### Video Transcript

The Net Change Theorem

In this video, weβll learn how to
apply the net change theorem and what this tells us about the value of
integrals.

The net change theorem tells us
that the integral of a rate of change is equal to the net change. But what does this actually
mean? Mathematically, we can write this
in the following way. The integral between π and π of
πΉ prime of π₯ with respect to π₯ is equal to πΉ of π minus πΉ of π. Letβs examine each of our
terms. πΉ of π₯ is some function. This means that πΉ prime of π₯,
which is the first derivative of πΉ of π₯, can be thought off as the rate of change
of πΉ of π₯.

When we integrate this between the
limits of π and π, we get our original function evaluated at π subtract our
original function evaluated at π. If we think of πΉ π as our initial
value of πΉ and πΉ π as our final value of πΉ, then we start to get a better
understanding that weβre looking at the net change, or the difference, between two
values of our original function. If our final value πΉ π is greater
than our initial value πΉ π, weβll see a positive net change. And the reverse is true if πΉ π is
smaller than πΉ π.

Here is where the quick mention
that the net change theorem is closely related to the second part of the fundamental
theorem of calculus. For this video, we might be going
into much detail on this, but you may already be familiar with some of the concepts
involved. Letβs look at an example to get a
feel for our theorem.

Suppose that πΉ of π‘ is equal to
π‘ squared. Evaluate the integral between one
and three of πΉ prime of π‘ with respect to π‘ using the net change theorem.

For this question, and many others,
itβs always helpful to write out the theorem that youβll be using. Here, we have the net change
theorem. Although the variable in our
question is π‘ and the variable in the theorem weβve written out is π₯, things apply
in exactly the same way. We take the lower limit of
integration, or π, as one and the upper limit of integration, or π, as three. Using our theorem, weβre able to
say that the integral between one and three of πΉ prime of π‘ with respect to π‘ is
equal to πΉ of three minus πΉ of one.

Again, hereβre the limits of
integration that weβve used. Notice, that weβre able to apply
our theorem because we have a definite integral of a derivative πΉ prime of π‘. And this can be thought of as the
rate of change of πΉ of π‘. Now, the function itself πΉ of π‘
has been given in the question. And itβs simply π‘ squared. We can, therefore, evaluate the
right-hand side of our function as three squared minus one squared. Of course, this is nine minus one,
which is simply equal to eight. Following this simplification, we
have answered our question. We have used the net change theorem
to evaluate that the given integral is equal to eight.

Okay, now that weβve seen an
example, letβs return to our theorem to get a better understanding of whatβs going
on. Here, we have a graph of some
function πΉ of π₯. And below, we have the
corresponding graph of πΉ prime of π₯. Taking the left-hand side of the
net change theorem, the integral between π and π of πΉ prime of π₯ with respect to
π₯ gives us the area under this line. The net change theorem tells us
that this gives us the difference between the value πΉ of π and the value πΉ of
π.

Here, we see that πΉ of π is
larger than πΉ of π. So, we expect this to be a positive
number. And of course, we know that when
looking at integrals, if we see an area above the π₯-axis, this evaluates to a
positive number, which agrees with what we see on our correspondent graph. Putting these two things together
gives us a visual understanding of the net change theorem.

Another interesting point to note
is that our quantity πΉ could change in both directions between π and π. Here, we see that as we move right
from point π to π, πΉ of π₯ first decreases then increases. The corresponding graph of πΉ prime
might look something like this. Again, performing our integration,
we see that we have a small area below the π₯-axis, which would evaluate to
negative, and a much larger area above the π₯-axis, which would evaluate to
positive. Again, these matches our
observational expectations. Since πΉ of π is larger than πΉ of
π, we would expect our net change to be a positive number.

Before we move on, itβs worth
highlighting a couple of common mistakes using this graphical example. The net change theorem only gives
us the difference between the value πΉ of π and the value πΉ of π. We should not get confused and
think that it is just giving us the value πΉ of π nor should we make any
conclusions as to the behavior of our function in-between the values of π and
π. For example, a positive net change
does not necessarily mean that our function will only increase between π and
π. Letβs look at another example which
illustrates one possible mistake.

True or false? If β of π‘ represents the rate of
change of the height of a baby in centimetres per month, when it is π‘ months old,
then the integral between zero and six of β of π‘ with respect to π‘ is equal to the
babyβs height when it is six months old.

