Suppose we are given a function and would like to determine the area underneath its graph over an interval. We could guess, but how could we figure out the exact area? Below, using a few clever ideas, we actually define such an area and show that by using what is called the definite integral we can indeed determine the exact area underneath a curve.
Definition of the Definite Integral[edit]
Figure 1: Approximation of the area under the curve

from

to

.
Figure 2: Rectangle approximating the area under the curve from

to

with sample point

.
The rough idea of defining the area under the graph of
is to approximate this area with a finite number of rectangles. Since we can easily work out the area of the rectangles, we get an estimate of the area under the graph. If we use a larger number of smaller-sized rectangles we expect greater accuracy with respect to the area under the curve and hence a better approximation. Somehow, it seems that we could use our old friend from differentiation, the limit, and "approach" an infinite number of rectangles to get the exact area. Let's look at such an idea more closely.
Suppose we have a function
that is positive on the interval
and we want to find the area
under
between
and
. Let's pick an integer
and divide the interval into
subintervals of equal width (see Figure 1). As the interval
has width
, each subinterval has width
. We denote the endpoints of the subintervals by
. This gives us

Figure 3: Riemann sums with an increasing number of subdivisions yielding better approximations.
Now for each
pick a sample point
in the interval
and consider the rectangle of height
and width
(see Figure 2). The area of this rectangle is
. By adding up the area of all the rectangles for
we get that the area
is approximated by

A more convenient way to write this is with summation notation:

For each number
we get a different approximation. As
gets larger the width of the rectangles gets smaller which yields a better approximation (see Figure 3). In the limit of
as
tends to infinity we get the area
.
It is a fact that if
is continuous on
then this limit always exists and does not depend on the choice of the points
. For instance they may be evenly spaced, or distributed ambiguously throughout the interval. The proof of this is technical and is beyond the scope of this section.
Notation
When considering the expression,

(read "the integral from

to

of the

of

"), the function

is called the
integrand and the interval
![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
is the interval of integration. Also

is called the
lower limit and

the
upper limit of integration.
Figure 4: The integral gives the signed area under the graph.
One important feature of this definition is that we also allow functions which take negative values. If
for all
then
so
. So the definite integral of
will be strictly negative. More generally if
takes on both positive and negative values then
will be the area under the positive part of the graph of
minus the area above the graph of the negative part of the graph (see Figure 4). For this reason we say that
is the signed area under the graph.
Independence of Variable[edit]
It is important to notice that the variable
did not play an important role in the definition of the integral. In fact we can replace it with any other letter, so the following are all equal:

Each of these is the signed area under the graph of
between
and
. Such a variable is often referred to as a dummy variable or a bound variable.
Left and Right Handed Riemann Sums[edit]
Figure 5: Right-handed Riemann sum
Figure 6: Left-handed Riemann sum
The following methods are sometimes referred to as L-RAM and R-RAM, RAM standing for "Rectangular Approximation Method."
We could have decided to choose all our sample points
to be on the right hand side of the interval
(see Figure 5). Then
for all
and the approximation that we called
for the area becomes

This is called the right-handed Riemann sum, and the integral is the limit

Alternatively we could have taken each sample point on the left hand side of the interval. In this case
(see Figure 6) and the approximation becomes

Then the integral of
is

The key point is that, as long as
is continuous, these two definitions give the same answer for the integral.
Examples[edit]
Example 1
In this example we will calculate the area under the curve given by the graph of
for
between 0 and 1. First we fix an integer
and divide the interval
into
subintervals of equal width. So each subinterval has width

To calculate the integral we will use the right-handed Riemann sum. (We could have used the left-handed sum instead, and this would give the same answer in the end). For the right-handed sum the sample points are

Notice that
. Putting this into the formula for the approximation,

Now we use the formula

to get

To calculate the integral of
between
and
we take the limit as
tends to infinity,

Example 2
Next we show how to find the integral of the function
between
and
. This time the interval
has width
so

Once again we will use the right-handed Riemann sum. So the sample points we choose are

Thus
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We have to calculate each piece on the right hand side of this equation. For the first two,


For the third sum we have to use a formula

to get

Putting this together

Taking the limit as
tend to infinity gives
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Exercises[edit]
Solutions
Basic Properties of the Integral[edit]
From the definition of the integral we can deduce some basic properties. For all the following rules, suppose that
and
are continuous on
.
The Constant Rule[edit]
Constant Rule

When
is positive, the height of the function
at a point
is
times the height of the function
. So the area under
between
and
is
times the area under
. We can also give a proof using the definition of the integral, using the constant rule for limits,

Example
We saw in the previous section that

Using the constant rule we can use this to calculate that
,
.
Example
We saw in the previous section that

We can use this and the constant rule to calculate that

There is a special case of this rule used for integrating constants:
Integrating Constants
If

is constant then

When
and
this integral is the area of a rectangle of height
and width
which equals
.
Example



The addition and subtraction rule[edit]
Addition and Subtraction Rules of Integration

As with the constant rule, the addition rule follows from the addition rule for limits:
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The subtraction rule can be proved in a similar way.
Example
From above
and
so


Example

Exercise[edit]
3. Use the subtraction rule to find the area between the graphs of

and

between

and



Solution
The Comparison Rule[edit]
Figure 7: Bounding the area under

on
![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
If
then each of the rectangles in the Riemann sum to calculate the integral of
will be above the
axis, so the area will be non-negative. If
then
and by the first property we get the second property. Finally if
then the area under the graph of
will be greater than the area of rectangle with height
and less than the area of the rectangle with height
(see Figure 7). So

Linearity with respect to endpoints[edit]
Additivity with respect to endpoints
Suppose
. Then

Again suppose that
is positive. Then this property should be interpreted as saying that the area under the graph of
between
and
is the area between
and
plus the area between
and
(see Figure 8).
Figure 8: Illustration of the property of additivity with respect to endpoints
Exercise[edit]
Solution
Even and odd functions[edit]
Recall that a function
is called odd if it satisfies
and is called even if
.
Suppose
is an odd function and consider first just the integral from
to
. We make the substitution
so
. Notice that if
then
and if
then
. Hence
.
Now as
is odd,
so the integral becomes
.
Now we can replace the dummy variable
with any other variable. So we can replace it with the letter
to give
.
Now we split the integral into two pieces
.
The proof of the formula for even functions is similar.
5. Prove that if

is a continuous even function then for any

,
.