Mathematics Learning Centre

Introduction to Integration
Part 2: The Denite Integral
Mary Barnes
c 1999 University of Sydney
Contents
1 Introduction 1
1.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Finding Areas 2
3 Areas Under Curves 4
3.1 What is the point of all this? . . . . . . . . . . . . . . . . . . . . . . . . . 6
3.2 Note about summation notation . . . . . . . . . . . . . . . . . . . . . . . . 6
4 The Denition of the Denite Integral 7
4.1 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
5 The Fundamental Theorem of the Calculus 8
6 Properties of the Denite Integral 12
7 Some Common Misunderstandings 14
7.1 Arbitrary constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
7.2 Dummy variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
8 Another Look at Areas 15
9 The Area Between Two Curves 19
10 Other Applications of the Denite Integral 21
11 Solutions to Exercises 23
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1 Introduction
This unit deals with the denite integral. It explains how it is dened, how it is calculated
and some of the ways in which it is used.
We shall assume that you are already familiar with the process of nding indenite inte-
grals or primitive functions (sometimes called anti-dierentiation) and are able to anti-
dierentiate a range of elementary functions. If you are not, you should work through
Introduction to Integration Part I: Anti-Dierentiation, and make sure you have
mastered the ideas in it before you begin work on this unit.
1.1 Objectives
By the time you have worked through this unit you should:
Be familiar with the denition of the denite integral as the limit of a sum;
Understand the rule for calculating denite integrals;
Know the statement of the Fundamental Theorem of the Calculus and understand
what it means;
Be able to use denite integrals to nd areas such as the area between a curve and
the x-axis and the area between two curves;
Understand that denite integrals can also be used in other situations where the
quantity required can be expressed as the limit of a sum.
A
B C D
A
B D C
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2 Finding Areas
Areas of plane (i.e. at!) gures
are fairly easy to calculate if they are
bounded by straight lines.
The area of a rectangle is clearly the
length times the breadth.
The area of a right-angled triangle can
be seen to be half the area of a rectangle
(see the diagram) and so is half the base
times the height.
Area of rectangle Area of triangle
=length breadth =
1
2
area of rectangle
=
1
2
length breadth
The areas of other triangles can be found
by expressing them as the sum or the dif-
ference of the areas of right angled trian-
gles, and from this it is clear that for any
triangle this area is half the base times
the height.
Area of ABC Area of ABC
= area of ABD = area of ABD
+ area of ACD area of ACD
Using this, we can nd the area of any
gure bounded by straight lines, by di-
viding it up into triangles (as shown).
Areas bounded by curved lines are a much more dicult problem, however. In fact,
although we all feel we know intuitively what we mean by the area of a curvilinear gure,
it is actually quite dicult to dene precisely. The area of a gure is quantied by asking
how many units of area would be needed to cover it? We need to have some unit of
area in mind (e.g. one square centimetre or one square millimetre) and imagine trying to
cover the gure with little square tiles. We can also imagine cutting these tiles in halves,
quarters etc. In this way a rectangle, and hence any gure bounded by straight lines, can
be dealt with, but a curvilinear gure can never be covered exactly.
We are therefore forced to rely on the notion of limit in order to dene areas of curvilinear
gures.
To do this, we make some simple assumptions which most people will accept as intuitively
obvious. These are:-
1. If one gure is a subset of a second gure, then the area of the rst will be less than
or equal to that of the second.
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2. If a gure is divided up into non-overlapping pieces, the area of the whole will be the
sum of the areas of the pieces.
Using these assumptions, we can approximate to curved gures by means of polygons
(gures with straight line boundaries), and hence dene the area of the curved gure as
the limit of the areas of the polygons as they approach the curved gure (in some sense
yet to be made precise).
x
y
a b
Area
y = f (x)
x
subdivisions
y
a b
y = f (x)
x
lower sum
y
a b
y
a b x
upper sum
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3 Areas Under Curves
Let us suppose that we are given a positive function
f(x) and we want to nd the area enclosed between
the curve y = f(x), the x-axis and the lines x = a
and x = b. (The shaded area in the diagram.)
If the graph of y = f(x) is not a straight line we do
not, at the moment, know how to calculate the area
precisely.
We can, however, approximate to the area as follows:
First we divide the area up into strips as shown, by
dividing the interval from a to b into equal subinter-
vals, and drawing vertical lines at these points.
Next we choose the least value of f(x) in each subin-
terval and construct a rectangle with that as its height
(as in the diagram). The sum of the areas of these
rectangles is clearly less than the area we are trying
to nd. This sum is called a lower sum.
Then we choose the greatest value of f(x) in each
subinterval and construct a rectangle with that as its
height (as in the diagram opposite). The sum of the
areas of these rectangles is clearly greater than the
area we are trying to nd. This sum is called an upper
sum.
Thus we have sandwiched the area we want to nd in between an upper sum and a lower
sum. Both the upper sum and the lower sum are easily calculated because they are sums
of areas of rectangles.
y
x
division into narrower strips
f(x')
a b x
y
x
y = f (x)
x
f(x*)
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Although we still cant say precisely what the area under the curve is, we know between
what limits it lies.
If we now increase the number of strips the area is
divided into, we will get new upper and lower sums,
which will be closer to one another in size and so closer
to the area which we are trying to nd. In fact, the
larger the number of strips we take, the smaller will
be the dierence between the upper and lower sums,
and so the better approximation either sum will be to
the area under the curve.
It can be shown that if f(x) is a nice function (for example, a continuous function) the
dierence between the upper and lower sums approaches zero as the number of strips the
area is subdivided into approaches innity.
We can thus dene the area under the curve to be:
the limit of either the upper sum or the lower sum, as the number of subdivi-
sions tends to innity (and the width of each subdivision tends to zero).
Thus nding the area under a curve boils down to nding the limit of a sum.
Now let us introduce some notation so that we can talk more precisely about these con-
cepts.
Let us suppose that the interval [a, b] is divided into
n equal subintervals each of width x. Suppose also
that the greatest value of f(x) in the ith subinterval
is f(x

