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In this section, we will learn about: Integrals with limits that represent a definite quantity.
2.
INTEGRALS Equation 1
We saw in Section 5.1 that a limit of the form
n
lim f ( xi *)x
n
i 1
lim[ f ( x1*)x f ( x2 *)x ... f ( xn *) x]
n
arises when we compute an area.
We also saw that it arises when we try to find
the distance traveled by an object.
3.
It turns out that this same type of limit
occurs in a wide variety of situations
even when f is not necessarily a positive
4.
In Chapters 6 and 8, we will see that
limits of the form Equation 1 also arise
in finding:
Lengths of curves
Volumes of solids
Centers of mass
Force due to water pressure
Work
5.
Therefore, we give this
type of limit a special name
and notation.
6.
5.2
The Definite Integral
In this section, we will learn about:
Integrals with limits that represent
a definite quantity.
7.
DEFINITE INTEGRAL Definition 2
If f is a function defined for a ≤ x ≤ b,
we divide the interval [a, b] into n subintervals
of equal width ∆x = (b – a)/n.
We let x0(= a), x1, x2, …, xn(= b) be the endpoints
of these subintervals.
We let x1*, x2*,…., xn* be any sample points in
these subintervals, so xi* lies in the i th subinterval.
8.
DEFINITE INTEGRAL Definition 2
Then, the definite integral of f from a to b is
b n
f ( x) dx lim f ( x *)x
a n
i 1
i
provided that this limit exists.
If it does exist, we say f is integrable on [a, b].
9.
DEFINITE INTEGRAL
The precise meaning of the limit that
defines the integral is as follows:
For every number ε > 0 there is an integer N
such that n
b
f ( x)dx f ( x *)x
a
i 1
i
for every integer n > N and for every choice
of xi* in [xi-1, xi].
10.
INTEGRAL SIGN Note 1
The symbol ∫ was introduced by Leibniz
and is called an integral sign.
It is an elongated S.
It was chosen because an integral is
a limit of sums.
11.
b
NOTATION f ( x) dx
a
b
Note 1
In the notation f ( x) dx ,
a
f(x) is called the integrand.
a and b are called the limits of integration;
a is the lower limit and b is the upper limit.
For
b
now, the symbol dx has no meaning by itself;
a f ( x) dx is all one symbol. The dx simply indicates
that the independent variable is x.
12.
INTEGRATION Note 1
The procedure of
calculating an integral
is called integration.
13.
b
DEFINITE INTEGRAL f ( x) dx Note 2
a
b
The definite integral f ( x)dx is a number.
a
It does not depend on x.
In fact, we could use any letter in place of x
without changing the value of the integral:
b b b
f ( x)dx f (t )dt f (r )dr
a a a
14.
RIEMANN SUM Note 3
The sum n
f ( x *)x
i 1
i
that occurs in Definition 2 is called
a Riemann sum.
It is named after the German mathematician
Bernhard Riemann (1826–1866).
15.
RIEMANN SUM Note 3
So, Definition 2 says that the definite integral
of an integrable function can be approximated
to within any desired degree of accuracy by
a Riemann sum.
16.
RIEMANN SUM Note 3
We know that, if f happens to be positive,
the Riemann sum can be interpreted as:
A sum of areas of approximating rectangles
17.
RIEMANN SUM Note 3
Comparing Definition 2 with the definition
of area in Section 5.1, we see that the definite
b
integral f ( x ) dx can be interpreted as:
a
The area under the curve y = f(x) from a to b
18.
RIEMANN SUM Note 3
If f takes on both positive and negative values, then the Riemann
sum is:
The sum of the areas of the rectangles that lie
above the x-axis and the negatives of the areas
of the rectangles that lie below the x-axis
That is, the areas of
the gold rectangles
minus the areas of
the blue rectangles
19.
RIEMANN SUM Note 3
When we take the
limit of such
Riemann sums, we
get the situation
illustrated here.
