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Chapter 4
Integrals
4.1
Sigma Notation
This notation is very important. You will use it in this chapter as well as later
on, when you study sequences and series.
4.1.1
Theory
De…nition 225 (sigma notation) Let am ; am+1 ; :::; an be real numbers and
let m and n be integers such that m n.
1. We represent am + am+1 + ::: + an
1
+ an by
n
X
ai and we say ”The sum
i=m
as i ranges from m to n of a sub i”.
2. i is called the index of summation.
3.
n
X
ai is the sum written in ”sigma” notation. am + am+1 + ::: + an
1 + an
i=m
is the sum written in ”expanded” form.
Remark 226 Let us make the following remarks.
1. When writing
n
X
ai , the index of summation takes on all integer values
i=m
between m and n:
2. The name of the index of summation is not important as it is being replaced
by integers when expanding the sum. In other words
n
X
ai =
i=m
n
X
aj
j=m
Usually, we use the letters i, j or k for index of summation.
155
156
CHAPTER 4. INTEGRALS
Example 227
10
X
i2 = 12 + 22 + 32 + ::: + 102
i=1
Example 228
10
X
j 2 = 12 + 22 + 32 + ::: + 102
j=1
Example 229
10
X
2i = 2 + 22 + 23 + ::: + 210
i=1
Example 230
10
X
i
( 1) 2i =
2 + 22
23 + ::: + 210
i=1
Remark 231 If you compare the last two examples, you will see that the difi
ference between them is the presence of the term ( 1) . The e¤ ect of this term
is the minus sign which appears every other term. You should remember this.
i
Whenever you need to generate a minus sign every other term, use ( 1) .
Example 232
5
X
i
( 1) x2i =
x2 + x4
x6 + x8
x10
i=1
Remark 233 To generate only even powers, use 2i instead of i.
Example 234
n
X
xi
i=0
4.1.2
i!
=1+x+
x2
x3
xn
+
+ ::: +
2!
3!
n!
Some Results Worth Remembering
Proposition 235 If C is any constant, then
n
X
Cai = C
n
X
ai
i=1
i=1
Proof. This is simply factoring C.
n
X
Cai
= Ca1 + Ca2 + ::: + Can
i=1
= C (a1 + a2 + ::: + an )
n
X
= C
ai
i=1
4.1. SIGMA NOTATION
Proposition 236
n
X
157
ai +
i=1
n
X
bi =
i=1
n
X
(ai + bi )
i=1
Proof. See problems at the end of the section.
n
X
n (n + 1)
2
i=1
Proof. Let us …rst remark that the above formula simply says that 1+2+:::+n =
n (n + 1)
. To prove it, we write
2
Proposition 237 (Sum of the …rst n integers)
S = 1 + 2 + ::: + (n
i=
1) + n
which we can also write as
S = n + (n
1) + ::: + 2 + 1
If we add the two equalities, we obtain
S + S = (1 + n) + (2 + (n
1)) + ::: + ((n
1) + 2) + n + 1
2S = (n + 1) + (n + 1) + ::: + (n + 1) + (n + 1)
We are adding n terms, which are all equal to n + 1, therefore
2S = n (n + 1)
S=
n (n + 1)
2
Proposition 238 There are similar results for the sum of the …rst n squares
and n cubes which we give without proof.
1. Sum of the …rst n squares:
n
X
i2 =
i=1
n (n + 1) (2n + 1)
6
2. Sum of the …rst n cubes:
n
X
i=1
4.1.3
i3 =
n (n + 1)
2
2
Things to know:
Given a sum in sigma notation, know how to write it in expanded form.
Given a sum in expanded form, know how to write it in sigma notation.
Be able to do problems like the ones below.
158
4.1.4
CHAPTER 4. INTEGRALS
Problems
1. Write the sum in expanded form
(a)
5
X
p
i
i=1
(b)
5
X
3i
i=4
(c)
4
X
2k
k=0
(d)
n
X1
1
2k + 1
( 1)
j
j=0
2. Write the sum in sigma notation
1 2 3 4
19
+ + + + ::: +
2 3 4 5
20
(b) 2 + 4 + 6 + 8 + ::: + 2n
(a)
(c) 1 + 3 + 5 + ::: + (2n
1)
(d) x + x2 + x3 + ::: + xn
(e) 1
x + x2
3. Prove that
n
X
i=1
4.1.5
n
x3 + ::: + ( 1) xn
ai +
n
X
bi =
i=1
n
X
(ai + bi ).