i
) and the least value is f(x

i
).
Then the upper sum can be written as:
f(x

1
)x + f(x

2
)x + ... + f(x

n
)x
or, using summation notation:

n
i=1
f(x

i
)x.
Similarly, the lower sum can be written as:
f(x

1
)x + f(x

2
)x + ... + f(x

n
)x
or, using summation notation:

n
i=1
f(x

i
)x.
With this notation, and letting A stand for the area under the curve y = f(x) from x = a
to x = b, we can express our earlier conclusions in symbolic form.
The area lies between the lower sum and the upper sum and can be written as follows:
n

i=1
f(x

i
)x A
n

i=1
f(x

i
)x.
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The area is equal to the limit of the lower sum or the upper sum as the number of
subdivisions tends to innity and can be written as follows:
A = lim
n
n

i=1
f(x

i
)x
or
A = lim
n
n

i=1
f(x

i
)x.
3.1 What is the point of all this?
Well, rstly it enables us to dene precisely what up till now has only been an impre-
cise intuitive concept, namely, the area of a region with curved lines forming part of its
boundary.
Secondly it indicates how we may calculate approximations to such an area. By taking
a fairly large value of n and nding upper or lower sums we get an approximate value
for the area. The dierence between the upper and lower sums tells us how accurate
this approximation is. This, unfortunately, is not a very good or very practical way of
approximating to the area under a curve. If you do a course in Numerical Methods you
will learn much better ways, such as the Trapezoidal Rule and Simpsons Rule.
Thirdly it enables us to calculate areas precisely, provided we know how to nd nite
sums and evaluate limits. This however can be dicult and tedious, so we need to look
for better ways of nding areas. This will be done in Section 5.
At this stage, many books ask students to do exercises calculating upper and lower sums
and using these to estimate areas. Frequently students are also asked to nd the limits
of these sums as the number of subdivisions approaches innity, and so nd exact areas.
We shall not ask you to do this, as it involves a great deal of computation.
3.2 Note about summation notation
The symbol

(pronounced sigma) is the capital letter S in the Greek alphabet, and
stands for sum.
The expression

4
i=1
f(i) is read the sum of f(i) from i = 1 to i = 4, or sigma from
i = 1 to 4 of f(i).
In other words, we substitute 1, 2, 3 and 4 in turn for i and add the resulting expressions.
Thus,

4
i=1
x
i
stands for x
1
+ x
2
+ x
3
+ x
4
,

5
i=1
i
2
stands for 1
2
+ 2
2
+ 3
2
+ 4
2
+ 5
2
,
and

2
i=1
f(x
1
)x stands for f(x
1
)x + f(x
2
)x.
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4 The Denition of the Denite Integral
The discussion in the previous section led to an expression of the form
A = lim
n
n

i=1
f(x
i
)x (1)
where the interval [a, b] has been divided up into n equal subintervals each of width x
and where x
i
is a point in the ith subinterval. This is a very clumsy expression, and
mathematicians have developed a simpler notation for such expressions. We denote them
by
_
b
a
f(x)dx
which is read as the integral from a to b of f(x)dx.
The
_
sign is an elongated s and stands for sum, just as the