20.
NET AREA Note 3
A definite integral can be interpreted as
a net area, that is, a difference of areas:
A1 is the area of the region
above the x-axis and below the graph of f. b
A2 is the area
of the region
below the
f ( x) dx A
a 1 A2
x-axis and
above the
graph of f.
21.
UNEQUAL SUBINTERVALS Note 4
b
Though we have defined a f ( x) dx by dividing
[a, b] into subintervals of equal width, there
are situations in which it is advantageous
to work with subintervals of unequal width.
In Exercise 14 in Section 5.1, NASA provided velocity
data at times that were not equally spaced.
We were still able to estimate the distance traveled.
22.
UNEQUAL SUBINTERVALS Note 4
There are methods for numerical
integration that take advantage of
unequal subintervals.
23.
UNEQUAL SUBINTERVALS Note 4
If the subinterval widths are ∆x1, ∆x2, …, ∆xn,
we have to ensure that all these widths
approach 0 in the limiting process.
This happens if the largest width, max ∆xi ,
approaches 0.
24.
UNEQUAL SUBINTERVALS Note 4
Thus, in this case, the definition of
a definite integral becomes:
b n
f ( x)dx
a
lim
max xi 0
f ( x *) x
i 1
i i
25.
INTEGRABLE FUNCTIONS Note 5
We have defined the definite integral
for an integrable function.
However, not all functions are integrable.
26.
INTEGRABLE FUNCTIONS
The following theorem shows that
the most commonly occurring functions
are, in fact, integrable.
It is proved in more advanced courses.
27.
INTEGRABLE FUNCTIONS Theorem 3
If f is continuous on [a, b], or if f has only
a finite number of jump discontinuities, then
f is integrable on [a, b].
b
That is, the definite integral f ( x) dx exists.
a
28.
INTEGRABLE FUNCTIONS
If f is integrable on [a, b], then the limit
in Definition 2 exists and gives the same
value, no matter how we choose the sample
points xi*.
29.
INTEGRABLE FUNCTIONS
To simplify the calculation of the integral,
we often take the sample points to be right
Then, xi* = xi and the definition of an integral
simplifies as follows.
30.
INTEGRABLE FUNCTIONS Theorem 4
If f is integrable on [a, b], then
b n
f ( x) dx lim f ( x ) x
a ni
i 1
i
b a
where x and xi a i x
n
31.
DEFINITE INTEGRAL Example 1
n
Express 3
lim ( xi xi sin xi )xi
n
i 1
as an integral on the interval [0, π].
Comparing the given limit with the limit
in Theorem 4, we see that they will be
identical if we choose f(x) = x3 + x sin x.
32.
DEFINITE INTEGRAL Example 1
We are given that a = 0 and b = π.
So, by Theorem 4, we have:
n
3 3
lim ( xi xi sin xi ) xi ( x x sin x) dx
n 0
i 1
33.
DEFINITE INTEGRAL
Later, when we apply the definite integral to
physical situations, it will be important to
recognize limits of sums as integrals—as we
did in Example 1.
34.
DEFINITE INTEGRAL
When Leibniz chose the notation for
an integral, he chose the ingredients
as reminders of the limiting process.
35.
DEFINITE INTEGRAL
In general, when we write
n b
lim f ( xi *) x f ( x) dx
n a
i 1
we replace:
lim Σ by ∫
xi* by x
∆x by dx
36.
EVALUATING INTEGRALS
When we use a limit to evaluate
a definite integral, we need to know
how to work with sums.
37.
EVALUATING INTEGRALS
The following three equations
give formulas for sums of powers
of positive integers.
38.
EVALUATING INTEGRALS Equation 5
Equation 5 may be familiar to you
from a course in algebra.
n
n(n 1)
i
i 1 2
39.
