In this section we will formally define the definite
integral and give many of the properties of definite integrals. Let’s start off with the definition of a
definite integral.
Definite Integral
The definite integral is defined to be exactly the limit and
summation that we looked at in the last section to find the net area between a
function and the x-axis. Also note that the notation for the definite
integral is very similar to the notation for an indefinite integral. The reason for this will be apparent
eventually.
There is also a little bit of terminology that we should get
out of the way here. The number “a” that is at the bottom of the integral
sign is called the lower limit of
the integral and the number “b” at
the top of the integral sign is called the upper
limit of the integral. Also, despite
the fact that a and b were given as an interval the lower
limit does not necessarily need to be smaller than the upper limit. Collectively we’ll often call a and b the interval of
integration.
Let’s work a quick example.
This example will use many of the properties and facts from the brief
review of summation notation in the Extras
chapter.
|
Example 1 Using
the definition of the definite integral compute the following.
Solution
First, we can’t actually use the definition unless we
determine which points in each interval that well use for . In order to make our life easier we’ll use
the right endpoints of each interval.
From the previous section we know that for a general n the width of each subinterval is,
The subintervals are then,
As we can see the right endpoint of the ith subinterval is
The summation in the definition of the definite integral
is then,
Now, we are going to have to take a limit of this. That means that we are going to need to
“evaluate” this summation. In other
words, we are going to have to use the formulas given in the summation notation review to eliminate the
actual summation and get a formula for this for a general n.
To do this we will need to recognize that n is a constant as far as the
summation notation is concerned. As we
cycle through the integers from 1 to n
in the summation only i changes and
so anything that isn’t an i will be
a constant and can be factored out of the summation. In particular any n that is in the summation can be factored out if we need to.
Here is the summation “evaluation”.
We can now compute the definite integral.
We’ve seen several methods for dealing with the limit in
this problem so I’ll leave it to you to verify the results.
|
Wow, that was a lot of work for a fairly simple
function. There is a much simpler way of
evaluating these and we will get to it eventually. The main purpose to this section is to get
the main properties and facts about the definite integral out of the way. We’ll discuss how we compute these in
practice starting with the next section.
So, let’s start taking a look at some of the properties of
the definite integral.
Properties
See the Proof
of Various Integral Properties section of the Extras chapter for the proof
of properties 1 4.
Property 5 is not easy to prove and so is not shown there. Property 6 is not really a property in the full
sense of the word. It is only here to
acknowledge that as long as the function and limits are the same it doesn’t
matter what letter we use for the variable.
The answer will be the same.
Let’s do a couple of examples dealing with these properties.
|
Example 2 Use
the results from the first example to evaluate each of the following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
All of the solutions to these problems will rely on the
fact we proved in the first example.
Namely that,
(a)
In this case the only difference between the two is that
the limits have interchanged. So,
using the first property gives,
[Return to Problems]
(b)
For this part notice that we can factor a 10 out of both
terms and then out of the integral using the third property.
[Return to Problems]
(c)
In this case the only difference is the letter used and so
this is just going to use property 6.
[Return to Problems]
|
Here are a couple of examples using the other properties.
|
Example 3 Evaluate
the following definite integral.
Solution
There really isn’t anything to do with this integral once
we notice that the limits are the same.
Using the second property this is,
|
|
Example 4 Given
that and determine the value of
Solution
We will first need to use the fourth property to break up
the integral and the third property to factor out the constants.
Now notice that the limits on the first integral are
interchanged with the limits on the given integral so switch them using the
first property above (and adding a minus sign of course). Once this is done we can plug in the known
values of the integrals.
|
|
Example 5 Given
that ,
,
and determine the value of .
Solution
This example is mostly an example of property 5 although
there are a couple of uses of property 1 in the solution as well.
We need to figure out how to correctly break up the
integral using property 5 to allow us to use the given pieces of
information. First we’ll note that
there is an integral that has a “-5” in one of the limits. It’s not the lower limit, but we can use
property 1 to correct that eventually.
The other limit is 100 so this is the number c that we’ll use in property 5.