did previously. The
dierence is that in this case it means the limit of a sum rather than a nite sum. The
dx comes from the x as we pass to the limit, just as happened in the denition of
dy
dx
.
Thus the denite integral is dened as the limit of a particular type of sum i.e. sums like
that given in (1) above, as the width of each subinterval approaches zero and the number
of subintervals approaches innity.
4.1 Notes
1. Although we used the area under a curve as the motivation for making this denition,
the denite integral is not dened to be the area under a curve but simply the limit
of the sum (1).
2. Initially, when discussing areas under curves, we introduced the restriction that f(x)
had to be a positive function. This restriction is not necessary for the denition of a
denite integral.
3. The denition can be made more general, by removing the requirement that all the
subintervals have to be of equal widths, but we shall not bother with such generali-
sations here.
4. Sums such as (1) are called Riemann sums after the mathematician Georg Riemann
who rst gave a rigorous denition of the denite integral.
5. The denition of a denite integral requires that f(x) should be dened everywhere
in the interval [a, b] and that the limit of the Riemann sums should exist. This will
always be the case if f is a continuous function.
c x
A(x)
y = f (t)
t
y
t
y
c x
P
Q
S R
x + x
f(x*)
T Q
U P
S R
x
f(x')
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5 The Fundamental Theorem of the Calculus
So far, we have dened denite integrals but have not given any practical way of calcu-
lating them. Nor have we shown any connection between denite integrals and dieren-
tiation.
Let us consider the special case where f(t) is a con-
tinuous positive function, and let us consider the area
under the curve y = f(t) from some xed point t = c
up to the variable point t = x. For dierent values of
x we will get dierent areas. This means that the area
is a function of x. Let us denote the area by A(x).
Clearly, A(x) increases as x increases. Let us try to
nd the rate at which it increases, that is, the deriva-
tive of A(x) with respect to x.
At this point, recall how we nd derivatives from rst
principles:
Given a function f(x), we let x change by an amount
x, so that f(x) changes to f(x+x). The derivative
of f(x) is the limit of
f(x + x) f(x)
x
as x 0.
We shall go through this process with A(x) in place
of f(x).
When we increase x by x, A(x) increases by the area
of the gure PQRS. That is, (see the diagram)
A(x + x) A(x) = area PQRS.
Now that the area PQRS is bounded by a curved line
at the top, but it can be seen to lie in between the
areas of two rectangles:
area PURS < area PQRS < area TQRS.
Both of these rectangles have width x. Let the
height of the larger rectangle be f(x

) and the height

of the smaller rectangle f(x

). (In other words, x

and x

are the values of x at which f(x) attains its

maximum and minimum values in the interval from x
to x + x.)
Thus area PURS = f(x

)x and area TQRS =

f(x

)x.
So, f(x

)x A(x + x) A(x) f(x

)x.
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Now if we divide these inequalities all through by x, we obtain
f(x

)
A(x + x) A(x)
x
f(x

).
Finally, if we let x 0, both f(x

) and f(x

) approach f(x), and so the expression in

the middle must also approach f(x), that is, the derivative of A(x),
dA
dx
= f(x).
This result provides the link we need between dierentiation and the denite integral.
If we recall that the area under the curve y = f(t) from t = a to t = x is equal to
_
x
a
f(t)dt,
the result we have just proved can be stated as follows:
d
dx
_
x
a
f(t)dt = f(x). (2)
This is the Fundamental Theorem of the Calculus.
In words
If we dierentiate a denite integral with respect to the upper limit of inte-
gration, the result is the function we started with.
You may not actually use this result very often, but it is important because we can derive
from it the rule for calculating denite integrals:
Let us suppose that F(x) is an anti-derivative of f(x). That is, it is a function whose
derivative is f(x). If we anti-dierentiate both sides of the equation (2) we obtain
_
x
a
f(t)dt = F(x) + c.
Now we can nd the value of c by substituting x = a in this expression.
Since
_
a
a
f(t)dt is clearly equal to zero, we obtain
0 = F(a) + c, and so c = F(a).
Thus
_
x
a
f(t)dt = F(x) F(a), or, letting x = b,
_
b
a
f(t)dt = F(b) F(a).
This tells us how to evaluate a denite integral
rst, nd an anti-derivative of the function
then, substitute the upper and lower limits of integration into the result and sub-
tract.
Note A convenient short-hand notation for F(b) F(a) is [F(x)]
b
a
.
y
y = x 2x+5
3
x
1 2
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To see how this works in practice, let us look at a few examples:
i Find
_
1
0
x
2
dx.
An anti-derivative of x
2
is
1
3
x
3
, so we write
_
1
0
x
2
dx =
_
1
3
x
3
_
1
0
=
1
3
(1)
3

1
3
(0)
3
=
1
3
.
ii Find
_

0
sin tdt.
_

0
sin tdt = [cos t]