EVALUATING INTEGRALS Equations 6 & 7
Equations 6 and 7 were discussed in
Section 5.1 and are proved in Appendix E.
n
2 n(n 1)(2n 1)
i
i 1
6
n 2
3 n(n 1)
ii 1
2
40.
EVALUATING INTEGRALS Eqns. 8, 9, 10 & 11
The remaining formulas are simple rules for
working with sigma notation:
n
c nc
i 1
n n
ca
i 1
i c ai
i 1
n n n
(a b ) a b
i 1
i i
i 1
i
i 1
i
n n n
(a b ) a b
i 1
i i
i 1
i
i 1
i
41.
EVALUATING INTEGRALS Example 2
a.Evaluate the Riemann sum for f(x) = x3 – 6x
taking the sample points to be right endpoints
and a = 0, b = 3, and n = 6.
3
3
b.Evaluate ( x
0
6 x) dx .
42.
EVALUATING INTEGRALS Example 2 a
With n = 6,
The interval width is: x
b a 3 0 1
n 6 2
The right endpoints are:
x1 = 0.5, x2 = 1.0, x3 = 1.5,
x4 = 2.0, x5 = 2.5, x6 = 3.0
43.
EVALUATING INTEGRALS Example 2 a
So, the Riemann sum is:
6
R6 f ( xi ) x
i 1
f (0.5) x f (1.0) x f (1.5) x
f (2.0) x f (2.5) x f (3.0) x
12 ( 2.875 5 5.625 4 0.625 9)
3.9375
44.
EVALUATING INTEGRALS Example 2 a
Notice that f is not a positive function.
So, the Riemann sum does not
represent a sum of areas of rectangles.
45.
EVALUATING INTEGRALS Example 2 a
However, it does represent the sum of the areas of
the gold rectangles (above the x-axis) minus the
sum of the areas of the blue rectangles (below the
46.
EVALUATING INTEGRALS Example 2 b
With n subintervals, we have:
b a 3
x
n n
Thus, x0 = 0, x1 = 3/n, x2 = 6/n, x3 = 9/n.
In general, xi = 3i/n.
47.
EVALUATING INTEGRALS Example 2 b
Since we are using right
endpoints, we can use Theorem 4,
as follows.
48.
EVALUATING INTEGRALS Example 2 b
3
3
6 x)dx
(
0
x
n
lim f ( xi )x
n
i 1
n
3i 3
lim f
n
i 1 nn
n 3
3 3i 3i
lim 6 (Eqn. 9 with c 3 / n)
n n
i 1 n n
3 n 27 3 18
lim 3 i i
n n
i 1 n n
49.
EVALUATING INTEGRALS Example 2 b
81 n 3 54 n
lim 4 i 2 i (Eqns. 11 & 9)
n n n i 1
i 1
81 n(n 1) 2 54 n(n 1)
lim 4 2 (Eqns. 7 & 5)
2
n n n 2
81 1 2 1
lim 1 27 1
n 4 n n
81 27
27 6.75
4 4
50.
EVALUATING INTEGRALS Example 2 b
This integral can’t be interpreted as
an area because f takes on both positive
and negative values.
51.
EVALUATING INTEGRALS Example 2 b
However, it can be interpreted as
the difference of areas A1 – A2, where
A1 and A2 are as shown.
52.
EVALUATING INTEGRALS Example 2 b
This figure illustrates the calculation by showing the
positive and negative terms
in the right Riemann sum Rn for n = 40.
53.
EVALUATING INTEGRALS Example 2 b
The values in the table
show the Riemann sums
approaching the exact
value of
the integral, -6.75, n →
54.
EVALUATING INTEGRALS
A much simpler method for
evaluating the integral in Example 2
will be given in Section 5.3
55.
EVALUATING INTEGRALS Example 3
3
x
a.Set up an expression for e dx as
1
a limit of sums.
b.Use a computer algebra system (CAS)
to evaluate the expression.
56.
