### Video Transcript

In this video, weβll learn how to
use properties of definite integration, such as the order of integration limits,
zero with limits, sums, and differences. Weβll also learn how these
properties can help us to simplify problems involving definite integrals.

On defining the definite integral,
the integral between π and π of π of π₯ with respect to π₯, we implicitly assumed
that π was less than π. But thinking about the definition
of a definite integral as the limit of Reimann sums, we see that this still holds if
π is greater than π. Notice also that if we reverse π
and π, then Ξπ₯ changes to π minus π over π. With that, we find that the
definite integral between π and π of π of π₯ with respect to π₯ is equal to the
negative definite integral between π and π of π of π₯ with respect to π₯. Now, if π is equal to π, then Ξπ₯
becomes zero. And so, the definite integral
between π and π of π of π₯ with respect to π₯ is itself equal to zero. Weβll now look at some other
important properties of integrals that weβre able to recall.

Now, itβs outside of the scope of
this video to fully derive these. But we will have a think about
where they might come from. Letβs assume we have some
functions, π and π, which are continuous. The second property weβre
interested in is that the definite integral between π and π of π of π₯ with
respect to π₯ is equal to the definite integral between π and π of π of π₯ with
respect to π₯ plus the definite integral between π and π of π of π₯ with respect
to π₯. And geometrically, this is really
to be expected. We think about the definite
integral as the area between the curve and the π₯-axis. And so, the total area between the
curve and the π₯-axis, bounded by the lines π₯ equals π and π₯ equals π, must be
equal to the area bounded by the lines between π₯ equals π and π₯ equals π and the
area bounded between the lines π₯ equals π and π₯ equals π.

For constant value π, the definite
integral between π and π of π with respect to π₯ is π times π minus π. In other words, the integral of a
constant function π is that constant times the length of the integral. Using π greater than zero and π
less than π, then this is to be expected. Since π times π minus π is the
area of the rectangle shown. The next property is that the
integral of a sum of two functions is equal to the sum of the integrals of those
functions. So for positive functions, the area
under π plus π is equal to the area under π plus the area under π.

Similarly, the integral of a
constant times a function is equal to that constant times the integral of the
function. Then, by combining the previous two
properties, we find that the integral of the difference of two functions is equal to
the integral of π of π₯ with respect to π₯ plus negative one times the integral of
π of π₯ with respect to π₯. But, of course, that must be equal
to the difference of the integrals of these two functions. So letβs now have a look at an
example of some of these properties.

The function π is continuous on
the closed interval negative four to four and satisfies the definite integral
between zero and four of π of π₯ with respect to π₯ is equal to nine. Determine the definite integral
between zero and four of π of π₯ minus six with respect to π₯.

To answer this question, weβll
begin by recalling some basic properties of definite integrals. Firstly, we know that the integral
of the sum of two functions is equal to the sum of the integral of each respective
function. Similarly, the integral of the
difference of two functions is equal to the difference of the integrals of those
functions. This means we can begin by
rewriting our integral as the difference between the integral between zero and four
of π of π₯ with respect to π₯ and the integral between zero and four of six with
respect to π₯. We also recall that for constant
function π, the definite integral between π and π of π with respect to π₯ is
equal to π times π minus π.

Now, our π is equal to six, and π
is zero, π is four. So the definite integral between
zero and four of six with respect to π₯ is simply six times four minus zero, which
is equal to 24. As per the question, we now replace
the integral between zero and four of π of π₯ with respect π₯ with nine. And we find that the integral of π
of π₯ minus six between the limits of zero and four with respect to π₯ is nine minus
24. Thatβs equal to negative 15. So, despite not knowing the
function π, we see that the definite integral between zero and four of π of π₯
minus six with respect to π₯ is negative 15.

Letβs now have a look at another
example involving simple properties of definite integrals.

If the definite integral between
negative seven and eight of π of π₯ with respect to π₯ is equal to 10, determine
the value of the definite integral between eight and negative seven of seven π of
π₯ with respect to π₯.

To answer this question, weβre
going to need to recall two specific properties of definite integrals. The first is that the definite
integral between π and π of π of π₯ with respect to π₯ is equal to the negative
definite integral between π and π of that same function with respect to π₯. We also know that the definite
integral between π and π of some constant π times π of π₯ with respect to π₯ is
equal to π times the definite integral of π of π₯ with respect to π₯. And this second oneβs great because
it means we can simply take the constant outside of the integral and focus on
integrating the function itself.

