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MATH_125_online_L5__5.3___5.4__Lecture_5
[00:00:00.12] Hello, everyone, and welcome back for Lecture 5. Today's lecture will focus on
part two of the fundamental theorem of calculus, which, recall is in Section 5.3 of your textbook,
and on a related result from Section 5.4, which is called the net change theorem. Let's start by
recalling part one of the fundamental theorem of calculus, which we discussed last time.
[00:00:26.25] The idea was that if we have a continuous function lowercase f on a closed interval
from a to b, for example, this could be our lowercase f of t, which we're going to take to be sin of
t. This is our t-axis, y-axis. And in this case, this function is from a is equal with zero to b is
equal to 2 pi. So if we have such a continuous function on a closed interval from a to b, and we
define this other function called the accumulation function, or the area so far-- by letting the
upper bound of the definite integral of f vary, and keeping the lower bound fixed at whatever our
starting point a happens to be-- so we're taking this definite integral from a to x.
[00:01:30.53] x in this case in the first picture is over here, and it varies between a and b. So it
can be moved forward and backward. And we're defining uppercase F of x to be the integral
from a to x. So in particular, for this picture that we have right here, the function uppercase F of
x would be the blue area under the graph. And with the second picture, it would be the blue area
above the graph, area 1. Subtract area 2 over here. And that would be when x is past pi, but
before 2 pi.
[00:02:23.57] So let's look at an animation quickly. This is from a website called Visual
Calculus. And they have a number of different interesting animations and visualizations. In this
one, they are trying to help us visualize this uppercase F of x function based on sin of t. I copied
this picture for my lecture notes. But the advantage of going to the website itself is that you can
actually drag the x across. So we're starting at 0. The function is sin of t. And as x goes up from 0
towards pi over 2 here, the red graph you're seeing that they're building for us, this represents
uppercase F of x.
[00:03:14.15] So for example, at this point, like we had in our lecture notes, this is uppercase F
of x. And our x is right over here. So the value at that point of uppercase F is this blue area under
the graph.
[00:03:35.41] Going back to the visualization here. Let's keep dragging the x, moving the x,
moving the x. You notice that the red function, our uppercase F, our accumulation function is
increasing. How is it increasing? At first, the area grows, but doesn't-- the beginning here, it
doesn't grow very fast. And as we're getting farther here, the area grows faster and faster because
the y-coordinate is bigger. And then it grows a little bit slower. And what's going to happen
when we reach this point?
[00:04:08.62] Well from this point on, to compute uppercase F, so the definite integral up to x,
these x's correspond to negative values of the original function, sin of t. So that means we're now
starting to subtract areas from the area so far function. Which means that the graph of uppercase
F of x, which is the accumulated area, or the area so far, is now decreasing. By how much? By
the amount of the blue area under the x-axis. So by this amount.
[00:04:44.80] And on this region of the graph, we also have that the y value of the red graph,
which is uppercase F of x, is going to be equal to the integral so far. So it's going to be the area
above the x-axis, minus the area below the x-axis. And it keeps going down until-- once these
two areas compensate for each other, the accumulated area function becomes zero because we
have the same area added and subtracted.
[00:05:20.27] So going back to these pictures, as I said, this is now-- if we take x to be on the
other side of the pi-- on the pi, this is the point where the accumulated area graph would be the
highest. Why? Because we have finished adding all the areas we have from 0 to 2 pi. And now
we're subtracting the areas below the x-axis. So this is as big as it gets at this point right here.
And then from that point on, it starts decreasing. And as you saw, it's going to decrease for
steeper than [INAUDIBLE] all the way on this way.
[00:06:00.15] And once again, the value of the uppercase F for x is the y-coordinate at this point
corresponding to this x. And that's going to be area 1 minus area 2. Where this is area 1, and
we're subtracting area 2.
[00:06:21.97] OK, so I'm insisting a little bit on this because I would like you to make sure to
understand what this accumulation function does, and how it corresponds to various values of x
along the t-axis. In particular, once you understand this function, you should also understand that
this function differentiated corresponds to the original function, lowercase f of x for all x of an
interval from a to b. So in other words, uppercase F is an antiderivative of lowercase f. And vice
versa, lowercase f f is the derivative of uppercase F.
[00:07:03.91] And you can kind of see this on some of these pictures. So what we're claiming is
that lowercase f is the derivative of uppercase F. So that would be the tangent line at that point.