EVALUATING INTEGRALS Example 3 a
Here, we have f(x) = ex, a = 1, b = 3,
and b a 2
x
n n
So, x0 = 1, x1 = 1 + 2/n, x2 = 1 + 4/n,
x3 = 1 + 6/n, and
xi = 1 + 2i / n
57.
EVALUATING INTEGRALS Example 3 a
From Theorem 4, we get:
3 n
x
e dx lim f ( x ) x
1 n
i 1
i
n
2i 2
lim f 1
n
i 1 n n
2 n 12i / n
lim e
n n
i 1
58.
EVALUATING INTEGRALS Example 3 b
If we ask a CAS to evaluate the sum
and simplify, we obtain:
n (3 n 2) / n ( n 2) / n
1 2 i / n e e
e
i 1
2/ n
e 1
59.
EVALUATING INTEGRALS Example 3 b
Now, we ask the CAS to evaluate
the limit:
(3 n 2) / n ( n 2) / n
3
x 2 e e
1 e dx lim
n n
2/ n
e 1
3
e e
60.
EVALUATING INTEGRALS Example 3 b
We will learn a much easier
method for the evaluation of integrals
in the next section.
61.
EVALUATING INTEGRALS Example 4
Evaluate the following integrals by interpreting
each in terms of areas.
1
a. 1 x 2 dx
0
3
b. ( x 1) dx
0
62.
EVALUATING INTEGRALS Example 4 a
2
Since f ( x) 1 x 0 ,
we can interpret this integral as
2
the area under the curve y 1 x
from 0 to 1.
63.
EVALUATING INTEGRALS Example 4 a
However, since y2 = 1
- x2, we get:
x2 + y2 = 1
This shows that
the graph of f is
the quarter-circle
with radius 1.
64.
EVALUATING INTEGRALS Example 4 a
1
2 2
1 x dx (1)
1
4
0 4
In Section 7.3, we will be able to prove that
the area of a circle of radius r is πr2.
65.
EVALUATING INTEGRALS Example 4 b
The graph of y = x – 1
is the line with slope 1
shown here.
We compute the integral
as the difference of the
areas of the two triangles:
3
( x 1) dx A1 A2 1
2 (2 2) 2 (1
1
1) 1.5
0
66.
MIDPOINT RULE
We often choose the sample point xi*
to be the right endpoint of the i th subinterval
because it is convenient for computing
the limit.
67.
MIDPOINT RULE
However, if the purpose is to find
an approximation to an integral, it is usually
better to choose xi* to be the midpoint of
the interval.
We denote this by xi .
68.
MIDPOINT RULE
Any Riemann sum is an approximation
to an integral.
However, if we use midpoints, we get
the following approximation.
69.
THE MIDPOINT RULE
b n
f ( x) dx f ( x ) x
i 1
i
x f ( x1 ) ... f ( x n )
b a
where x
n
and x i 12 ( xi 1 xi ) midpoint of xi 1 , xi
70.
MIDPOINT RULE Example 5
Use the Midpoint Rule with n = 5
to approximate
1 2
1 x dx
The endpoints of the five subintervals
are: 1, 1.2, 1.4, 1.6, 1.8, 2.0
So, the midpoints are: 1.1, 1.3, 1.5, 1.7, 1.9
71.
MIDPOINT RULE Example 5
The width of the subintervals is:
∆x = (2 - 1)/5 = 1/5
So, the Midpoint Rule gives:
2 1
1 x dx x f (1.1) f (1.3) f (1.5) f (1.7) f (1.9)
1 1 1 1 1 1
5 1.1 1.3 1.5 1.7 1.9
0.691908
72.
MIDPOINT RULE Example 5
As f(x) = 1/x for 1 ≤ x ≤ 2,
the integral represents
an area, and the
approximation given by
the rule is the sum of the
areas of
the rectangles shown.
73.
MIDPOINT RULE
At the moment, we don’t know
how accurate the approximation in
Example 5 is.