0
= cos() + cos 0
= (1) + 1
= 2.
iii Find the area enclosed between the x-axis, the curve y = x
3
2x+5 and the ordinates
x = 1 and x = 2.
In a question like this it is always a good idea to draw a rough sketch of the graph of
the function and the area you are asked to nd. (See below)
If the required area is A square units, then
A =
_
2
1
_
x
3
2x + 5
_
dx
=
_
x
4
4
x
2
+ 5x
_
2
1
= (4 4 + 10)
_
1
4
1 + 5
_
= 5
3
4
.
Exercises 5
1. a.
_
2x
3
_
4
2
b.
_
1
x
2
_
3
1
c.
_

x
_
16
9
d. [ln x]
4
2
2. a.
_
9
4
1

x
dx
b.
_
2

2
cos tdt
x
y
1 2
1
y = x +1
2
1 3
y = 1/x
x
y

2 4
u
2
v
v = (4 u)
y
0
y = 2sint
2
t
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c.
_
2
1
1
y
2
dy
d.
_
1
2
(s
2
+ 2s + 2)ds
3. Find the area of the shaded region in each of the diagrams below:
a. b.
c. d.
4. Evaluate
a.
_
1
0
xe
x
2
dx
b.
_
1
2
1
3 x
dx
c.
_
2
0
sin 2ydy
d.
_
5
1
t
4 + t
2
dt
a c x
y
b
b x
y
y = g(x)
y = f (x)
a
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6 Properties of the Denite Integral
Some simple properties of denite integrals can be derived from the basic denition, or
from the Fundamental Theorem of the Calculus. We shall not give formal proofs of these
here but you might like to think about them, and try to explain, to yourself or someone
else, why they are true.
a.
_
a
a
f(x)dx = 0.
If the upper and lower limits of the integral are the same, the integral is zero. This
becomes obvious if we have a positive function and can interpret the integral in terms
of the area under a curve.
b. If a b c,
_
c
a
f(x)dx =
_
b
a
f(x)dx +
_
c
b
f(x)dx.
This says that the integral of a function over the union
of two intervals is equal to the sum of the integrals over
each of the intervals. The diagram opposite helps to
make this clear if f(x) is a positive function.
c.
_
b
a
cf(x)dx = c
_
b
a
f(x)dx for any constant c.
This tells us that we can move a constant past the integral sign, but beware: we can
only do this with constants, never with variables!
d.
_
b
a
(f(x) + g(x))dx =
_
b
a
f(x)dx +
_
b
a
g(x)dx.
That is, the integral of a sum is equal to the sum of the integrals.
e. If f(x) g(x) in [a, b] then
_
b
a
f(x)dx
_
b
a
g(x)dx.
That is, integration preserves inequalities between
functions. The diagram opposite explains this result
if f(x) and g(x) are positive functions.
c
a b x
y
c
b a
y = f (x)
M
m
a b x
y
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f.
_
b
a
cdx = c(b a).
This tells us that the integral of a constant is equal to
the product of the constant and the range of integra-
tion. It becomes obvious when we look at the diagram
with c > 0, since the area represented by the integral
is just a rectangle of height c and width b a.
g. We can combine (e) and (f) to give the result that,
if M is any upper bound and m any lower bound for
f(x) in the interval [a, b], so that m f(x) M,
then
m(b a)
_
b
a
f(x)dx M(b a).
This, too, becomes clear when f(x) is a positive func-
tion and we can interpret the integral as the area un-
der the curve.
h. Finally we extend the denition of the denite integral slightly, to remove the restric-
tion that the lower limit of the integral must be a smaller number than the upper
limit. We do this by specifying that
_
a
b
f(x)dx =
_
b
a
f(x)dx.
For example,
_
1
2
f(x)dx =
_
2
1
f(x)dx.
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7 Some Common Misunderstandings
7.1 Arbitrary constants
When you rst learned how to nd indenite integrals (anti-derivatives), you probably
also learned that it was important to remember always to add an arbitrary constant to
the answer.
There is no arbitrary constant in a denite integral.
If we interpret a denite integral as an area, it is clear that its value is a xed number
(the number of units of area in the region). There is no ambiguity, and so no need to add
an arbitrary constant - in fact, it is wrong to do so.
When we apply the Fundamental Theorem of the Calculus to nding a denite integral,
however, the possibility of an arbitrary constant appears to arise.
For example, in calculating
_
2
1
x
2
dx, we have to nd an anti-derivative for x
2
. The most
natural choice would be
1
3
x
3
, but instead of that we could choose
1
3
x
3
+ c, where c is any
constant.
Then,
_
1
0
x
2
dx =
_
1
3
x
3
+ c
_
1
0
=
_
1
3
(1)
3
+ c
_