So going back here, the slope of the tangent line at this point to uppercase F would actually
correspond to the value of lowercase f. And you can see that the best right here are the top.
Because the rate of change of uppercase F at that point is 0. It's got the horizontal tangent line.
And you can see that the value of lowercase f at that point is 0 as well.
[00:07:49.95] And as an example, if we take the derivative with respect to x of integral from
your favorite constant, let's say 7, to x of t squared tangent t, we can make this function as crazy
as we want. We don't have to worry about anything because by the time we differentiate it, we
don't have to actually integrate this function. If we had to integrate it first, that would be
complicated and difficult.
[00:08:24.86] But what this theorem is telling us, part 1, is that all we have to do is take the
integrand, this would be our lowercase f of t, and plug in the upper bound when we're computing
the derivative. So the integrand is a function of t, like sin of t here. By the time we take the
definite integral from 7 to x, we get the function of x, where x varies along the t-axis from a to b.
In this case, from 7 to whatever, pick a number you want. And the derivative with respect to x at
that particular x of this function is going to be just f of x. So in other words, it's going to be x
squared tangent of x.
[00:09:12.37] We don't need to integrate, and we do not need to add plus c because we are not
doing a indefinite integral here. Rather, we're doing the derivative of a definite integral with a
variable upper bound.
[00:09:25.33] The second part of the fundamental theorem of calculus tells us what happens
when we integrate a derivative. It says that if f is a continuous function on a closed interval from
a to b, and uppercase is any antiderivative of lowercase f-- that means, by the way, if you
remember, that lowercase f is the derivative of uppercase f. So that's what I mean by integrating
a derivative. Lowercase f of x is actually the derivative of uppercase f.
[00:10:04.36] So if we have any such antiderivative of lowercase f, then we can compute the
definite integral from a to b of lowercase f of x dx by computing the difference between the
values of uppercase F at the upper bound b. So notice the order in which the upper bound and the
lower bound get plugged into the antiderivative. First we compute the antiderivative uppercase f
at the upper bound b. And then we subtract the value of the antiderivative at the lower bound a.
[00:10:52.58] So another way to write this is integral from a to b of uppercase f prime of x with
respect to x is-- now remember and notice here that we're talking about an actual definite
integral. The upper bound and the lower bound are both constant. So the integral from a to b in
this case is a number corresponding to the sined area under or over the graph of lowercase f of x
as x ranged from a to b.
[00:11:27.27] So this definite integral is just a number. What this theorem is saying is that we
can compute that number via this difference between the values of the any antiderivative of it at
the upper and lower limits.
[00:11:45.70] Now the second part of fundamental theorem of calculus, this result we're looking
at right now, is actually the most useful part of FTC because it does us a great service. It allows
us to compute definite integrals if we just can find an antiderivative with of the integrand. So if
you're trying to compute integral from a to b-- for example, let's say we're trying to compute
integral from 0 to pi over 2 of cosine of x, with respect to x. If we can find any antiderivative of
this-- and what would be an antiderivative of cosine?
[00:12:37.14] That would be a function whose derivative would be equal to cosine of x. Do we
know any such function? Well, from Math 124, we know that the derivative of sine of x is equal
exactly to cosine of x. So one antiderivative of cosine of x would be sine of x. And to denote the
difference uppercase F of b minus uppercase F or a-- the shorthand notation we use for that is,
we write the antiderivative, for example, sine of x. Then we put a vertical bar, and we put the
bounds. In this case, that would b from 0 to pi over 2 lower bound and upper bound.
[00:13:20.27] Don't confuse this bar with an integral. Here we're talking about the integral of
cosine. Here we're just meaning we're going to take this function, which is sine of x, and we're
going to plug in the upper bound first, and then the lower bound, then compute the difference. So
this means that we're doing sine of pi over 2 minus sine of 0. And right here, this was FTC part
2. That's what we used.
[00:13:50.29] Sine of pi over 2 is 1, and sine of 0 is 0. So that means that this integral, its value is
equal to-- let me put it here. Its value is equal to 1. So what just happened? We took the function
y is equal to cosine of x between x is 0 and x is pi over 2. And we computed the definite integral
over this interval, which corresponds to that area right there. So we found out that this area here
is exactly 1.