i=1
Answers
1. Write the sum in expanded form
(a)
5
X
p
i
i=1
5
X
p
i=
p
1+
p
2+
p
3+
i=1
(b)
5
X
3i
i=4
5
X
i=4
3i = 34 + 35
p
4+
p
5
4.1. SIGMA NOTATION
(c)
4
X
2k
k=0
159
1
2k + 1
4
X
2k
k=0
(d)
n
X1
( 1)
1 1 3 5 7
1
=
+ + + +
2k + 1
1
3 5 7 9
j
j=0
n
X1
j
0
1
2
( 1) = ( 1) + ( 1) + ( 1) + ::: + ( 1)
n 1
j=0
2. Write the sum in sigma notation
(a)
19
1 2 3 4
+ + + + ::: +
2 3 4 5
20
19
19 X i
1 2 3 4
+ + + + ::: +
=
2 3 4 5
20
i+1
i=1
(b) 2 + 4 + 6 + 8 + ::: + 2n
2 + 4 + 6 + 8 + ::: + 2n =
n
X
2i
i=1
(c) 1 + 3 + 5 + ::: + (2n
1)
1 + 3 + 5 + ::: + (2n
1) =
n
X
(2i
1)
i=1
(d) x + x2 + x3 + ::: + xn
x + x2 + x3 + ::: + xn =
n
X
xi
i=1
(e) 1
n
x + x2
x3 + ::: + ( 1) xn
1
x + x2
n
x3 + ::: + ( 1) xn =
n
X
i=0
3. No answer to write.
i
( 1) xi
160
CHAPTER 4. INTEGRALS
Figure 4.1: Partition with 6 subintervals
4.2
Area and Riemann Sums
4.2.1
Preliminary De…nitions
De…nition 239 (partition) In this section, unless stated otherwise, we assume that a and b are …nite real numbers.
1. Given an interval [a; b] ; a partition P is a subdivision of the interval into
n subintervals. If we denote the subintervals [x0 ; x1 ], [x1 ; x2 ], :::, [xn 1 ; xn ]
then P = fx0 ; x1 ; :::; xn g. With this notation, we always have a = x0
and b = xn , and n is the number of subintervals. Figure 4.1 shows an
interval [a; b] and a partition P = fx0 ; x1 ; x2 ; x3 ; x4 ; x5 ; x6 g of the interval
with n = 6. You will note that the number of points is always one more
than the number of subintervals that is if there are n intervals, there will
be n + 1 points . You will also note that the …rst subinterval is always
[x0 ; x1 ]. The second subinterval is [x1 ; x2 ]. In general, the ith subinterval
will be [xi 1 ; xi ]. The last subinterval will be the nth subinterval, that is
[xn 1 ; xn ]. It is important to understand this notation in order to be able
to understand what follows.
2. If all the subintervals have the same length, the partition is called a regular
partition. The length of each subinterval, denoted x is then
x=
b
a
n
Figure 4.2 shows an interval and a regular partition P = fx0 ; x1 ; x2 ; x3 ; x4 ; x5 ; x6 g
of the interval with n = 6.
Example 240 A possible partition of the interval [0; 4] with n = 4 is P =
f0; 0:5; 1; 3:5; 4g. The subintervals obtained are [0; 0:5] ; [0:5; 1] ; [1; 3:5] ; [3:5; 4].
This is not a regular partition since not all the subintervals have the same length.
There are many other ways of partitioning [0; 4].
4.2. AREA AND RIEMANN SUMS
161
Figure 4.2: Regular partition with 6 subintervals
Figure 4.3: Area below a graph
Example 241 A regular partition of [0; 4] with n = 4 is P = f0; 1; 2; 3; 4g. The
subintervals obtained are [0; 1] ; [1; 2] ; [2; 3] ; [3; 4]. This is a regular partition
since all the intervals have the same length.
4.2.2
Area Below a Graph
Let f be a positive function (its graph is above the x-axis), assume that f
is continuous on the interval [a; b]. The goal is to …nd the area of the region
bounded by the graph of y = f (x), the x-axis, the vertical lines x = a and
x = b. In other words, we want to …nd the area of the shaded region shown on
Figure 4.3. We often say that we want to …nd the area below the graph of f
between a and b.In the case that the graph of y = f (x) is a straight line, the
corresponding region may be a rectangle, or a polygon. In this case, we know
how to compute its area. In general though, the region will not correspond to a
region for which we have a formula for its area. A purpose of this section is to
devise a way to compute its area. We do it in two steps. First, we will describe
a procedure to approximate the area. Then, we see how to …nd the exact value
162
CHAPTER 4. INTEGRALS
Figure 4.4: Approximating an area with rectangles
of the area.
4.2.3
Approximation of the Area below a Graph
To better understand the procedure we are about to describe, you can refer to
Figure 4.4. It illustrates the procedure in the case n = 4.