_
1
3
(0)
3
+ c
_
=
1
3
.
Note that the constants cancel one another out, and we get the same answer as we
did before. Thus we might as well take the simplest course, and forget about arbitrary
constants when we are calculating denite integrals.
7.2 Dummy variables
What is the dierence between
_
b
a
f(x)dx and
_
b
a
f(t)dt?
Lets work them both out in a special case.
_
4
2
1
x
dx = [ln x]
4
2
= ln 4 ln 2.
_
4
2
1
t
dt = [ln t]
4
2
= ln 4 ln 2.
So both integrals give the same answer.
It is clear that the value of a denite integral depends on the function and the limits of
integration but not on the actual variable used. In the process of evaluating the integral,
we substitute the upper and lower limits for the variable and so the variable doesnt
appear in the answer. For this reason we call the variable in a denite integral a dummy
variable - we can replace it with any other variable without changing a thing.
Thus,
_
b
a
f(x)dx =
_
b
a
f(y)dy =
_
b
a
f(t)dt =
_
b
a
f()d.
0
x
y
x
f(x)
y = f (x)
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8 Another Look at Areas
We have dened the denite integral
_
b
a
f(x)dx as the limit of a particular type of sum,
without placing any restrictions on whether the function f(x) is positive or negative.
We know that, if f(x) is positive,
_
b
a
f(x)dx is equal
to the area between the curve y = f(x), the x-axis
and the ordinates x = a and x = b, (which we refer to
as the area under the curve). The natural question
to ask now is: what does
_
b
a
f(x)dx equal if f(x) is
negative? Can we represent it as an area in this case
too; perhaps the area above the curve ?
If we go back to the denition of
_
b
a
f(x)dx as the limit
of a sum, we can see clearly that if f(x) is always
negative then each of the terms f(x
i
)x will also be
negative (since x is positive).
So the sum
n

i=1
f(x
i
)x will be a sum of negative terms and so will be negative too. And
when we let n approach innity and pass to the limit, that will be negative also.
Thus, if f(x) is negative for x between a and b,
_
b
a
f(x)dx will also be negative.
Now areas are, by denition, positive. Remember that, in section 1, we explained that we
can measure the area of a region by counting the number of little square tiles (each of unit
area) needed to cover it. Since we cant cover a region with a negative number of tiles
(it doesnt make sense to talk of it) we cant have a negative area. On the other hand, if
we ignore the fact that each of the terms f(x)x is negative, and consider its numerical
value only, we can see that it is numerically equal to the area of the rectangle shown.
And, if we go through the usual process, adding up the areas of all the little rectangles
and taking the limit, we nd that
_
b
a
f(x)dx is numerically equal to the area between
the curve and the x-axis.
So to nd the area, we calculate
_
b
a
f(x)dx, which will turn out to be negative, and then
take its numerical (i.e. absolute) value.
x
0 1 1
Mathematics Learning Centre, University of Sydney 16
To see this more clearly, lets look at an example. Consider the curve, y = x(x
2
1). This
is a cubic curve, and cuts the x-axis at 1, 0 and 1. A sketch of the curve is shown below.
Let us nd the shaded area. First we calculate the denite integral
_
1
0
x(x
2
1)dx.
_
1
0
x(x
2
1)dx =
_
1
0
(x
3
x)dx
=
_
1
4
x
4

1
2
x
2
_
1
0
=
_
1
4

1
2
_
(0 0)
=
1
4
.
Since x(x
2
1) is negative when x lies between 0 and 1, the denite integral is also
negative, as expected. We can conclude that the area required is
1
4
square units.
As a check, let us nd the area of the other loop of the curve, i.e. the area between the
curve and the x-axis from 1 to 0. Since x(x
2
1) is positive for this range of values of
x, the area will be given by
_
0
1
x(x
2
1)dx =
_
1
4
x
4

1
2
x
2
_
0
1
= (0 0)
_
1
4

1
2
_
=
1
4
.
This is the answer we would expect, since a glance at the diagram shows that the curve
has point symmetry about the origin. If we were to rotate the whole graph through
180

, the part of the curve to the left of the origin would t exactly on top of the part
to the right of the origin, and the unshaded loop would t on top of the shaded loop. So
the areas of the two loops are the same.
Now let us calculate
_
1
1
x(x
2
1)dx.
_
1
1
x(x
2
1)dx =
_
1
4
x
4

1
2
x
2
_
1
1
=
_
1
4

1
2
_

_
1
4

1
2
_
= 0.
This makes it very clear that
a denite integral does not always represent the area under a curve.
x
y
0 1 1 2
A
B
Mathematics Learning Centre, University of Sydney 17
We have found that
1. If f(x) is positive between a and b, then
_
b
a
f(x)dx does represent the area under
the curve.
2. If f(x) is negative between a and b, then