[00:14:39.68] So this is what FTC part 2 does for us. It allows us to compute an integral such as
this, as long as we can find one antiderivative. And then we can plug in the upper and the lower
bounds, and take the difference. The other important part of this is that this is a difference. It's a
minus sign.
[00:15:00.05] Well, that's great, but why in the world would this be true? In fact, it follows pretty
easily from the fundamental theorem of calculus part 1. We want to show this result, that the
integral from a to b of lowercase f of t with respect to t is uppercase F of b minus uppercase F of
a for any antiderivative uppercase F of lowercase f.
[00:15:30.95] And I just claimed that this is an easy consequence of FTC 1. Why? Because FTC
1 gave us an antiderivative. It gave us a way to construct the antiderivative of lowercase f, albeit
not its mathematical expression. But still, we can construct an antiderivative for any continuous
function by doing what? By taking the accumulated area so far.
[00:15:59.79] And that's exactly what FTC 1 told us. It told us that we can do this construction,
and that its derivative is lowercase f, which makes this an antiderivative of lowercase f. Not only
do we have an antiderivative of any lowercase f you want by this construction, but if you use a in
the upper bound to compute what the value of this function is when x is equal to the starting
point a, then we have integral a to a, the same lower and upper bound. So that means we're going
to get that uppercase F of a is 0. That was one of the properties of intervals. If you have the same
upper and lower bound, the integral is 0.
[00:16:43.60] What is uppercase F of b here? According to the definition of uppercase F, that
would be the integral a to b of lowercase f of t dt. So uppercase F of b is exactly the definite
integral we are trying to compute. But now we're looking at a specific antiderivative here. And
we want to prove that it holds for all. We're going to start by proving that it works for this
particular one.
[00:17:18.27] So integral from a to b of lowercase f of t dt. If we use the area so far
antiderivative, that's going to be exactly F of b. But that's the same as F of b minus zero, because
subtracting 0 doesn't change anything. And remember that uppercase F of a was exactly 0. So
that means that FTC part 2 holds for this particular antiderivative, which we constructed via the
accumulation function.
[00:18:00.96] Now, we show that it works for an antiderivative, the accumulation function. Now
we need to extend it to any antiderivative. So what happens if you have another antiderivative?
How do any two antiderivatives of a function differ from each other? We saw at the beginning of
the term when we discussed Section 4.9 that if we have any two antiderivatives of a function,
they differ by a constant.
[00:18:35.31] So if we take any other antiderivative of lowercase f, let's call it an uppercase G
because we're running out of cases for f, then G of x have to be equal to the accumulation area
function uppercase F, plus a constant. So this means that if you're computing G of b minus G of
a, that would be plugging into the expression for G of x is F of x plus c. We get F of b plus c,
minus F of a plus the same constant c.
[00:19:22.12] Now when we distribute the sign, this c cancels out with this c. So the constant
doesn't matter. And we get that the difference between the antiderivative at a and b is the same
regardless which antiderivative use. So in particular, since we already showed that this difference
is equal to the definite integral, then G of b minus G of a is indeed equal to integral from a to b of
f of t dt for any antiderivative G of f.
[00:20:06.35] And this proves part 2 based on part 1. This proof is optional. But as I said before
in the previous lecture, it's always good to understand why things work, how they work, rather
than just trying to memorize the formulas. It helps your deeper understanding of the material.
And it helps you familiarize yourself with the concepts a little bit more.
[00:20:33.91] And for some people, like me for example, some people are not satisfied until they
understand why things work a certain way. If you're not concerned with that aspect, then feel free
to skip over the proof. But again, there would be advantages to understanding the proofs, even if
you don't particularly care for proofs.
[00:20:54.51] OK, what is not optional is being able to actually apply this theorem to compute
definite integrals. So here are a couple of relatively simple examples of this.
[00:21:04.61] Let's suppose we want to compute integral from 0 to 10 of t squared with respect
to t. What FTC 2 is telling us is, integral from a to b of f of t dt is equal to any antiderivative of
the function inside the integral with the bounds plugged in. First the upper one, then subtract the
value at the lower one.
[00:21:35.36] So we need to find an antiderivative of t squared. And by what we were doing in
Section 4.9 at the beginning, we know that we can take 1/3 t cubed. Why? Because 1/3 t cubed
differentiates to t squared. So this is one antiderivative of the function. It doesn't matter which
one we pick, so might as well pick the simplest one. And then plug in the bounds from 0 to 10.