Problem: Given a positive function f and an interval [a; b], approximate
the area (denoted A) below the graph of f between a and b. We will approximate
the region with rectangles and the area of the region by the sum of the areas of
the rectangles. Since a rectangle is de…ned by its base and its height, we describe
how to derive the base and the height of each rectangle used. We proceed as
follows:
1. Deriving the base of each rectangle. Divide the interval [a; b] into n
subintervals (we will see later which number to choose for n). Though the
intervals do not have to have equal length, to simplify the procedure, we
will only consider subintervals of equal length. Thus, we obtain a regular
partition P = fa = x0 ; x1 ; :::; xn = bg. Also, the length of each subinterval
b a
will be x =
. The subintervals obtained are [x0 ; x1 ], [x1 ; x2 ], :::,
n
[xn 1 ; xn ]. In other words, the subintervals are of the form [xi 1 ; xi ] for
i = 1; 2; :::; n. The points xi can be generated by the formula
a
n
Though these formulae are not needed to carry out the procedure, they
are useful if one were to write a program to simulate the procedure. These
subintervals [xi 1 ; xi ] will form the base of our rectangles.
xi = a + i
b
4.2. AREA AND RIEMANN SUMS
163
2. Deriving the height of each rectangle. In each interval [xi 1 ; xi ],
select a point denoted xi . In theory, xi can be selected anywhere in
[xi 1 ; xi ]. In practice, it will usually be one of the end points, or the
midpoint, or the point at which f is either maximum or minimum. Thus,
in the …rst interval, the interval [x0 ; x1 ] we select a point that we call x1 .
In the second interval, the interval [x1 ; x2 ], we select a point we call x2 ,
and so on. The height of our rectangles will be f (xi ).
If xi is the right endpoint of [xi
If xi is the left endpoint of [xi
If xi is the midpoint of [xi
1 ; xi ]
1 ; xi ]
then xi = xi
then xi = xi
1 ; xi ] then xi =
xi +xi
2
1
1
3. Forming the rectangles. In each subinterval [xi 1 ; xi ], consider the
rectangle Ri with base the width of the interval, and height f (xi ). Let
Ai denote its area. We can compute Ai easily.
Ai = width height
= (xi
xi
1) f
(xi )
=
xf (xi )
b a
f (xi )
=
n
4. Approximating the area. We approximate the area below the graph
(denoted A) by adding the area of all the rectangles. Thus,
A
A1 + A2 + ::: + An
n
X
Ai
i=1
n
X
f (xi ) x
i=1
x
n
X
f (xi )
i=1
b
n
n
aX
f (xi )
i=1
Example 242 Approximate the area below the graph of y = x2 between x = 0
and x = 4, using 4 subintervals and by selecting xi to be the right end point of
each subinterval. Repeat the procedure by selecting xi to be the left end point.
In this case, a = 0, b = 4 and n = 4.
If we select xi to be the right end point of each subinterval, then x1 = 1,
x2 = 2, x3 = 3 and x4 = 4. The function f is f (x) = x2 . We can now
164
CHAPTER 4. INTEGRALS
Figure 4.5: Rectangles using right end points
apply the formula.
A
n
b
n
4
aX
f (xi )
i=1
0
(f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ))
4
f (1) + f (2) + f (3) + f (4)
1 + 4 + 9 + 16
30
Figure 4.5 illustrates this procedure. You will note that the exact value of
the area is less than the approximation. Thus, we know that A < 30.
If we select xi to be the left end point of each subinterval, then x1 = 0,
x2 = 1, x3 = 2 and x4 = 3. The function f is f (x) = x2 . We can now
apply the formula.
A
n
b
n
4
aX
f (xi )
i=1
0
(f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ))
4
f (0) + f (1) + f (2) + f (3)
0+1+4+9
14
Figure 4.6 illustrates this procedure. You will note that the exact value of
the area is more than the approximation. Thus, we know that A > 14.
4.2. AREA AND RIEMANN SUMS
165
Figure 4.6: Rectangles using left end points
Remark 243 From the two methods above used to approximate the area, we
see that 14 < A < 30. Therefore, we have an upper bound as well as a lower
bound on the area. They are not very good, since they are far apart. The next
example will show that we can do better.
Example 244 Same as the previous example, using 8 subintervals.
In this case, a = 0, b = 4, n = 8.
If we select xi to be the right end point of each subinterval, then x1 = 0:5,
x2 = 1, x3 = 1:5, x4 = 2, x5 = 2:5, x6 = 3, x7 = 3:5 and x8 = 4. The
function f is f (x) = x2 . We can now apply the formula.
A
n
b
n
4
aX
0
8
f (xi )
i=1
(f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f (x5 ) + f (x6 ) + f (x7 ) + f (x8 ))
1
(f (0:5) + f (1) + f (1:5) + f (2) + f (2:5) + f (3) + f (3:5) + f (4))
2
1 1
9
25
49
+1+ +4+
+9+
+ 16
2 4
4
4
4
25:5
Figure 4.7 illustrates this procedure. You will note that the exact value of
the area is less than the approximation. Thus, we know that A < 25:5.
If we select xi to be the left end point of each subinterval, then x1 = 0,
x2 = 0:5, x3 = 1, x4 = 1:5, x5 = 2, x6 = 2:5, x7 = 3, x8 = 3:5. The
166
CHAPTER 4. INTEGRALS
Figure 4.7: Rectangles using right end points
function f is f (x) = x2 . We can now apply the formula.