_
b
a
f(x)dx

represents the area above the

curve, since the value of
_
b
a
f(x)dx is negative.
3. If f(x) is sometimes positive and sometimes negative between a and b, then
_
b
a
f(x)dx measures the dierence in area between the part above the x-axis and
the part below the x-axis. (In the example above, the two areas were equal, and so
the dierence came out to be zero.)
Lets look at another example.
Consider the function y = (x + 1)(x 1)(x 2) = x
3
2x
2
x + 2.
This is a cubic function, and the graph crosses the x-axis at 1, 1 and 2. A sketch of the
graph is shown.
The area marked A is given by
_
1
1
(x
3
2x
2
x + 2)dx =
_
1
4
x
4

2
3
x
3

1
2
x
2
+ 2x
_
1
1
=
_
1
4

2
3

1
2
+ 2
_

_
1
4
+
2
3

1
2
2
_
=
4
3
+ 4 = 2
2
3
.
So the area of A is 2
2
3
square units.
The area marked B can be found by evaluating
_
2
1
_
x
3
2x
2
x + 2
_
dx.
This works out as
5
12
. (The details of the calculation are left to you.)
So the area of B is
5
12
square units.
y
x a c
b
Mathematics Learning Centre, University of Sydney 18
If we calculate
_
2
1
(x
3
2x
2
x + 2)dx the answer will be the dierence between the area
of A and the area of B, that is, 2
1
4
square units. (Check it out for yourself.)
If we want the total area enclosed between the curve and the x-axis we must add the
area of A and the area of B.
i.e. 2
2
3
+
5
12
= 3
1
12
square units.
WARNING In working out area problems you should always sketch the curve rst. If
the function is sometimes positive and sometimes negative in the range you are interested
in, it may be necessary to divide the area into two or more parts, as shown below.
The area between the curve and the x-axis from a to b is NOT equal to
_
b
a
f(x)dx.
Instead, it is
_
c
a
f(x)dx +|
_
b
c
f(x)dx|.
Before you can calculate this, you must nd the value of c, i.e. nd the point where the
curve y = f(x) crosses the x-axis.
Exercises 8
1. Find the area enclosed by the graph of y = 3x
2
(x 4) and the x-axis.
2. i Find the value of
_
2
0
sin xdx.
ii Find the area enclosed between the graph of y = sin x and the x-axis from x = 0
to x = 2.
3. Find the total area enclosed between the graph of y = 12x(x + 1)(2 x) and the
x-axis.
x
y
y = f (x)
y = g(x)
x
y
y = f (x)
y = g(x)
f(x' ) g(x*)
i i
x
y
(2,2)
(2,3)
(4,2)
(4,3)
Mathematics Learning Centre, University of Sydney 19
9 The Area Between Two Curves
Sometimes we want to nd, not the area between a
curve and the x-axis, but the area enclosed between
two curves, say between y = f(x) and y = g(x).
We can approach this problem in the same way as
before by dividing the area up into strips and approx-
imating the area of each strip by a rectangle. The
lower sum is found by calculating the area of the in-
terior rectangles as shown in the diagram.
The height of each interior rectangle is equal to the
dierence between the least value of f(x), f(x

), and
the greatest value of g(x), g(x

), in the rectangle. The

area of the ith rectangle is (f(x

i
) g(x

i
))x.
The lower sum =
n

i=1
(f(x

i
) g(x

i
))x.
The upper sum can be found in the same way. The
area enclosed between the curves is sandwiched be-
tween the lower sum and the upper sum.
When we pass to the limit as x 0, we get
Area enclosed between the curves =
_
b
a
(f(x) g(x))dx.
Note that the height is always f(x) g(x), even when
one or both of the curves lie below the x-axis.
For example, if for some value of x, f(x) = 2 and
g(x) = 3, the distance between the curves is f(x)
g(x) = 2 (3) = 5, or, if f(x) = 2 and g(x) = 3,
the distance between the curves is (2) (3) = 1
(see the diagram).
So, to nd the area enclosed between two curves, we must:
1. Find where the curves intersect.
2. Find which is the upper curve in the region we are interested in.
Mathematics Learning Centre, University of Sydney 20
3. Integrate the function (upper curve lower curve) between the appropriate limits.
In other words, if two curves f(x) and g(x) intersect at x = a and x = b, and f(x) g(x)
for a x b, then
Area enclosed between the curves =
_
b
a
(f(x) g(x))dx.
Exercises 9
(Remember to draw a diagram rst, before beginning any problem.)
1. Find the area enclosed between the parabola y = x(x 2) and the line y = x + 2.
2. Find the area enclosed between the two parabolas y = x
2
4x + 2 and y = 2 x
2
.
3. Check that the curves y = sin x and y = cos x intersect at