For this question, we first
recognise that weβve been given a function which represents a rate of change. We have then been asked about a
definite integral involving this rate of change. When evaluating definite integrals
of rates of change, the tool that we use is the net change theorem. Now, we may be used to seeing this
prime notation inside our integral to tell us that this is a rate of change or a
derivative. But here, the question tells us
that the function β of π‘ is already a rate of change of some quantity. We can, therefore, define capital
π» of π‘ to be the quantity itself, the babyβs height at π‘ months old. In other words, it is the
antiderivative of lowercase β of π‘.

Since the net change theorem uses
the antiderivative of the rate that weβre integrating, weβre now in a position to
form an equation. The integral given in our question
is equal to capital π» of six, the babyβs height at six months old, minus capital π»
of zero, the babyβs height at zero months old. At this point, we may recognise a
potential trap in our question.

Upon first reading our question, we
may notice that our integral has a lower limit of zero and an upper limit of
six. This might lead us to conclude that
our lower limit can be ignored, since it is equal to zero. We would, therefore, conclude that
the integral is indeed equal to π» of six, the babyβs height at six months old. And we might think that the
statement in the question is true.

This would be a mistake. Our lower limit cannot be ignored,
since π» of zero is not equal to zero. π» of zero is actually the babyβs
height at zero months old, which is its birth. We know that when it is born, a
baby is very small but doesnβt have a height of zero. In actual fact, what our integral
gives us is the net change between the babyβs height at zero months old and its
height at six months old. Since weβve just reasoned that π»
of zero is not equal to zero, this net change is not the same as π» of six. This means that we have just proved
that the answer to our question is false.

The example weβve just seen
illustrates that the net change theorem can be applied to many physical
processes. A common example is the
relationship between position, velocity, and acceleration. Letβs imagine an object moving in
one dimension, say, along the π₯-axis. We can represent its position at
some time π‘ as the function π₯ of π‘. Its velocity is then the rate of
change of its position with respect to time. In other words, velocity is the
derivative of position with respect to time. Similarly, acceleration is the rate
of change of velocity with respect to time. Again, itβs a derivative with
respect to time.

Given these relationships, we
should, therefore, understand that velocity is the antiderivative of acceleration
and position is the antiderivative of velocity. Since the net change theorem
involves some function, which is a rate of change and its antiderivative function,
we can, therefore, use it to evaluate definite integrals involving velocity, giving
us the net change between two positions, and definite integrals involving
acceleration, giving us the net change between two velocities. Letβs see an example of this.

A particle is moving along the
π₯-axis. Its velocity in metres per second
as a function of time is π£ of π‘ equals six π‘ squared minus eight π‘. Find the displacement of the
particle between π‘ equals one and π‘ equals five.

For this question, the first thing
we can do is to pay attention to this word, displacement. The displacement of a particle is
the distance between its final position and its initial position. Here, our particle is at its
initial position when π‘ equals one, and in its final position when π‘ equals
five. Given that the particleβs velocity
is defined as π£ of π‘, letβs define its position along the π₯-axis as π₯ of π‘. The displacement that weβre trying
to find is, therefore, π₯ of five, its final position, minus π₯ of one, its initial
position.

Okay, now, letβs recall the
relationship between velocity and position. If we differentiate position with
respect to time, we get velocity. We could, in fact, represent
velocity as π₯ prime of π‘. Doing so might help us recognise
that we can find our displacement using the net change theorem. Since velocity is the derivative of
position with respect to time, the net change theorem allows us to form the
following equation.

The integral between one and five
of π₯ prime of π‘ with respect to π‘ is equal to π₯ of five minus π₯ of one. And weβll notice that the
right-hand side of this equation is exactly the displacement that we need. Of course, π₯ prime of π‘, which is
the rate of change of position with respect to time, is velocity.

Now, the question has given us the
function for velocity, which is six π‘ squared minus eight π‘. Using standard rules of
integration, we raise each of our powers of π‘ by one and divide by the new
power. We can then simplify slightly. Six over three is two. And eight over two is four. We now work on the limits of our
definite integral. Inputting these limits, weβre left
with the following.

And after a few steps of working,
weβre left with our answer, which is 152 metres. Finally, we remember that the
integral that weβve just evaluated is equal to the displacement of the particle in
metres between π‘ equals one and π‘ equals five. This means that we have answered
our question. And the displacement that we were
looking for is 152 metres.

