In this section we are going to concentrate on how we
actually evaluate definite integrals in practice. To do this we will need the Fundamental
Theorem of Calculus, Part II.
Fundamental Theorem of Calculus, Part II
To see the proof of this see the Proof of Various Integral
Properties section of the Extras chapter.
Recall that when we talk about an anti-derivative for a
function we are really talking about the indefinite integral for the
function. So, to evaluate a definite
integral the first thing that we’re going to do is evaluate the indefinite
integral for the function. This should
explain the similarity in the notations for the indefinite and definite
integrals.
Also notice that we require the function to be continuous in
the interval of integration. This was
also a requirement in the definition of the definite integral. We didn’t make a big deal about this in the
last section. In this section however,
we will need to keep this condition in mind as we do our evaluations.
Next let’s address the fact that we can use any
anti-derivative of 
in the evaluation. Let’s take a final look at the following
integral.
Both of the following are anti-derivatives of the integrand.
Using the FToC to evaluate this integral with the first
anti-derivatives gives,
Much easier than using the definition wasn’t
it? Let’s now use the second
anti-derivative to evaluate this definite integral.
The constant that we tacked onto the second anti-derivative
canceled in the evaluation step. So,
when choosing the anti-derivative to use in the evaluation process make your
life easier and don’t bother with the constant as it will only end up canceling
in the long run.
Also, note that we’re going to have to be very careful with
minus signs and parenthesis with these problems. It’s very easy to get in a hurry and mess
them up.
Let’s start our examples with the following set designed to
make a couple of quick points that are very important.
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Example 1 Evaluate
each of the following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
This is the only indefinite integral in this section and
by now we should be getting pretty good with these so we won’t spend a lot of
time on this part. This is here only
to make sure that we understand the difference between an indefinite and a
definite integral. The integral is,
[Return to Problems]
(b)
Recall from our first example above that all we really
need here is any anti-derivative of the integrand. We just computed the most general
anti-derivative in the first part so we can use that if we want to. However, recall that as we noted above any
constants we tack on will just cancel in the long run and so we’ll use the
answer from (a) without the “+c”.
Here’s the integral,
Remember that the evaluation is always done in the order
of evaluation at the upper limit minus evaluation at the lower limit. Also be very careful with minus signs and
parenthesis. It’s very easy to forget
them or mishandle them and get the wrong answer.
Notice as well that, in order to help with the evaluation,
we rewrote the indefinite integral a little.
In particular we got rid of the negative exponent on the second
term. It’s generally easier to
evaluate the term with positive exponents.
[Return to Problems]
(c)
This integral is here to make a point. Recall that in order for us to do an
integral the integrand must be continuous in the range of the limits. In this case the second term will have
division by zero at and since is in the interval of integration, i.e. it is between the lower and upper
limit, this integrand is not continuous in the interval of integration and so
we can’t do this integral.
Note that this problem will not prevent us from doing the
integral in (b) since is not in the interval of integration.
[Return to Problems]
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So what have we learned from this example?
First, in order to do a definite integral the first thing
that we need to do is the indefinite integral.
So we aren’t going to get out of doing indefinite integrals, they will
be in every integral that we’ll be doing in the rest of this course so make
sure that you’re getting good at computing them.
Second, we need to be on the lookout for functions that
aren’t continuous at any point between the limits of integration. Also, it’s important to note that this will
only be a problem if the point(s) of discontinuity occur between the limits of
integration or at the limits themselves.
If the point of discontinuity occurs outside of the limits of
integration the integral can still be evaluated.
In the following sets of examples we won’t make too much of
an issue with continuity problems, or lack of continuity problems, unless it
affects the evaluation of the integral.
Do not let this convince you that you don’t need to worry about this
idea. It arises often enough that it can
cause real problems if you aren’t on the lookout for it.
Finally, note the difference between indefinite and definite
integrals. Indefinite integrals are
functions while definite integrals are numbers.
Let’s work some more examples.
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Example 2 Evaluate
each of the following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
There isn’t a lot to this one other than simply doing the
work.
[Return to Problems]
(b)
Recall that we can’t integrate products as a product of
integrals and so we first need to multiply the integrand out before
integrating, just as we did in the indefinite integral case.
In the evaluation process recall that,
Also, don’t get excited about the fact that the lower
limit of integration is larger than the upper limit of integration. That will happen on occasion and there is
absolutely nothing wrong with this.
[Return to Problems]
(c)
First, notice that we will have a division by zero issue
at ,
but since this isn’t in the interval of integration we won’t have to worry
about it.
Next again recall that we can’t integrate quotients as a
quotient of integrals and so the first step that we’ll need to do is break up
the quotient so we can integrate the function.
Don’t get excited about answers that don’t come down to a
simple integer or fraction. Often
times they won’t. Also don’t forget
that .
[Return to Problems]
(d)
This one is actually pretty easy. Recall that we’re just integrating 1!.
[Return to Problems]
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The last set of examples dealt exclusively with integrating
powers of x. Let’s work a couple of examples that involve
other functions.
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Example 3 Evaluate
each of the following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
Solution
(a) .
This one is here mostly here to contrast with the next
example.