4
and
5
4
, and nd the
area enclosed by the curves between these two point.
4. i Sketch the graphs of the function y = 6 x x
2
and y = x
3
7x + 6.
ii Find the points of intersection of the curves.
iii Find the total area enclosed between them.
|
|
x
Area
A(x)
P
i1
P
i
P
n
l
i
P
0
Mathematics Learning Centre, University of Sydney 21
10 Other Applications of the Denite Integral
The problem with which we introduced the idea of the denite integral was that of nding
the area under a curve. As a result, most people tend to think of denite integrals always
in terms of area. But it is important to remember that the denite integral is actually
dened as the limit of a sum:
lim
n
n

i=1
f(x
i
)x
and that any other problem which can be approximated by a similar sum will give rise to
a denite intregral when we take the limit.
Examples
1. Volume of a solid
If we want to nd the volume of a solid, we
can imagine it being put through a bread
slicer, and cut into slices of thickness x.
If A(x) is the cross sectional area at distance
x along the x-axis, the volume of the slice
will be approximately A(x)x, and the total
volume of the solid will be approximately
n

i=1
A(x
i
)x.
When we pass to the limit as x 0 and
n , this becomes the denite integral
_
b
a
A(x)dx.
2. Length of a curve
We can approximate to the length of a curve
by dividing it up into segments, as shown,
and approximating the length of each seg-
ment by replacing the curved line with a
straight line joining the end points. If the
length of the ith straight line segment is l
i
,
the total length of the curve will be approx-
imately
n

i=1
l
i
.
If we take the limit of this sum as the length of each segment approaches zero and the
number of segments approaches innity, we again get a denite integral. The details
are rather complicated and are not given here.
Mathematics Learning Centre, University of Sydney 22
3. Mass of a body of varying density
Suppose we have a bar, rope or chain whose linear density (mass per unit length)
varies. Let the density at distance x along the x-axis be d(x). If we subdivide the
object into small sections of length x, the total mass can be approximated by the
sum
n

i=1
d(x
i
)x.
When we take the limit as n , we obtain the denite integral
_
b
a
d(x)dx.
4. Work done by a variable force
In mechanics, the work done by a constant force is dened to be the product of the
magnitude of the force and the distance moved in the direction of the force. If the
force F(x) is varying, we can approximate to the work by dividing up the distance into
small subintervals. If these are small enough, we can regard the force as eectively
constant throughout each interval and so the work done in moving through distance
x is approximately F(x)x.
The total work is thus approximately

n
i=1
F(x
i
)x and when we take the limit as
n , we nd that the work done in moving the force from x = a to x = b is
_
b
a
F(x)dx.
Many other examples could be given, but these four should be sucient to illustrate the
wide variety of applications of the denite integral.
Mathematics Learning Centre, University of Sydney 23
11 Solutions to Exercises
Exercises 5
1. a. 2(4
3
) 2(2
3
) = 112
b.
1
9

1
1
=
8
9
c.

16

9 = 1
d. ln 4 ln 2 = ln
4
2
= ln 2
2. a.
_
9
4
x

1
2
dx =
_
2x
1
2
_
9
4
= 2

9 2

4 = 2
b.
_
2

2
cos tdt = [sin t]

2
= sin

2
sin
_

2
_
= 1 (1) = 2
c.
_
2
1
y
2
dy =
_
y
1
_
2
1
=
1
2

_

1
1
_
=
1
2
d.
_
1
2
(s
2
+ 2s + 2)ds =
_
1
3
s
3
+ s
2
+ 2s
_
1
2
=
_

1
3
+ 1 2
_

8
3
+ 4 4
_
= 1
1
3
3. a. Area =
_
2
1
(x
2
+ 1)dx =
_
1
3
x
3
+ x
_
2
1
=
_
8
3
+ 2
_

_
1
3
+ 1
_
= 3
1
3
b. Area =
_
3
1
1
x
dx = [ln x]
3
1
= ln 3 ln 1 = ln 3
c. Area =
_
4
0
_
(4 u)du =
_
4
0
(4 u)
1
2
(1)du =
_
2
3
(4 u)
3
2
_
4
0
= 5
1
3
d. Area =
_

0
2 sin tdt = [2 cos t]