A
n
b
n
4
aX
0
8
f (xi )
i=1
(f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f (x5 ) + f (x6 ) + f (x7 ) + f (x8 ))
1
(f (0) + f (0:5) + f (1) + f (1:5) + f (2) + f (2:5) + f (3) + f (3:5))
2
1 1
9
25
49
+1+ +4+
+9+
2 4
4
4
4
17:5
Figure 4.8 illustrates this procedure. You will note that the exact value of
the area is more than the approximation. Thus, we know that A > 17:5.
Remark 245 This time, we see that 17:5 < A < 25:5, which is slightly better
than what we had before. Therefore, increasing the number of subintervals seems
to give a better approximation. In fact, it can be proven (this is done in a more
advanced class such as Real Analysis) that as n gets larger, our approximation
becomes closer and closer to the exact value of the area.
4.2.4
Exact Value of the Area
From the remark made above, we de…ne the area below a graph as follows:
De…nition 246 (area below a graph) Let f be a positive function and a and
b two …nite real numbers such that a < b.
4.2. AREA AND RIEMANN SUMS
167
Figure 4.8: Rectangles using left end points
1. The area A below the graph of y = f (x) between x = a and x = b is
A = lim
n!1
n
X
f (xi ) x
i=1
providing the limit exists. In the above formula, xi is a point chosen in
[xi 1 ; xi ].
2. The sum in the above formula is called a Riemann sum. If xi corresponds to a maximum of f on [xi 1 ; xi ] then the sum is called an upper
Riemann sum. If xi corresponds to a minimum of f on [xi 1 ; xi ] then
the sum is called an lower Riemann sum.
In practice, …nding A using the de…nition is very di¢ cult. First, the expression resulting from the sum is usually very complicated. Then, we have to take
its limit as n approaches in…nity. We will not spend any more time on it, but
you should remember the formula.
4.2.5
Use of Technology
Carrying out the computations involved in the procedure described above is very
tedious. I have designed a Java applet to assist students in this procedure. It
can be found at http://science.kennesaw.edu/~plaval/tools/index.html
under integration.
4.2.6
Things to know
Know how to approximate an area using Riemann sums.. If you are given
a partition, use it, otherwise make up your own partition.
168
4.2.7
CHAPTER 4. INTEGRALS
Problems
Following the procedure described in this section, approximate the area below
the graph of the given function f (x) in the given interval [a; b] using the given
number of subdivisions n and the given rule to determine the height of the
rectangle for each situation below.
1. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be
the left endpoint of each subinterval.
2. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be
the right endpoint of each subinterval.
3. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be
the midpoint of each subinterval.
4. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be
the left endpoint of each subinterval.
5. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be
the right endpoint of each subinterval.
6. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be
the midpoint of each subinterval.
7. Using the applet mentioned in the notes, approximate the area below the
graph of f (x) = sin x between x = 0 and x = by using upper and lower
Riemann sums and various values for the number of subintervals.
8. Using the applet mentioned in the notes, approximate the area below the
graph of f (x) = ex between x = 0 and x = 1 by using upper and lower
Riemann sums and various values for the number of subintervals.
4.2.8
Answers
Following the procedure described in this section, approximate the area below
the graph of the given function f (x) in the given interval [a; b] using the given
number of subdivisions n and the given rule to determine the height of the
rectangle for each situation below.
1. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be
the left endpoint of each subinterval.
A
= 50
2. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be
the right endpoint of each subinterval.
A
34
4.3. THE DEFINITE INTEGRAL
169
3. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be
the midpoint of each subinterval.
A
43
4. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be
the left endpoint of each subinterval.
A
46:5
5. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be
the right endpoint of each subinterval.
A
38:5
6. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be
the midpoint of each subinterval.
A
42:75
7. Using the applet mentioned in the notes, approximate the area below the
graph of f (x) = sin x between x = 0 and x = by using upper and lower
Riemann sums and various values for the number of subintervals.
The area is approximately 2.
8. Using the applet mentioned in the notes, approximate the area below the
graph of f (x) = ex between x = 0 and x = 1 by using upper and lower
Riemann sums and various values for the number of subintervals.
The area is approximately 1:7.
4.3
4.3.1
The De…nite Integral
Theory
De…nition 247 (de…nite integral) The de…nite integral of f from a to b
is de…ned by
Zb
n
X
f (x) dx = lim
f (xi ) x
n!1
a
i=1
if this limit exists. When the limit exists, f is said to be integrable. The
number a is called the lower limit of integration, b is the upper limit of
integration. f is called the integrand.
Remark 248 You may ask what functions are integrable. It turns out that the
answer is not simple. You will study this question in a more advanced class
such as Real Analysis. For now, we will only say that if f is continuous then f
is integrable. Keep in mind that this does not mean if f is not continuous it is
not integrable.
170
CHAPTER 4. INTEGRALS
Regarding areas, as we discussed in the previous section, we see that if f
happens to be a positive function, then the integral is the area below the graph.
If the function is entirely negative, then the above sum will be negative since the
term f (xi ) will be negative while x > 0. However, an area is always positive.
Therefore, the integral will be the negative of the area. The proposition below
summarizes the relation between integral and area.
