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The Net Change Theorem
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In this video, weβll learn how to apply the net change theorem and what this tells us about the value of integrals.
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The net change theorem tells us that the integral of a rate of change is equal to the net change.
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But what does this actually mean?
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Mathematically, we can write this in the following way.
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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.
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Letβs examine each of our terms.
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πΉ of π₯ is some function.
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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₯.
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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.
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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.
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If our final value πΉ π is greater than our initial value πΉ π, weβll see a positive net change.
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And the reverse is true if πΉ π is smaller than πΉ π.
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Here is where the quick mention that the net change theorem is closely related to the second part of the fundamental theorem of calculus.
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For this video, we might be going into much detail on this, but you may already be familiar with some of the concepts involved.
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Letβs look at an example to get a feel for our theorem.
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Suppose that πΉ of π‘ is equal to π‘ squared.
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Evaluate the integral between one and three of πΉ prime of π‘ with respect to π‘ using the net change theorem.
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For this question, and many others, itβs always helpful to write out the theorem that youβll be using.
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Here, we have the net change theorem.
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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.
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We take the lower limit of integration, or π, as one and the upper limit of integration, or π, as three.
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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.
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Again, hereβre the limits of integration that weβve used.
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Notice, that weβre able to apply our theorem because we have a definite integral of a derivative πΉ prime of π‘.
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And this can be thought of as the rate of change of πΉ of π‘.
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Now, the function itself πΉ of π‘ has been given in the question.
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And itβs simply π‘ squared.
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We can, therefore, evaluate the right-hand side of our function as three squared minus one squared.
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Of course, this is nine minus one, which is simply equal to eight.
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Following this simplification, we have answered our question.
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We have used the net change theorem to evaluate that the given integral is equal to eight.
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Okay, now that weβve seen an example, letβs return to our theorem to get a better understanding of whatβs going on.
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Here, we have a graph of some function πΉ of π₯.
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And below, we have the corresponding graph of πΉ prime of π₯.
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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.
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The net change theorem tells us that this gives us the difference between the value πΉ of π and the value πΉ of π.
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Here, we see that πΉ of π is larger than πΉ of π.
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So, we expect this to be a positive number.
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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.
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Putting these two things together gives us a visual understanding of the net change theorem.
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Another interesting point to note is that our quantity πΉ could change in both directions between π and π.
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Here, we see that as we move right from point π to π, πΉ of π₯ first decreases then increases.
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The corresponding graph of πΉ prime might look something like this.
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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.
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Again, these matches our observational expectations.
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Since πΉ of π is larger than πΉ of π, we would expect our net change to be a positive number.
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Before we move on, itβs worth highlighting a couple of common mistakes using this graphical example.
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The net change theorem only gives us the difference between the value πΉ of π and the value πΉ of π.
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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.
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For example, a positive net change does not necessarily mean that our function will only increase between π and π.
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Letβs look at another example which illustrates one possible mistake.
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True or false?
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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.
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For this question, we first recognise that weβve been given a function which represents a rate of change.
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We have then been asked about a definite integral involving this rate of change.
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When evaluating definite integrals of rates of change, the tool that we use is the net change theorem.
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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.
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But here, the question tells us that the function β of π‘ is already a rate of change of some quantity.
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We can, therefore, define capital π» of π‘ to be the quantity itself, the babyβs height at π‘ months old.
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In other words, it is the antiderivative of lowercase β of π‘.
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Since the net change theorem uses the antiderivative of the rate that weβre integrating, weβre now in a position to form an equation.
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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.
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At this point, we may recognise a potential trap in our question.
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Upon first reading our question, we may notice that our integral has a lower limit of zero and an upper limit of six.
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This might lead us to conclude that our lower limit can be ignored, since it is equal to zero.
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We would, therefore, conclude that the integral is indeed equal to π» of six, the babyβs height at six months old.
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And we might think that the statement in the question is true.
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This would be a mistake.
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Our lower limit cannot be ignored, since π» of zero is not equal to zero.
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π» of zero is actually the babyβs height at zero months old, which is its birth.
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We know that when it is born, a baby is very small but doesnβt have a height of zero.
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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.
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Since weβve just reasoned that π» of zero is not equal to zero, this net change is not the same as π» of six.
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This means that we have just proved that the answer to our question is false.
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The example weβve just seen illustrates that the net change theorem can be applied to many physical processes.
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A common example is the relationship between position, velocity, and acceleration.
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Letβs imagine an object moving in one dimension, say, along the π₯-axis.
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We can represent its position at some time π‘ as the function π₯ of π‘.
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Its velocity is then the rate of change of its position with respect to time.
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In other words, velocity is the derivative of position with respect to time.