Another useful application of the
net change theorem is to find the value of a function at a given point when certain
information is available. We can see how this works by
rearranging the familiar form of the net change theorem. Adding πΉ of π to both sides of
the equation, we are left with πΉ of π plus our integral is equal to πΉ of π. This form of the net change theorem
can be used when we have the value for πΉ at π₯ equals π and weβve been given the
function πΉ prime of π₯, which is the rate of change of πΉ. We can use these two pieces of
information to find the value of πΉ at π₯ equals π. And here, itβs worth noting that we
can choose π to be whatever we like.

One way to think about this is we
start with the value of πΉ at π, add the net change of πΉ between π and π, and
weβre left with πΉ of π. A logically equivalent statement to
our first rearrangement is the following. πΉ of π minus our integral is
equal to πΉ of π. And of course, this makes
sense. πΉ of π subtract the net change of
πΉ between π and π is equal to πΉ of π. As mentioned, both of these are
equivalent statements and can be used to solve the same types of problems.

However, here we have always
treated πΉ of π as our initial value of πΉ and πΉ of π as our final value of
πΉ. Remember to pay attention to this,
as it will dictate which way round the limits of your integral will go. Let us look at one final example
which uses this rearrangement.

A barrel is filled with water at a
rate of ππ‘ equals three π‘ squared over four plus a half liters per day, where π‘
is the number of days. Given that the barrel contains 10
liters of water when π‘ equals two, find the volume of water in the barrel when π‘
equals six.

For this question, weβre told that
a barrel is filled with water at a certain rate, ππ‘. In other words, ππ‘ is the rate of
change of the water in the barrel. To move forward, let us define the
volume of water in the barrel as capital π΅π‘. Note that capital π΅π‘ is the
antiderivative of our rate of change, lowercase ππ‘. Now, alongside the rate of change
for the volume of water in a barrel, the question has given us the volume itself at
a known time, in this case when π‘ equals two. Weβve then been asked to find the
volume of water at some other time when π‘ equal six.

With the information that weβve
been given, we can use a rearrangement of the net change theorem. Letβs express the information given
in our question in this form. The net change theorem tells us
that the integral between two and six of the rate of change of the volume of water
in the barrel with respect to time is equal to the volume of water in the barrel
when π‘ equals six subtract the volume of water in the barrel when π‘ equals
two.

Given the relationship between
lowercase π of π‘ and capital π΅ of π‘, we can express lowercase π of π‘ in the
following way. Of course, itβs just the derivative
of π of π‘. If we replace this in our equation,
we see that we have two known pieces of information and one unknown piece of
information. Letβs rearrange this equation to
isolate the thing that weβre trying to find, capital π΅ of six, on one side.

Adding capital π΅ of two to both
sides, weβre left with the following. We know capital π΅ of two. The amount of water in the barrel
when π‘ equals two is 10 liters. We also have the function lowercase
π of π‘. So, letβs substitute these into the
equation. Letβs stop and take a moment to
understand our equation. If we take the amount of water in
the barrel when π‘ equals two, and add the amount of water that enters the barrel
between π‘ equals two and π‘ equals six, weβll be left with the amount of water in
the barrel when π‘ equal six. Logically, this should make a lot
of sense to us. So, letβs continue with our
calculations.

Using the standard rules of
integration, we raise the power of π‘ by one and divide by the new power. We then input the limits of our
integral and continue to simplify. With a few more simplification
steps, we reach an answer, which is that capital π΅ of six is equal to 64
liters. Of course, capital π΅ of six is the
volume of water in the barrel when π‘ equals six. So, in reaching this line, we have
answered our question. We reached this answer using a
rearrangement of the net change theorem and inputting the rate of change and the
known value given by the question.

Following this final example, letβs
run through a few key points. Taking a definite integral of a
rate of change gives us a net change. And this is expressed
mathematically by the net change theorem shown here. The theorem gives us a formula to
calculate the change that occurs between something we can interpret as the initial
value of πΉ, πΉ of π, and something we interpret as the final value of πΉ, πΉ of
π. We do this using the rate of change
of πΉ with respect to π₯, which is πΉ prime of π₯.

The net change theorem gives us a
useful tool to evaluate many physical systems, such as those involving position,
velocity and acceleration, volume of liquid and flow rate, or perhaps even
population and population growth rate. Finally, the net change theorem can
be used to find an unknown value of πΉ. This can be done when youβre given
a so-called initial or final value of πΉ and a function for the rate of change of πΉ
with respect to π₯.