Letβs begin by taking that constant
factor of seven outside the integral. When we do, we see that the
definite integral weβre looking to evaluate is equal to seven times the definite
integral between eight and negative seven of π of π₯ with respect to π₯. Now, in the question, weβve been
told that the definite integral between negative seven and eight of our function is
equal to 10. So we use the first property we
recall to rewrite the definite integral between eight and negative seven of π of π₯
with respect to π₯ as negative the definite integral between negative seven and
eight of π of π₯ with respect to π₯.

This can, of course, be
simplified. And we can say that our integral is
equal to negative seven times the definite integral between negative seven and eight
of π of π₯ with respect to π₯. But, of course, we know that that
definite integral is equal to 10. So our definite integral is
negative seven times 10, which is simply negative 70.

Now, whatβs really nice about the
properties weβve considered so far is that they hold for values of π less than π,
values of π greater than π, and values of π equal to π. Now, there are some properties that
we can consider if, and only if, π is less than or equal to π. We call these comparison properties
of the integral. And this is because they allow us
to compare the general size of definite integrals.

The first says that if π of π₯ is
greater than or equal to zero, where π₯ is in the closed interval π to π, then the
definite integral between π and π of π of π₯ with respect to π₯ will also be
greater than zero. And, of course, this makes a lot of
sense if we think about the definite integral as the area between the graph of π
and the π₯-axis. The geometric interpretation is
simply that the area here is positive. It sits above the π₯-axis. The second simply says that a
bigger function has a bigger integral. This also makes sense geometrically
but can also be inferred from the first property because π minus π must be greater
than or equal to zero.

The third takes a little more
thinking about. If π is continuous, we can take
lowercase π and capital π to be the absolute minimum and maximum values of π on
the closed interval π to π, respectively. This property tells us that the
area under the graph of π is greater than the area of the rectangle with a height
of lowercase π but less than the area of a rectangle with a height of capital
π.

Letβs now have a look at a question
that involves some of these properties.

Suppose that on the closed interval
negative two to five, the values of π lie in the closed interval lowercase π to
capital π. Between which bounds does the
definite integral between negative two and five of π of π₯ with respect to π₯
lie?

We recall one of the comparison
properties of integrals. It tells us that if π of π₯ is
greater than or equal to lowercase π and less than or equal to capital π for
values of π₯ greater than or equal to π and less than or equal to π. Then π times π minus π is less
than or equal to the definite integral between π and π of π of π₯ with respect to
π₯, which in turn is less than or equal to capital π times π minus π. In other words, given that lower
case π is the absolute minimum of π and uppercase π is the absolute maximum of
π, the area under the graph of π is greater than the area of the rectangle with a
height lower case π. But itβs less than the area of the
rectangle with a height uppercase π.

In this example, weβre going to let
π be equal to negative two and π be equal to five. Then, we see that π times five
minus negative two is less than or equal to the definite integral between negative
two and five of π of π₯ with respect to π₯ which, in turn, is less than or equal to
capital π times five minus negative two. Five minus negative two is
seven. And so, we see that our definite
integral must be greater than or equal to seven π and less than or equal to seven
uppercase π.

The final properties weβre
interested in involve integration of odd and even functions. We recall that if a function is
odd, π of negative π₯ is equal to negative π of π₯. And a function is said to be even
if π of negative π₯ equals π of π₯. Geometrically, an odd function has
rotational symmetry about the origin. For example, π¦ equals sin of
π₯. Whereas an even function has
reflectional symmetry about the π¦-axis, the graph of π¦ equals cos of π₯, for
example. For the closed interval negative π
to π, when the function is odd, we say that the definite integral between negative
π and π of π of π₯ with respect to π₯ is equal to zero. And when itβs even, we find that
itβs equal to two times the definite integral between zero and π of π of π₯ with
respect to π₯.

Once again, geometrically, this
makes a lot of sense. We saw that an even function has
reflectional symmetry about the π¦-axis. So integrating between negative π
and π would provide an area double the size of that between zero and π. For an odd function, the areas
would be equal in size but on either side of the π₯-axis, thereby cancelling one
another out.

Letβs have a look at a couple of
examples of this.