Proposition 249 The relation between an integral and the area of the region
bounded by y = f (x), the x-axis, the lines x = a and x = b is as follows:
1. If f (x)
0 then
Rb
f (x) dx is the area of the region between the graph of
a
f and the x-axis and between x = a and x = b.
2. If f (x)
0 then
Rb
f (x) dx is the negative of the area of the region between
a
the graph of f and the x-axis and between x = a and x = b.
3. In general,
Rb
f (x) dx = A1
A2 where A1 is the area of the region above
a
the x-axis, below the graph of f and A2 is the area of the region below the
x-axis, above the graph of f .
If the relationship between integral and area is still not entirely clear, Figure
4.9 along with the explanations below should clarify everything.
1. Assuming A1 , A2 , and A3 are known, then we could use this knowledge
to compute the following integrals:
(a)
R1
f (x) dx = A1
0
(b)
R3
f (x) dx =
A2 (remember, the numbers Ai represent areas, they
1
are positive).
R5
(c) f (x) dx = A3
3
(d)
R3
f (x) dx = A1
A2
0
(e)
R5
f (x) dx =
A2 + A3
1
(f)
R5
f (x) dx = A1
A2 + A3
0
2. If, on the other hand we could compute
Rb
f (x) dx for any a and b, then
a
we could use this knowledge to compute areas as follows:
4.3. THE DEFINITE INTEGRAL
171
Figure 4.9: Relationship between integral and area
(a) A1 =
R1
f (x) dx
0
(b) A2 =
R3
f (x) dx
1
(c) A3 =
R5
f (x) dx
3
(d) area of the shaded region =
R1
0
f (x) dx
R3
1
f (x) dx +
R5
f (x) dx
3
Thus, we see that our knowledge of areas can be used to compute integrals.
Our ability to compute integrals can also be used to …nd the area of certain
regions. Once we know how to compute integrals e¢ ciently, we will use integrals
to compute areas.
Integrals have the following properties:
Proposition 250 Assuming all the integrals below exist, and a
1.
Rb
f (x) dx =
a
2.
Ra
a
Ra
b
f (x) dx = 0
f (x) dx
b we have:
172
3.
CHAPTER 4. INTEGRALS
Rb
cdx = c (b
a)
a
4.
Rb
[f (x)
g (x)] dx =
a
5.
Rb
a
6.
Rb
Rb
Rb
f (x) dx
a
g (x) dx
a
Rb
cf (x) dx = c f (x) dx
a
f (x) dx =
a
7. If f (x)
Rc
f (x) dx +
Rb
f (x) dx
c
a
0 for x in [a; b] then
Rb
f (x) dx
0
a
8. If f (x)
g (x) for x in [a; b] then
Rb
f (x) dx
a
Rb
g (x) dx
a
9. If f has a maximum value M on [a; b] and a minimum value m on [a; b]
Rb
then m (b a)
f (x) dx M (b a)
a
Proof. Though we will not prove these properties, they are not very hard.
Most of them follow from the de…nition.
For the moment, we have two ways of computing integrals. We can either
approximate them using Riemann sums, or we can computing them using the
area interpretation of integrals. Both methods are very limited, we illustrate
them with a few examples. In the next section (4.4), we will see an easier way
of computing integrals.
Example 251 Approximate
R4
x2 dx using Riemann sums. You will use 4 subin-
0
tervals, and select xi to be the right end point of each subinterval.
This is precisely the work we did for one of the examples on area above. The
answer we found was 30.
Example 252 Approximate
R
cos xdx using Riemann sums. You will use 4
0
subintervals, and select xi to be the right end point of each subinterval. Figure
4.10 illustrates this. The general formula we use is
Z
0
cos xdx
4
X
f (xi ) x
i=1
4
4
0X
i=1
cos xi
4.3. THE DEFINITE INTEGRAL
Figure 4.10: Approximating
R
173
cos xdx using 4 rectangles and right end points
0
We need to …gure out what the xi are. For this, we need to …gure out what the
xi are. Since we divide the interval [0; ] into four subintervals, we will have
3
,
…ve points: x0 ; x1 ; :::; x4 . These points are: x0 = 0, x1 = , x2 = , x3 =
4
2
4
3
x4 = . Since we pick right end points, we have x1 = , x2 = , x3 =
and
4
2
4
x4 = . It follows that
Z
0
cos xdx
4
cos
4
+ cos
2
+ cos
3
+ cos
4
0
=
Example 253 Compute
4
R2
xdx using the area interpretation of the integral.
0
For this, we begin by sketching the graph of f (x) = x (see Figure 4.11). Between
0 and 2, we see that the integrand is positive, therefore the integral is exactly
the area of the region below the graph between 0 and 2. The region is a triangle,
174
CHAPTER 4. INTEGRALS
Figure 4.11: Integral and area
its base is 2, its height is 2, therefore
Z2
xdx = area of shaded region
0
1
2
2
=2
2
=
Example 254 Compute
R2
xdx using the area interpretation of the integral.