[00:22:00.11] So we get 1/3 times 10 cubed minus 1/3 times 0 cubed. This is FTC here, part 2.
[00:22:14.46] Now this would be 1/3 times 1,000, and the second part is 0. So the answer would
be 1,000 over 3. Or if you prefer, this would be 333 and 1/3. Now in calculus, we typically prefer
exact answers to approximations via couple of decimals. So in general, it's always safer to leave
your answer in simplified but exact form. So instead of writing 333.33 and putting an equals
sign, these two quantities are not exactly equal. This one is an approximation to two digits of this
more precise number.
[00:22:56.55] So it's generally better to just leave it in the exact form, not to round it to various
number of decimals. Unless otherwise instructed, of course.
[00:23:06.18] OK, second example. We have integral from 2 to e, you remember is the special
constant 2.718281828, keeps going forever. It's one of those irrational numbers. And we're
integrating from 2 to e the function of 1 over x. Again, by FTC part 2, all we need is an
antiderivative of the integrand 1 over x. So do we know any function whose derivative is 1 over
x?
[00:23:42.24] We do. That was the derivative of natural log of x. And in fact, and it's going to be
important to remember this later in the quarter, we can extend it. If we just put the natural log of
x like this, this function is only defined for values of x greater than 0. However, you can show
with a little bit of effort that we can look at natural log of absolute value of x, which is defined
everywhere but at x is equal to zero. And its derivative will be 1 over x, regardless of whether x
is positive or negative.
[00:24:23.30] So when we do integral of 1 over x, we don't get natural log of x, we get natural
log of absolute value of x. And that's one antiderivative. And we had the bounds from 2 to e.
[00:24:39.51] Notice a couple of things. While we're still computing the integral, there should be
an integral at the front, this part, and there should be a dx at the end. Once we're done computing
the integral because we found an antiderivative and we're just plugging in bounds, the integral
sign should go away, the dx part should go away. We should just have the value of the
antiderivative. And then if you have bounds, then you should put the bounds here with a vertical
bar.
[00:25:07.29] Then we're going to plug in the bounds. So we get natural log of e. That's a
positive number, so I don't have to worry about absolute value, minus natural log of 2. Natural
log of e, that's 1. You should know these special values. And natural log of 2, that's nothing
special. So we're going to leave it in exact form as 1 minus natural log of 2. Simplified but in
exact form. If you want to put this in a calculator, this would be approximately equal to 0.31.
[00:25:47.99] So what we computed here is that if we're looking at the function 1 over x from 2
to e, which is slightly bigger than 2, this area here, or the function is positive throughout the
interval. So the area under the graph is exactly the definite integral. This area here is equal to 1
minus natural log of 2.
[00:26:20.38] One last example. We'd like to compute the integral from 0 to 1/2 of 6 divided by
the square root of 1 minus z squared, with respect to z. So here z the name of our variable. Can
we simplify the integrand a little bit? Yes. What we have inside the integral is really a constant
multiple 6, multiplied by square root of 1 minus z squared.
[00:26:52.75] And one of our properties for integrals told us that if we have a constant multiple
of a function-- so if there is a constant multiplied by everything else inside the integral, we can
factor it outside of the integral So we get 6 times the integral from 0 to 1/2 of 1 over square root
1 minus z squared integral with respect to z.
[00:27:14.82] Now the integrand here, 1 over square root of 1 minus z squared, cannot be
simplified any further. At least not in any useful way. So it's time to ask ourselves if we can
recognize this as the derivative of anything we know. In other words, do we know an
antiderivative of this function? And we should know. This is one of the standard formulas from
Math 124.
[00:27:40.85] This function is the derivative of arcsine, or sine inverse of z. So applying FTC
part 2, we can write this integral as arcsine of z where we plug in the bounds from 0 to 1/2.
[00:28:11.76] And remember, quick reminder here, arcsine of z is the same sine inverse of z.
Both notations mean exactly the same thing. This is the inverse trigonometric function of sine of
z. And it's not equal in particular. Be very careful. It's a common mistake. It's not the reciprocal
of sine of z. Rather, it's the function that is the mathematical inverse of sine.