[Return to Problems]
(b)
Be careful with signs with this one. Recall from the indefinite integral
sections that it’s easy to mess up the signs when integrating sine and
cosine.
Compare this answer to the previous answer, especially the
evaluation at zero. It’s very easy to
get into the habit of just writing down zero when evaluating a function at
zero. This is especially a problem
when many of the functions that we integrate involve only x’s raised to positive integers; these evaluate is zero of course.
After evaluating many of these kinds of definite integrals it’s easy
to get into the habit of just writing down zero when you evaluate at
zero. However, there are many
functions out there that aren’t zero when evaluated at zero so be careful.
[Return to Problems]
(c)
Not much to do other than do the integral.
For the evaluation, recall that
and so if we can evaluate cosine at these angles we can
evaluate secant at these angles.
[Return to Problems]
(d)
In order to do this one will need to rewrite both of the
terms in the integral a little as follows,
For the first term recall we used the following fact about
exponents.
In the second term, taking the 3 out of the denominator will
just make integrating that term easier.
Now the integral.
Just leave the answer like this. It’s messy, but it’s also exact.
Note that the absolute value bars on the logarithm are
required here. Without them we
couldn’t have done the evaluation.
[Return to Problems]
(e)
This integral can’t be done. There is division by zero in the third term
at and lies in the interval of integration. The fact that the first two terms can be
integrated doesn’t matter. If even one
term in the integral can’t be integrated then the whole integral can’t be
done.
[Return to Problems]
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So, we’ve computed a fair number of definite integrals at
this point. Remember that the vast
majority of the work in computing them is first finding the indefinite
integral. Once we’ve found that the rest
is just some number crunching.
There are a couple of particularly tricky definite integrals
that we need to take a look at next.
Actually they are only tricky until you see how to do them, so don’t get
too excited about them. The first one
involves integrating a piecewise function.
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Example 4 Given,
Evaluate each of the following integrals.
(a) [Solution]
(b) [Solution]
Solution
Let’s first start with a graph of this function.
The graph reveals a problem. This function is not continuous at and we’re going to have to watch out
for that.
(a)
For this integral notice that is not in the interval of integration and so
that is something that we’ll not need to worry about in this part.
Also note the limits for the integral lie entirely in the
range for the first function. What
this means for us is that when we do the integral all we need to do is plug
in the first function into the integral.
Here is the integral.
[Return to Problems]
(b)
In this part is between the limits of integration. This means that the integrand is no longer
continuous in the interval of integration and that is a show stopper as far
we’re concerned. As noted above we
simply can’t integrate functions that aren’t continuous in the interval of
integration.
Also, even if the function was continuous at we would still have the problem that the
function is actually two different equations depending where we are in the
interval of integration.
Let’s first address the problem of the function not
being continuous at . As we’ll see, in this case, if we can find
a way around this problem the second problem will also get taken care of at
the same time.
In the previous examples where we had functions that
weren’t continuous we had division by zero and no matter how hard we try we
can’t get rid of that problem.
Division by zero is a real problem and we can’t really avoid it. In this case the discontinuity does not
stem from problems with the function not existing at . Instead the function is not continuous
because it takes on different values on either sides of . We can “remove” this problem by recalling Property 5 from the
previous section. This property tells
us that we can write the integral as follows,
On each of these intervals the function is
continuous. In fact we can say
more. In the first integral we will
have x between -2 and 1 and this
means that we can use the second equation for and likewise for the second integral x will be between 1 and 3 and so we
can use the first function for . The integral in this case is then,
[Return to Problems]
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So, to integrate a piecewise function, all we need to do is
break up the integral at the break point(s) that happen to occur in the
interval of integration and then integrate each piece.
Next we need to look at is how to integrate an absolute
value function.
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Example 5 Evaluate
the following integral.
Solution
Recall that the point behind indefinite integration (which
we’ll need to do in this problem) is to determine what function we
differentiated to get the integrand.
To this point we’ve not seen any functions that will differentiate to
get an absolute value nor will we ever see a function that will differentiate
to get an absolute value.
The only way that we can do this problem is to get rid of
the absolute value. To do this we need
to recall the definition of absolute value.
Once we remember that we can define absolute value as a
piecewise function we can use the work from Example 4 as a guide for doing
this integral.
What we need to do is determine where the quantity on the
inside of the absolute value bars is negative and where it is positive. It looks like if the quantity inside the absolute
value is positive and if the quantity inside the absolute
value is negative.
Next, note that is in the interval of integration and so, if
we break up the integral at this point we get,
Now, in the first integrals we have and so in this interval of integration. That means we can drop the absolute value
bars if we put in a minus sign.
Likewise in the second integral we have which means that in this interval of
integration we have and so we can just drop the absolute value
bars in this integral.
After getting rid of the absolute value bars in each
integral we can do each integral. So,
doing the integration gives,
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Integrating absolute value functions isn’t too bad. It’s a little more work than the “standard”
definite integral, but it’s not really all that much more work. First, determine where the quantity inside
the absolute value bars is negative and where it is positive. When we’ve determined that point all we need
to do is break up the integral so that in each range of limits the quantity inside
the absolute value bars is always positive or always negative. Once this is done we can drop the absolute
value bars (adding negative signs when the quantity is negative) and then we
can do the integral as we’ve always done.