1
Again, we sketch the graph of the function, see Figure 4.12. According to the
area interpretation of the integral, we have:
Z2
xdx = A2
A1
1
1
2
2
3
=
2
=
Example 255 Compute
R2 p
4
2
1
1
2
1
x2 dx using the area interpretation of the inte-
0
gral.
p
The function f (x) = 4
x2 is the upper half of the circle of radius 2, centered
R2 p
at the origin. Therefore, the integral
4 x2 dx corresponds to the area of the
0
4.3. THE DEFINITE INTEGRAL
175
Figure 4.12: Integral and area
shaded region shown on Figure 4.13, that is one fourth of the area of a circle of
radius 2. It follows that
Z2 p
4
x2 dx =
1
2
(2)
4
0
=
1
4
4
=
Example 256 Compute
R2
x+
p
4
x2 dx using the area interpretation of the
0
integral.
Using the properties of the integral, we have:
Z2
0
p
x+ 4
x2
dx =
Z2
0
xdx +
Z2 p
4
x2 dx
0
We computed each integral in the examples above, so
Z2
x+
0
4.3.2
p
4
x2 dx = 2 +
Things to Know
Given a partition and a function, be able to evaluate
Rb
a
Riemann sums.
f (x) dx with
176
CHAPTER 4. INTEGRALS
Figure 4.13: Integral and area
Be able to select your own partition to evaluate
f (x) dx with Riemann
a
sums.
Be able to evaluate
Rb
Rb
f (x) dx by interpreting it in terms of areas.
a
Know and be able to use the properties of the de…nite integral.
4.3.3
Problems
R2
R4
R4
1. Assuming that 1 f (x) dx = 1, 1 f (x) dx = 3 and 1 g (x) dx = 2, compute the de…nite integrals below using the properties in integrals. State
which properties are being used.
R1
(a) 1 f (x) dx
R1
(b) 4 f (x) dx
R1
(c) 4 5g (x) dx
R4
(d) 1 (f (x) g (x)) dx
R4
(e) 1 (2f (x) + 3g (x)) dx
R4
(f) 2 f (x) dx
2. Use the area interpretation of the integral to evaluate the integrals below.
R3
(a) 0 xdx
R3
(b)
xdx
1
4.3. THE DEFINITE INTEGRAL
(c)
177
R3
jxj dx
2
R3 p
(d)
9 x2 dx
3
R3p
(e) 0 9 x2 dx
p
R3
(f) 0 x + 9 x2 dx
3. Write the area of each region below as a de…nite integral or a sum of
de…nite integrals. You do not need to evaluate the integrals.
(a) The region between the x-axis and the graph of sin x between x = 0
and x = .
2
(b) The region between the x-axis and the graph of sin x between x =
and x = 2 .
(c) The region between the x-axis and the graph of sin x between x = 0
and x = 2 .
(d) The upper half of a circle of radius 16.
4.3.4
Answers
R2
R4
R4
1. Assuming that 1 f (x) dx = 1, 1 f (x) dx = 3 and 1 g (x) dx = 2, compute the de…nite integrals below using the properties in integrals. State
which properties are being used.
(a)
(b)
(c)
(d)
(e)
(f)
R1
1
f (x) dx = 0
4
f (x) dx =
4
5g (x) dx = 10
1
(f (x)
1
(2f (x) + 3g (x)) dx = 0
2
f (x) dx = 2
R1
R1
R4
R4
R4
3
g (x)) dx = 5
2. Use the area interpretation of the integral to evaluate the integrals below.
(a)
R3
0
xdx
Z
3
xdx =
0
(b)
R3
1
xdx
Z
9
2
3
xdx = 4
1
178
CHAPTER 4. INTEGRALS
(c)
(d)
(e)
R3
2
jxj dx
R3 p
3
R3p
0
Z
x2 dx
9
Z
x2 dx
9
3
3
0
(f)
R3
0
x+
p
9
jxj dx =
2
p
13
2
9
x2 dx =
9
2
p
9
x2 dx =
9
4
p
x2 dx =
3
Z
3
x2 dx
Z
0
3
x+
9
9 9
+
2
4
3. Write the area of each region below as a de…nite integral or a sum of
de…nite integrals. You do not need to evaluate the integrals.
(a) The region between the x-axis and the graph of sin x between x = 0
and x = .
2
Z 2
Area =
sin xdx
0
(b) The region between the x-axis and the graph of sin x between x =
and x = 2 .
Z 2
Area =
sin xdx
(c) The region between the x-axis and the graph of sin x between x = 0
and x = 2 .
Z
Z 2
Area =
sin xdx
sin xdx
0
(d) The upper half of a circle of radius 16.
Area =
4.4
4.4.1
Z
4
4
p
16
x2 dx
The Fundamental Theorem of Calculus
Theory
De…nition 257 An antiderivative of a function f is a function F such that
F 0 (x) = f (x).
4.4. THE FUNDAMENTAL THEOREM OF CALCULUS
179
1
Example 258 Since the derivative of ln x is , it means that an antiderivative
x
1
of
is ln x
x
0
Example 259 Since (sin x) = cos x, it means that an antiderivative of cos x
is sin x.