[00:28:43.80] Here is sine between negative pi over 2 and pi over 2. And this would be sine of,
let's say x, and this would be z. So arcsine of z is the inverse function. It takes a value between
negative 1 and 1, and it returns the angle between negative pi over 2 and pi over 2 that
corresponds to that value.
[00:29:13.38] So here we're going to get 6 times arcsine of 1/2 half minus arcsine of 0. This
would be the angle between negative pi over 2 and pi over 2 whose sine is equal to 1/2. So we're
looking for an angle, sine of x is equal to 1/2. And x should be between negative pi over 2 and pi
over 2.
[00:29:45.66] So which angle has sine equal to 1/2? Again, you should know the special values
of trig functions, exponential function, logarithmic functions. And this is one of them. Sine of 30
degrees, or pi over 6, is equal to 1/2. So arcsine of 1/2 is equal to pi over 6. And arcsine of 0,
well, which angle has sine equal to zero? Which angle between negative pi over 2 and pi over 2?
That would be the 0 angle. So altogether we get 6 times pi over 6. So we get pi.
[00:30:29.77] So this integral from 0 to 1/2 of 6 divided by square root of 1 minus z squared is
equal to pi.
[00:30:41.52] A related result to the fundamental theorem of calculus part 2 is this theorem that
actually is presented in Section 5.4 in your textbook. It's called the net change theorem because it
tells us how we can compute the net change of a function by integrating its rate. Or backwards, it
says the definite integral of the rate of change of a function uppercase F over an interval from a
to b equals the net change in F over that interval.
[00:31:18.97] Which, in mathematical notation, would mean the integral from a to b of F prime
of x, the rate of change of F-- that's the derivative, with respect to x is equal to the net change.
Which means delta F, a change in F from a to b. So that would be exactly F of b minus F of a.
[00:31:41.74] Now notice something interesting, here. Haven't we seen this before? Hmm, that
sure looks a whole lot like this result, right? That would be fundamental theorem of calculus part
2. And the reason it looks a lot like it is because it is actually it. So putting down here as a note,
the net change theorem is really saying the same thing as the fundamental theorem of calculus
part 2. The difference between them is just one of perspective. So it's simply in the way we
interpret the two theorems.
[00:32:52.63] In particular, when we're talking about the net change theorem, we're thinking
about the change in some specific function that we worry about, and computing that via an
integral.
[00:33:08.00] One other note. The units for the net change in F will be the units for rate the
prime of x times the units for the variable x. This is because-- we talked about last time briefly
the units for an integral of any function really, as long as your function inside the integral has
units are the units for the function inside the integral times the units for the variable x.
[00:33:45.52] And once again, what this theorem is saying is that the change in F is integral from
a to b of its rate, f prime of x with respect to x. This is called a net change. And this would be the
integral from a to b of the rate of F.
[00:34:24.86] OK, let's look at a couple of examples of applying this theorem to various
situations. So let's start with a simple one. Suppose I have a situation in which the population of
a country grows at a constant rate of, let's say, 50,000 people per year. In this context, let's use
the fundamental theorem of calculus part 2 to compute the integral from 0 to 5 of 50,000 with
respect to t. And after we do so, let's figure out what the value we computed actually represents
using the net change theorem.
[00:35:05.32] Why the net change? Because 50,000, the quantity inside the integral, is a rate. In
particular, it's the rate of the population. So let's set up some notation. Let P of t denote the
population of this country after t years. And one thing that you're going to see over and over
again with a lot of the problems, if they don't give us a name for the functions or the variables
we're trying to use in a problem, you need to feel free to set up your own notation and your own
definitions. Just make sure that they are consistent, and that they actually give you what you need
to. But it's very important to actually write these things down so you remind yourself, and the
grader when you're writing on an exam, what in the world each letter means, what are the units,
and so on.
[00:36:13.49] So that's why I wrote uppercase P. That will be my name for the population of the
country. And let's also write down what we know. We know that the rate of change of P with
respect to time-- notice it's people per year, and they told us it's a rate. So the rate of change of P
with respect with time t is 50,000. And that's people per year.
[00:36:43.40] Now let's compute this integral by FTC part 2. All we need is one antiderivative of
50,000. So that would be 50,000 times t. t is the name of our variable here. That's what the dt
tells us. That time. So we get 50,000 times t. That indeed-- let's check it differentiates to 50,000.