However, in Section 7.7, we will learn a method
for estimating the error involved in using the rule.
At that time, we will discuss other methods
for approximating definite integrals.
74.
MIDPOINT RULE
If we apply the rule to
the integral in
Example 2, we get
this picture.
75.
MIDPOINT RULE
The approximation
M40 = -6.7563
is much closer to
the true value -6.75 than
the right endpoint
R40 = -6.3998,
in the earlier figure.
76.
PROPERTIES OF DEFINITE INTEGRAL
When we defined the definite integral
b
f ( x) dx , we implicitly assumed that a < b.
However, the definition as a limit of Riemann
sums makes sense even if a > b.
77.
PROPERTIES OF DEFINITE INTEGRAL
Notice that, if we reverse a and b, then ∆x
changes from (b – a)/n to (a – b)/n.
a b
Therefore, f ( x) dx f ( x) dx
b a
a
If a = b, then ∆x = 0, and so f ( x) dx 0
b
78.
PROPERTIES OF INTEGRALS
We now develop some basic
properties of integrals that will
help us evaluate integrals in a simple
79.
PROPERTIES OF THE INTEGRAL
We assume f and g are continuous functions.
b
1. c dx c(b a ), where c is any constant
a
b b b
2. f ( x) g ( x) dx f ( x) dx g ( x) dx
a a a
b b
3. c f ( x) dx c f ( x) dx, where c is any constant
a a
b b b
4. f ( x) g ( x) dx f ( x) dx g ( x) dx
a a a
80.
PROPERTY 1
b
c dx c(b a),
a
where c is any constant
Property 1 says that the integral of a constant
function f(x) = c is the constant times the
length of the interval.
81.
PROPERTY 1
If c > 0 and a < b, this
is to be expected,
because c(b – a) is the
area of the shaded
rectangle here.
82.
PROPERTY 2
b b b
a f ( x ) g ( x ) dx a f ( x ) dx a g ( x ) dx
Property 2 says that the integral of a sum
is the sum of the integrals.
83.
PROPERTY 2
For positive functions, it says that
the area under f + g is the area under
f plus the area under g.
84.
PROPERTY 2
The figure helps us
understand why
this is true.
In view of how graphical
addition works,
the corresponding
vertical line segments
have equal height.
85.
PROPERTY 2
In general, Property 2 follows from Theorem 4
and the fact that the limit of a sum is the sum
ofb the limits: n
f ( x) g ( x) dx lim f ( x ) g ( x ) x
a n
i 1
i i
n n
lim f ( xi ) x g ( xi )x
n
i 1 i 1
n n
lim f ( xi ) x lim g ( xi ) x
n n
i 1 i 1
b b
f ( x) dx g ( x) dx
a a
86.
PROPERTY 3
b b
c f ( x) dx c f ( x) dx,
a a
where c is any constant
Property 3 can be proved in a similar manner
and says that the integral of a constant times
a function is the constant times the integral
of the function.
That is, a constant (but only a constant) can be taken
in front of an integral sign.
87.
PROPERTY 4
b b b
a f ( x ) g ( x) dx f ( x) dx g ( x ) dx
a a
Property 4 is proved by writing f – g = f + (-g)
and using Properties 2 and 3 with c = -1.
88.
PROPERTIES OF INTEGRALS Example 6
Use the properties of integrals to
1
2
evaluate (4 3x ) dx
0
Using Properties 2 and 3 of integrals,
we have: 1 2
1 1
2
(4 3 x ) dx 4 dx dx
3 x
0 0 0
1 1
4 dx 3x 2 dx
0 0
89.
PROPERTIES OF INTEGRALS Example 6
We know from Property 1 that:
1
4 dx 4(1 0) 4
0
We found in Example 2 in Section 5.1
that: 1
2
x
0
dx 1
3
90.
