
Explain, geometrically, why
1
π
1 − x2 dx =
.
−1
2
2.5.2
Volumes
Consider a three-dimensional body B. Given a line, which we will call the z-axis,
let V (a, b) be the volume of B which lies between planes which are perpendicular
to the z-axis and pass through z = a and z = b. Clearly, for any a < c < b,
V (a, b) = V (a, c) + V (c, b).
(2.5.14)
Now suppose that, for any a ≤ x ≤ b, R(z) is a cross section of B perpendicular
to the z-axis. See Figure 2.5.6. Let A(z) be the area of R(z). We assume A is a continuous function of z. For a positive infinitesimal dz, let A have, on the
interval [z, z + dz], a minimum value at c and a maximum value at d. Then
A(c)dz ≤ V (z, z + dz) ≤ A(d)dz.
(2.5.15)
Since we also have A(c) ≤ A(z) ≤ A(d), it follows that
|V (z, z + dz) − A(z)dz| ≤ (A(d) − A(c))dz.
(2.5.16)
Thus
|V (z, z + dz) − A(z)dz| ≤ A(d) − A(c).
(2.5.17)
dz
2.5. APPLICATIONS OF DEFINITE INTEGRALS
85
b
z