And we're going to plug in the bounds from 0 to 5. So we get 50,000 times 5, minus 50,000
times 0. That gives us 0.
[00:37:21.04] Always plugged the lower bound in, even if it's zero. Because occasionally you
will actually be unfortunate enough to have a function that doesn't cancel out at 0. And we
should assume that just because the lower bound is 0 you have nothing here, you are going to be
wrong in some of the cases. So 5 times 50,000, that gives us 250,000. OK, 250,000 what? Well
the 50,000 was in people per year. And we integrated, so we have to multiply by the units for t. t
was the time in years. So dividing by years or multiplying by years, that cancels out. So this is
250,000 people.
[00:38:12.11] OK, what does this represent, the 250,000 people? By the net change theorem, we
had integral from 0 to 5 of 50,000. That was the derivative of the rate of change of population
with respect to time. And we're integrating with respect to time. So that gives us the change in
population. So population at 5 minus population at 0. So 250,000 people represents the net
change in population during the first five years.
[00:39:21.62] Or in other words, we could say that the population of this particular country grew
by 250 people. Notice that it's a positive change, so that means it went up by that much. It grew
by 250,000 people over the first five years.
[00:39:40.71] OK, let's try another example. Suppose that water drains from a reservoir at a rate
dv over dt 1,000 gallons per minute. So here they came up with some letters for us. And they
using v for something that's in thousands of gallons. So volume rate with respect to time. So we
have a rate of volume with respect to time. They also tell us that no water comes into the
reservoir. So all the change in the water of the reservoir is given by this mystery rate. They didn't
tell us what it is. Just dv over dt.
[00:40:34.14] They are trying to figure out how we can compute the net change of the volume of
water in the reservoir. So by how much is the water in the reservoir-- well, they said that it
drains. So this rate is going to be a negative number, which means that the volume of water in
the reservoir goes down. We're trying to figure out by how much it goes down in thousands of
gallons from t1 to t2 minutes.
[00:41:01.33] So the way we would compute that, the change in the volume, would be the
integral from time 1 to time 2, whatever those times are, of the rate of change of the volume with
respect to time. Then we're to integrate with respect to time.
[00:41:23.88] The final and most complicated example has to do with displacement versus total
distance. And this is very important to understand. So let's suppose we have the situation where a
particle travels in a straight line with velocity t seconds given by this quadratic function of time.
So the velocity at time t is t squared minus 5t plus 6. And it's measured in meters per second, t's
in seconds.
[00:42:04.12] First question. Let's just compute the integral from 0 to 3 of d of t with respect to t.
Then list the units for this integral. And what exactly does that number we computed represent?
Before we do that, let me just draw a quick picture here. This is time in seconds. And we're given
the velocity in meters per second.
[00:42:33.37] We have a quadratic. So its graph is a concave parabola. And we can figure out its
roots, where the x intercepts, where it crosses the x-axis. We can do that by using the quadratic
formula, or by factoring this expression. It actually factors nicely into t minus 2 times t minus 3.
Because negative 2 plus negative 3 give us 5. And multiply together to give us 6. If you don't
like that, or you are not comfortable with factoring off the top of your head, then use the
quadratic formula.
[00:43:10.48] So what does the graph of the velocity look like? It's a concave parabola. Its yintercept is 6. So it's going to start here. That's when t is 0. And the x-intercepts are, let's say, 1,
2, 3. So it goes down, then it dips below the x-axis, then it goes back up. Something like this. So
this the velocity at time t. We're going to come back to these a bit later.
[00:43:48.41] Now let's go to our integral. Integral from 0 to 3 of the velocity with respect to
time. This would be integral from 0 to 3. We were asked to compute it, so let's do so. We have
the integral of t squared minus 5t plus 6, with respect for time t. We need one antiderivative. If
we compute an antiderivative of t squared by guess and check, that would be 130 cubed. And
minus 5. And antiderivative of t would be 1/2 t squared. And then an antiderivative of 6 would
be 6t.
[00:44:31.31] And we don't have to worry about constants because we're not computing an
indefinite integral with no bounds. Rather, we have bounds to plug in. And it doesn't matter
which antiderivative we use for that. So we're going to plug in 3 first. 1/3 3 cubed minus 5/2 3
squared plus 6 times 3 minus-- then we're going to plug in 0. But notice that there is a t in every
single term, so that means we're going to get 0 when we plug in the lower bound.