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Similarly, acceleration is the rate of change of velocity with respect to time.
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Again, itβs a derivative with respect to time.
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Given these relationships, we should, therefore, understand that velocity is the antiderivative of acceleration and position is the antiderivative of velocity.
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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.
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Letβs see an example of this.
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A particle is moving along the π₯-axis.
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Its velocity in metres per second as a function of time is π£ of π‘ equals six π‘ squared minus eight π‘.
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Find the displacement of the particle between π‘ equals one and π‘ equals five.
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For this question, the first thing we can do is to pay attention to this word, displacement.
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The displacement of a particle is the distance between its final position and its initial position.
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Here, our particle is at its initial position when π‘ equals one, and in its final position when π‘ equals five.
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Given that the particleβs velocity is defined as π£ of π‘, letβs define its position along the π₯-axis as π₯ of π‘.
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The displacement that weβre trying to find is, therefore, π₯ of five, its final position, minus π₯ of one, its initial position.
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Okay, now, letβs recall the relationship between velocity and position.
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If we differentiate position with respect to time, we get velocity.
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We could, in fact, represent velocity as π₯ prime of π‘.
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Doing so might help us recognise that we can find our displacement using the net change theorem.
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Since velocity is the derivative of position with respect to time, the net change theorem allows us to form the following equation.
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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.
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And weβll notice that the right-hand side of this equation is exactly the displacement that we need.
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Of course, π₯ prime of π‘, which is the rate of change of position with respect to time, is velocity.
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Now, the question has given us the function for velocity, which is six π‘ squared minus eight π‘.
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Using standard rules of integration, we raise each of our powers of π‘ by one and divide by the new power.
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We can then simplify slightly.
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Six over three is two.
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And eight over two is four.
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We now work on the limits of our definite integral.
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Inputting these limits, weβre left with the following.
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And after a few steps of working, weβre left with our answer, which is 152 metres.
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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.
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This means that we have answered our question.
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And the displacement that we were looking for is 152 metres.
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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.
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We can see how this works by rearranging the familiar form of the net change theorem.
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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.
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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.
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We can use these two pieces of information to find the value of πΉ at π₯ equals π.
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And here, itβs worth noting that we can choose π to be whatever we like.
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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.
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A logically equivalent statement to our first rearrangement is the following.
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πΉ of π minus our integral is equal to πΉ of π.
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And of course, this makes sense.
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πΉ of π subtract the net change of πΉ between π and π is equal to πΉ of π.
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As mentioned, both of these are equivalent statements and can be used to solve the same types of problems.
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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.
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Remember to pay attention to this, as it will dictate which way round the limits of your integral will go.
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Let us look at one final example which uses this rearrangement.
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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.
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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.
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For this question, weβre told that a barrel is filled with water at a certain rate, ππ‘.
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In other words, ππ‘ is the rate of change of the water in the barrel.
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To move forward, let us define the volume of water in the barrel as capital π΅π‘.
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Note that capital π΅π‘ is the antiderivative of our rate of change, lowercase ππ‘.
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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.
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Weβve then been asked to find the volume of water at some other time when π‘ equal six.
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With the information that weβve been given, we can use a rearrangement of the net change theorem.
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Letβs express the information given in our question in this form.
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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.
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Given the relationship between lowercase π of π‘ and capital π΅ of π‘, we can express lowercase π of π‘ in the following way.
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Of course, itβs just the derivative of π of π‘.
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If we replace this in our equation, we see that we have two known pieces of information and one unknown piece of information.
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Letβs rearrange this equation to isolate the thing that weβre trying to find, capital π΅ of six, on one side.
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Adding capital π΅ of two to both sides, weβre left with the following.
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We know capital π΅ of two.
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The amount of water in the barrel when π‘ equals two is 10 liters.
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We also have the function lowercase π of π‘.
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So, letβs substitute these into the equation.
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Letβs stop and take a moment to understand our equation.
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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.
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Logically, this should make a lot of sense to us.
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So, letβs continue with our calculations.
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Using the standard rules of integration, we raise the power of π‘ by one and divide by the new power.
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We then input the limits of our integral and continue to simplify.
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With a few more simplification steps, we reach an answer, which is that capital π΅ of six is equal to 64 liters.
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Of course, capital π΅ of six is the volume of water in the barrel when π‘ equals six.
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So, in reaching this line, we have answered our question.
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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.
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Following this final example, letβs run through a few key points.
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Taking a definite integral of a rate of change gives us a net change.
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And this is expressed mathematically by the net change theorem shown here.
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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.
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We do this using the rate of change of πΉ with respect to π₯, which is πΉ prime of π₯.
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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.
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Finally, the net change theorem can be used to find an unknown value of πΉ.
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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₯.