PROPERTIES OF INTEGRALS Example 6
Thus,
1 1 1
2 2
(4 3 x ) dx 4 dx 3 x dx
0 0 0
4 3 5
1
3
91.
PROPERTY 5
Property 5 tells us how to combine
integrals of the same function over
adjacent intervals:
c b b
f ( x ) dx f ( x ) dx f ( x ) dx
a c a
92.
PROPERTY 5
In general, Property 5 is
not easy to prove.
93.
PROPERTY 5
However, for the case where f(x) ≥ 0 and
a < c < b, it can be seen from the geometric interpretation in
the figure.
The area under y = f(x)
from a to c plus
the area from c to b
is equal to the total area
from a to b.
94.
PROPERTIES OF INTEGRALS Example 7
If it is known that
10 8
f ( x) dx 17 and f ( x) dx 12
0 0
10
find: f ( x) dx
8
95.
PROPERTIES OF INTEGRALS Example 7
By Property 5, we have:
8 10 10
f ( x) dx
0 8
f ( x) dx f ( x ) dx
0
So, 10 10 8
f ( x) dx f ( x) dx f ( x) dx
8 0 0
17 12
5
96.
PROPERTIES OF INTEGRALS
Properties 1–5 are true
a < b
a = b
a > b
97.
COMPARISON PROPERTIES OF THE INTEGRAL
These properties, in which we compare sizes
of functions and sizes of integrals, are true
only if a ≤ b.
b
6. If f ( x) 0 for a x b, then f ( x) dx 0
a
b b
7. If f ( x) g ( x) for a x b, then f ( x) dx g ( x) dx
a a
8. If m f ( x) M for a x b, then
b
m(b a) f ( x) dx M (b a )
a
98.
PROPERTY 6
b
If f ( x) 0 for a x b, then f ( x) dx 0
a
b
If f(x) ≥ 0, then f ( x) dx represents
a
the area under the graph of f.
99.
PROPERTY 6
Thus, the geometric interpretation of
the property is simply that areas are
However, the property can be proved
from the definition of an integral.
100.
PROPERTY 7
If f ( x) g ( x) for a x b,
b b
then f ( x) dx g ( x) dx
a a
Property 7 says that a bigger function has
a bigger integral.
It follows from Properties 6 and 4 because f - g ≥ 0.
101.
PROPERTY 8
Property 8 is illustrated
for the case where
f(x) ≥ 0. If m f ( x) M for a x b,
b
then m(b a ) f ( x) dx M (b a )
a
102.
PROPERTY 8
If f is continuous, we could take m and M
to be the absolute minimum and maximum values of
f on the interval [a, b].
103.
PROPERTY 8
In this case, Property 8 says that:
The area under the graph of f is greater than
the area of the rectangle with height m and less than the area
of the rectangle with height M.
104.
PROPERTY 8—PROOF
Since m ≤ f(x) ≤ M, Property 7 gives:
b b b
m dx f ( x) dx M dx
a a a
Using Property 1 to evaluate the integrals
on the left and right sides, we obtain:
b
m(b a) f ( x) dx M (b a )
a
105.
PROPERTY 8
Property 8 is useful when all we want is
a rough estimate of the size of an integral
without going to the bother of using
the Midpoint Rule.
106.
PROPERTY 8 Example 8
1
x2
Use Property 8 to estimate e dx
0
x2
f ( x ) e is a decreasing function on [0, 1].
So, its absolute maximum value is M = f(0) = 1
and its absolute minimum value is m = f(1) = e-1.
107.
PROPERTY 8 Example 8
Thus, by Property 8,
1
1 x2
e (1 0) e dx 1(1 0)
0
1
1 x2
or e e dx 1
0
As e-1 ≈ 0.3679, we can write:
1
x2
0.367 e dx 1
0
108.
PROPERTY 8
The result of
Example 8 is
illustrated here.
The integral is greater
than the area of the lower
rectangle and less than
the area of the square.
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