We’ll be able to get the value of the first integral, but
the second still isn’t in the list of know integrals. However, we do have second limit that has a
limit of 100 in it. The other limit
for this second integral is -10 and this will be c in this application of property 5.
At this point all that we need to do is use the property 1
on the first and third integral to get the limits to match up with the known
integrals. After that we can plug in
for the known integrals.
|
There are also some nice properties that we can use in
comparing the general size of definite integrals. Here they are.
More Properties
See the Proof
of Various Integral Properties section of the Extras chapter for the proof
of these properties.
Interpretations of
Definite Integral
There are a couple of quick interpretations of the definite
integral that we can give here.
First, as we alluded to in the previous section one possible
interpretation of the definite integral is to give the net area between the
graph of and the x-axis
on the interval [a,b]. So, the net area between the graph of and the x-axis
on [0,2] is,
If you look back in the last section this was the exact area
that was given for the initial set of problems that we looked at in this area.
Another interpretation is sometimes called the Net Change
Theorem. This interpretation says that
if is some quantity (so is the rate of change of ,
then,
is the net change in on the interval [a,b]. In other words, compute the definite integral
of a rate of change and you’ll get the net change in the quantity. We can see that the value of the definite
integral, ,
does in fact give use the net change in and so there really isn’t anything to prove
with this statement. This is really just
an acknowledgment of what the definite integral of a rate of change tells us.
So as a quick example, if is the volume of water in a tank then,
is the net change in the volume as we go from time to time .
Likewise, if is the function giving the position of some
object at time t we know that the
velocity of the object at any time t
is : . Therefore the displacement of the object time
to time is,
Note that in this case if is both positive and negative (i.e. the object moves to both the right
and left) in the time frame this will NOT give the total distance
traveled. It will only give the
displacement, i.e. the difference
between where the object started and where it ended up. To get the total distance traveled by an
object we’d have to compute,
It is important to note here that the Net Change Theorem
only really makes sense if we’re integrating a derivative of a function.
Fundamental Theorem
of Calculus, Part I
As noted by the title above this is only the first part to
the Fundamental Theorem of Calculus. We
will give the second part in the next section as it is the key to easily
computing definite integrals and that is the subject of the next section.
The first part of the Fundamental Theorem of Calculus tells
us how to differentiate certain types of definite integrals and it also tells
us about the very close relationship between integrals and derivatives.
Fundamental Theorem of Calculus, Part I
An alternate notation for the derivative portion of this is,
To see the proof of this see the Proof of Various Integral Properties
section of the Extras chapter.
Let’s check out a couple of quick examples using this.
|
Example 6 Differentiate
each of the following.
(a) [Solution]
(b) [Solution]
Solution
(a)
This one is nothing more than a quick application of the
Fundamental Theorem of Calculus.
[Return to Problems]
(b)
This one needs a little work before we can use the
Fundamental Theorem of Calculus. The
first thing to notice is that the FToC requires the lower limit to be a
constant and the upper limit to be the variable. So, using a property of definite integrals
we can interchange the limits of the integral we just need to remember to add
in a minus sign after we do that.
Doing this gives,
The next thing to notice is that the FToC also requires an
x in the upper limit of integration
and we’ve got x2. To do this derivative we’re going to need
the following version of the chain rule.
So, if we let u= x2
we use the chain rule to get,
The final step is to get everything back in terms of x.
[Return to Problems]
|
Using the chain rule as we did in the last part of this
example we can derive some general formulas for some more complicated problems.
First,
This is simply the chain rule for these kinds of problems.
Next, we can get a formula for integrals in which the upper
limit is a constant and the lower limit is a function of x. All we need to do here is
interchange the limits on the integral (adding in a minus sign of course) and
then using the formula above to get,
Finally, we can also get a version for both limits being
functions of x. In this case we’ll need to use Property 5
above to break up the integral as follows,
We can use pretty much any value of a when we break up the integral.
The only thing that we need to avoid is to make sure that exists.
So, assuming that exists after we break up the integral we can
then differentiate and use the two formulas above to get,
Let’s work a quick example.
|
Example 7 Differentiate
the following integral.
Solution
This will use the final formula that we derived above.
|