Remark 260 Antiderivatives are not unique. If F is an antiderivative of f
(that is if F 0 = f ), then F + C where C is any constant is also an antiderivative
0
of f . To verify this, we need to show that (F + C) = f .
0
0
(F + C) = F 0 + (C)
= F0 + 0
= F0
=f
Remark 261 Because of the previous remark, antiderivatives are always given
with a constant. For example, we say that an antiderivative for cos x is sin x +
C. In the remaining part of this document, C will always be used to denote a
constant.
Antiderivatives play an important in evaluating integrals as the next theorem
will show.
Theorem 262 (Fundamental Theorem of Calculus) Let f be a continuous function on [a; b].
1. The function g (x) =
Rx
f (t) dt for x in [a; b] is continuous and di¤ eren-
a
0
tiable. Furthermore, g (x) = f (x)
or
an antiderivative of f .
2. If F is any antiderivative of f then
Rb
a
d
dx
Rx
f (t) dt = f (x)
that is g is
a
b
f (x) dx = F (x)ja = F (b)
F (a).
Remark 263 The following follows from the theorem:
1. Part 1 of the Fundamental Theorem of Calculus says that the derivative
of the integral of a function is the function itself.
2. Part 2 of the Fundamental Theorem of Calculus provides a way to compute
integrals. We simply have to …nd an antiderivative of the integrand, and
plug in the limits of integration.
180
CHAPTER 4. INTEGRALS
R
De…nition 264 (inde…nite integral) f (x) dx is used to represent an antiderivative of f . That is
Z
f (x) dx = F (x) , F 0 (x) = f (x)
It is a function, not a number. It is called the inde…nite integral of f .
Remark
265 In the above de…nition, if we replace f (x) by F 0 (x), we obtain
R 0
F (x) dx = F (x). In other words, the integral of the derivative of a function is
the function itself. If we combine this with part 1 of the Fundamental Theorem
of Calculus which says that the derivative of the integral of a function is the
function itself, we see that integration and di¤ erentiation are inverse processes.
One undoes what the other one does.
At this point, you should know the following antiderivatives:
1.
R
un+1
+ C if n 6=
n+1
un du =
1
R 1
du = ln juj + C
u
R
3. eu du = eu + C
2.
4.
5.
6.
7.
8.
9.
10.
11.
12.
R
R
R
R
R
R
R
R
R
au du =
au
+C
ln a
sin udu =
cos u + C
cos udu = sin u + C
sec2 udu = tan u + C
csc2 udu =
cot u + C
sec u tan udu = sec u + C
csc u cot udu =
csc u + C
1
du = tan
u2 + 1
1
p
1
1
u2
du = sin
u+C
1
u+C
In addition, the following properties of the integral are also often used:
R
R
13. Cf (x) dx = C f (x) dx
R
R
R
14. [f (x) g (x)] dx = f (x) dx
g (x) dx
4.4. THE FUNDAMENTAL THEOREM OF CALCULUS
4.4.2
181
Part 2 of the Fundamental Theorem of Calculus
Let us …rst illustrate part 2 of the fundamental theorem of calculus. Recall it
says that if F is any antiderivative of f then
Zb
b
f (x) dx = F (x)ja = F (b)
F (a)
a
Thus, the problem of computing a de…nite integral is reduced to the problem of
…nding antiderivatives. Let us look at some examples.
Example 266 Find
R3
x2 dx
1
Since an antiderivative of x2 is
Z3
x3
(formula 1 above with n = 2), it follows that
3
x2 dx
=
x3
3
1
3
1
3
=
=
Example 267 Find
Z
13
3
3
3
26
3
sin xdx
0
Z
sin xdx = ( cos x)j0
0
=
cos
=
( 1) + 1
=2
( cos 0)
182
CHAPTER 4. INTEGRALS
Example 268 Find
Z2
x + x2 dx
0
Z2
x + x2 dx =
x3
x2
+
2
3
0
2
2
0
3
02
03
+
2
3
2
2
+
2
3
8
=2+
3
14
=
3
=
4.4.3
Part 1 of the Fundamental Theorem of Calculus
We now illustrate part 1 of the fundamental theorem of calculus. Recall, it says
Rx
Rx
0
d
that if g (x) = f (t) dt then g (x) = f (x) or dx
f (t) dt = f (x) . Note that
a
a
in order to be able to apply it, the lower limit of integration must be a constant,
we must be di¤erentiating with respect to the upper limit of integration.
If we combine this with the chain rule, then we have
d
dx
Zu
f (t) dt = f (u)
du
dx
(4.1)
a
where u is a function of x.
1
0 x
Z
d @
Example 269 Find
sin tdtA
dx
0 x
1
Z
d @
By part 1 of the Fundamental Theorem of Calculus,
sin tdtA = sin x.
dx
0
0
0 0
1
Z
d @
sin tdtA
Example 270 Find
dx
x
Before we can apply the Fundamental Theorem of Calculus, we must switch the
4.4. THE FUNDAMENTAL THEOREM OF CALCULUS
183
limits of integration, which we can do by one of the properties of the integral.