The function π is odd, continuous
on the closed interval negative one to seven, and satisfies the definite integral
between one and seven of π of π₯ with respect to π₯ equals negative 17. Determine the definite integral
between negative one and seven of π of π₯ with respect to π₯.

Weβre firstly told that the
function π is odd. So we recall the following property
for integrating odd functions. The definite integral between
negative π and π of π of π₯ with respect to π₯ is equal to zero. Weβre also told that the definite
integral between one and seven of π of π₯ with respect to π₯ is equal to negative
17. So we split the integral up. And we see that the integral that
weβre looking for between negative one and seven of π of π₯ with respect to π₯ is
equal to the definite integral between negative one and one of π of π₯ plus the
definite integral between one and seven of π of π₯.

Now, the function π is odd. So by the first property, we see
that the definite integral between negative one and one of π of π₯ with respect to
π₯ must be equal to zero. Then, we simply take the definite
integral between one and seven of π of π₯ from the question. Itβs negative 17. This means the definite integral
weβre looking for is equal to zero plus negative 17 which is negative 17.

Weβll now have a look at an example
that involves an even function.

The function π is even, continuous
on the closed interval negative eight to eight, and satisfies the definite integral
between negative eight and eight of π of π₯ with respect to π₯ is equal to 19 and
the definite integral between zero and four of π of π₯ with respect to π₯ is equal
to 13. Determine the definite integral
between negative eight and negative four of π of π₯ with respect to π₯.

We begin by recalling the property
of the integral of an even function That is, the definite integral between negative
π and π of that even function is equal to two times the definite integral between
zero and π of π of π₯ with respect to π₯. Now, in fact, weβre looking to find
the definite integral between negative eight and negative four of our even
function. So weβre going to do this in two
parts. Firstly, weβre going to split it up
and say that the definite integral must be equal to the integral between negative
eight and zero minus the integral between negative four and zero. Now, actually, weβll form an
equation using the first part of this integral and the fact that the function is
even.

The definite integral between
negative eight and eight of π of π₯ with respect to π₯ must be two times the
definite integral between zero and eight of πof π₯ with respect to π₯. Now, it also follows that this must
also be equal then to two times the definite integral between negative eight and
zero of π of π₯ with respect to π₯. Of course, in the question, we were
told that the definite integral between negative eight and eight of π of π₯ is
19. So we set 19 equal to two times the
definite integral that weβre looking for. And then, we divide both sides of
our equation by two. And we see that this is equal to 19
over two. The integral weβre looking for then
is equal to 19 over two minus the definite integral between negative four and zero
of π of π₯ with respect to π₯.

Now, once again, the function is
even. So this must, in turn, be equal to
19 over two minus the definite integral between zero and four of π of π₯ with
respect to π₯. Remember, this is because even
functions have reflection or symmetry about the π¦-axis. Now, weβre told in the question
that this definite integral is equal to 13. Then to evaluate 19 over two minus
13, we write 13 as 26 over two. So weβre looking to find 19 over
two minus 26 over two which is negative seven over two. And so, we found the definite
integral between negative eight and negative four of π of π₯ with respect to
π₯. Itβs negative seven over two.

In this video, weβve developed some
basic properties of integrals that help us to evaluate integrals in a simple
manner. For a continuous function π, the
definite integral between π and π of π of π₯ with respect to π₯ is equal to the
negative definite integral between π and π of π of π₯ with respect to π₯. We saw that the integral of a
constant function π is the constant times the length of the interval. And that the integral of the sum or
difference of two continuous functions is equal to this sum or difference of the
integrals of each of those functions. We saw that we can take constant
factors outside of the integral and focus on integrating the function itself.

We learned that comparison
properties of the integral can be used when π is greater than or equal to π. The first of these says that if π
of π₯ is greater than or equal to zero, then the definite integral between π and π
of π of π₯ with respect to π₯ is also greater or equal than zero. A bigger function has a bigger
integral. And if π and capital π are
absolute minimum and maximum values of π on some closed interval π to π, then the
area under the graph of π must be greater than the area of the rectangle with
height π and less than the area of the rectangle with height capital π.

Finally, we learned that if a
function is even or odd over some closed interval negative π to π, then we can
apply the following rules. If itβs odd, the definite integral
between negative π and π of π of π₯ with respect to π₯ is zero. And if itβs even, itβs two times
the definite integral between zero and π of π of π₯ with respect to π₯.