0
= 2 cos + 2 cos 0 = 4
4. a.
_
1
0
xe
x
2
dx =
1
2
_
1
0
e
x
2
2xdx =
1
2
_
e
x
2
_
1
0
=
1
2
(e 1)
b.
_
1
2
1
3 x
dx =
_
1
2
1
3 x
(1)dx = [ln(3 x)]
1
2
= (ln 4 ln 5) = ln 5 ln 4 = ln
5
4
c.
_
2
0
sin 2ydy =
1
2
_
2
0
sin 2y 2dy =
1
2
[cos 2y]

2
0
=
1
2
(cos + cos 0) = 1
d.
_
5
1
t
4 + t
2
dt =
1
2
_
5
1
2t
4 + t
2
dt =
1
2
_
ln(4 + t
2
)
_
5
1
=
1
2
ln
29
5
y
x
0 4
x
y
0 2
x
y
1 0 2
B
A
Mathematics Learning Centre, University of Sydney 24
Exercises 8
1. First, draw a graph.
The area is below the x-axis, so we rst cal-
culate
_
4
0
3x
2
(x 4)dx.
_
4
0
3x
2
(x 4)dx =
_
4
0
(3x
3
12x
2
)dx
=
_
3
4
x
4
4x
3
_
4
0
= 64.
The required area is therefore 64 units.
2. i
_
2
0
sin xdx = [cos x]
2
0
= cos 2 + cos 0 = 1 + 1 = 0
ii
From the graph we see that the area
Area =
_

0
sin xdx +|
_
2

sin xdx|
= [cos x]

0
+| [cos x]
2

|
= (cos + cos 0) +| cos 2 + cos |
= ((1) + 1) +| 1 + (1)|
= 4.
3. The graph of the curve cuts the x-axis at 1,
0 and 2.
The total area = area A + area B.
Area A = |
_
0
1
12x(x + 1)(2 x)dx|
= |
_
0
1
(12x
3
+ 12x
2
+ 24x)dx|
= |
_
3
4
+ 4x
3
+ 12x
_
0
1
|
= |0 (3 4 + 12)|
= | 5| = 5.
Area B =
_
2
0
12x(x + 1)(2 x)dx
=
_
3x
4
+ 4x
3
+ 12x
2
_
2
0
= (48 + 32 + 48) = 32.
x
y
(1,3)
(2,0)
(2,2)
(0,2)
y = 2x
2
y
x
x
y
/4
5/4
Mathematics Learning Centre, University of Sydney 25
Therefore the total area is 5 + 32 = 37 square units.
Exercises 9
1. The curves y = x
2
2x and y = x + 2
intersect where x
2
2x = x + 2. i.e. at
x = 1 or x = 2.
The upper curve is y = x + 2.
Area =
_
2
1
((x + 2) (x
2
2x))dx
=
_
2
1
(2 + x x
2
)dx
=
_
2x +
1
2
x
2

1
3
x
3
_
2
1
= (4 + 2
8
3
) (2 +
1
2
+
1
3
)
= 4
1
2
.
2. The curves intersect where x
2
4x +2 = 2 x
2
i.e. 2x
2
4x = 0 i.e. x = 0 or x = 2.
The upper curve is y = 2 x
2
(see sketch).
Area =
_
2
0
((2 x
2
) (x
2
4x + 2))dx
=
_
2
0
(4x 2x
2
)dx
=
_
2x
2
+
2
3
x
3
_
2
0
= (8
2
3
8) 0
= 2
2
3
.
3. When x =

4
, sin x =
1

2
and cos x =
1

2
.
When x =
5
4
, sin x =
1

2
and cos x =
1

2
.
So the curves y = sin x and y = cos x inter-
sect at

4
and
5
4
.
Area =
_ 5
4

4
(sin x cos x)dx
= [cos x sin x]
5
4

4
x
y
0 3 2
A
B
Mathematics Learning Centre, University of Sydney 26
= (cos
5
4
sin
5
4
) + (cos

4
+ sin

4
)
=
4

2
= 2

2.
4. (i) and (ii) The curves are easier to sketch if
we rst nd the points of intersection: they
meet where x
3
7x + 6 = 6 x x
2
.
That is,
x
3
+ x
2
6x = 0
or
x(x 2)(x + 3) = 0.
So the points of intersection are (0, 6); (2, 0);
and (3, 0).
The rst curve is an upside-down parabola,
and the second a cubic.
Total area = area A + area B.
Area A =
_
0
3
((x
3
7x + 6) (6 x x
2
))dx
=
_
0
3
(x
3
+ x
2
6x)dx
=
_
1
4
x
4
+
1
3
x
3
3x
2
_
0
3
= 15
3
4
.
Area B =
_
2
0
((6 x x
2
) (x
3
7x + 6))dx
=
_
2
0
(6x x
2
x
3
)dx
= 5
1
3
.
. . the total area = 15
3
4
+ 5
1
3
= 21
1
12
square units.