1
1
0 0
0 x
Z
Z
d
d @
@
sin tdtA =
sin tdtA (property of the integral)
dx
dx
x
0
11
0 0 x
Z
d
= @ @ sin tdtAA (property of the derivative)
dx
0
=
sin x (by the previous example)
1
0 2
Zx
d B
C
Example 271 Find
@ sec tdtA
dx
1
The upper limit of integration is not x but a function of x, so we must use
formula 4.1. Therefore
0 2
1
Zx
2
d B
C
2 dx
@ sec tdtA = sec x
dx
dx
1
=
Example 272 Find
0
d B
@
dx
Zx
x
2x sec x2
1
2
C
sin tdtA
Here, neither limit of integration is a constant. Using a property of integrals,
we can break the interval of integration by introducing a constant. We have
0 2
1
0 2
1
0 0
1
Zx
Zx
Z
d B
d @
d B
C
C
sin tdtA +
@ sin tdtA =
@ sin tdtA
dx
dx
dx
x
x
=
0
d @
dx
=
0
Zx
0
1
sin tdtA +
d B
@
dx
=
4.4.4
Zx
0
1
2
C
sin tdtA
0 2
1
0 x
11
Z
Zx
C
@ d @ sin tdtAA + d B
@ sin tdtA
dx
dx
0
0
=
0
0
dx2
sin x + sin x2
dx
sin x + 2x sin x2
Things to know
Know part 1 of the Fundamental Theorem, be able to apply it.
184
CHAPTER 4. INTEGRALS
Know part 2 of the Fundamental Theorem, be able to apply it to compute
integrals.
4.4.5
Problems
Evaluate the integrals below (problems 1-14)
R4
1. 0 x3 dx
R
2. 02 cos xdx
R1
3. 0 ex dx
R
4. 0 sin xdx
R4p
5. 1 xdx
R2
6. 0 x3 + 3x2 1 dx
R0
7.
(2x ex ) dx
1
R4 p
8. 0 x ( x + 1) dx
R
9. 04 3 sec2 tdt
R9 1
dx
1 2x
R2
2
11.
(3x + 1) dx
2
10.
R e x2 + x + 1
dx
1
x
R 1 + cos2 t
dt
13. 04
cos2 t
R2
14. 0 4x dx
12.
15. Find F 0 (x) for F (x) =
16. Find F 0 (x) for F (x) =
17. Find G0 (x) for G (x) =
18. Find G0 (x) for G (x) =
Rxp
1
R2
x
t2 sin tdt
R x3
1
R x2
x
1 + 2tdt
tan tdt
ln tdt
R x2
19. Find F 0 (x) for F (x) = 0 cos tdt two di¤erent ways. The …rst method,
you evaluate the integral …rst, then compute the derivative of the result.
The second method, you will use the Fundamental Theorem of Calculus.
20. Find the area of the region between the x-axis and the graph of y = x2
between x = 1 and x = 4.
4.4. THE FUNDAMENTAL THEOREM OF CALCULUS
4.4.6
Answers
Evaluate the integrals below (problems 1-14)
1.
2.
3.
4.
5.
6.
7.
8.
9.
R4
x3 dx = 64
0
R
cos xdx = 1
2
0
R1
ex dx = e
0
R
sin xdx = 2
0
R4p
1
R2
1
R4
0
R
xdx =
14
3
x3 + 3x2
0
R0
1
4
0
(2x
1 dx = 10
ex ) dx =
1
e
2
p
104
x ( x + 1) dx =
5
3 sec2 tdt = 3
R9 1
dx = ln 3
1 2x
R2
2
11.
(3x + 1) dx = 52
2
10.
12.
13.
14.
R e x2 + x + 1
1
dx = e + e2
1
x
2
R
4
0
R2
0
1
2
1 + cos2 t
1
dt =
+1
cos2 t
4
4x dx =
15
2 ln 2
15. Find F 0 (x) for F (x) =
Rxp
1
1 + 2tdt
F 0 (x) =
16. Find F 0 (x) for F (x) =
17. Find G0 (x) for G (x) =
R2
x
1 + 2x
t2 sin tdt
F 0 (x) =
R x3
1
p
x2 sin x
tan tdt
G0 (x) = 3x2 tan x3
185
186
CHAPTER 4. INTEGRALS
18. Find G0 (x) for G (x) =
R x2
x
ln tdt
G0 (x) = ln x (4x
1)
R x2
19. Find F 0 (x) for F (x) = 0 cos tdt two di¤erent ways. The …rst method,
you evaluate the integral …rst, then compute the derivative of the result.
The second method, you will use the Fundamental Theorem of Calculus.
(a) Method 1: F 0 (x) = 2x cos x2
(b) Method 2: F 0 (x) = 2x cos x2
20. Find the area of the region between the x-axis and the graph of y = x2
between x = 1 and x = 4.
Area = 21