[00:45:07.76] Computing further, three cubed divided by 3 is 3 squared. That's 9. This is 5 times
9 is 45 divided by 2, and 6 times 3 is 18. So we get 27-- 9 plus 18, minus 45 over 2. Bringing to
common denominator, we get 54 minus 45 divided by 2. 54 minus 45 is 9, divided by 2, so we
get 4.5. So this integral that we just computed is 4.5.
[00:45:49.61] What are its units? Well, they will be the units for velocity times the units for time.
So this would be meters per second times second, cancel out. So this is in meters.
[00:46:10.31] So this integral is equal to 4.5 meters. What does it represent? The velocity, you
remember from Math 124, or even this quarter, is the rate of change of position-- not distance,
but position, really, with respect to time. So S is the usual way to denote position in meters in
this case at t seconds.
[00:46:52.55] So what we computed when we did integral from 0 to 3 of the velocity with
respect to time, we computed the integral from 0 to 3 of the rate of change of position with
respect to time. So by the net change theorem, this gives us the change in the position S. And
that's going to be the position at time 3, minus the position at time 0. So this is the change in
position-- this or this of the particle. Or it's also known as displacement because it's how far it
traveled, how much it was displaced from time 0 to time 3 seconds.
[00:48:01.27] So the answer is integral from 0 to 3 of v of t with respect to t is equal to 4.5
meters. Which is the displacement, or the change in position for the first three seconds. So if we
integrate the velocity, we get the displacement.
[00:48:43.92] And I'd like to make a point here before we move on to the second part of this
example. And that's the last thing I'm going to talk about in this lecture. Remember the graph of
the velocity, it looks something like this. And we computed the definite integral of this function,
the velocity at time t from 0 to 3. So we have these area 1-- let me make it-- green is a good color
for it. And then the second area, area 2-- yeah, it's going to be too small. Let me put an arrow like
this, and write it here. Area 2.
[00:49:54.10] The second area corresponds to the portion between 0 and 3 where the velocity is
negative. So that gets subtracted. So my point is that if we take the integral 0 to 3 of V of t dt
which is this 4.5, another way to think about it is-- area 1 minus area 2. This area from 0 to 2,
minus the area from 2 to 3.
[00:50:31.18] OK, now we're going to come back to this in a little while. But let's take a look at
the second question. Second question is asking us to compute the total distance traveled by this
particle in the first three seconds. So the same time interval as in part a. Haven't we just
computed that? Didn't we just compute the integral from 0 to 3 of the velocity? And shouldn't
that be the distance traveled? Well, not exactly. You have to be a little bit careful.
[00:51:02.82] What we computed in part a with the integral was the displacement of the particle
over the first three seconds. Let's ask ourselves for a second, what does it mean when the
velocity is negative? Below the x-axis. So from two to three seconds. Let's assume that this is the
line on which the particle is moving. And let's assume that the particle starts here at time 0. And
let's assume that the positive direction of motion is this way.
[00:51:41.35] Velocity being positive from 0 to 2 seconds means that the position is increasing,
going in the positive direction, from 0 to 2 seconds. By how much? Well, when the velocity is
above the x-axis, or in this case the t-axis, the area under the graph is exactly the distance. So this
particle has moved in the positive direction for a while. By how much? This distance that it has
traveled here is exactly equal to area 1.
[00:52:26.84] Then at 2 seconds-- right here we're at two seconds, what happens? The velocity
becomes negative. What does that mean? It means that now the car is moving in the negative
direction, which is this way. And it also means that the position is decreasing because it's
derivative is negative. So as the position is getting smaller.
[00:52:51.27] So now the car is moving in this way up to, let's say, here. We get the time is equal
to 3. How big is this distance that the car has traveled backwards from time 2 to time 3? Well-the particle, not the car, story. This distance that the particle has traveled backwards is equal to
area 2, the area we're subtracting. So notice that if we take that distance the car has moved
forward and subtract the distance that the car has moved backwards-- so this way minus this
way, what we get here-- and then me use a third color. Why don't we use blue?
[00:53:45.75] So this here, this is the displacement or the change in position. So this was the 4.5
meters. Now how is that different from the total distance traveled? Well, the total distance travel
would be the green length, how far it traveled forward, plus the red length, the distance it
traveled backwards. So instead of wanting to subtract area 2 from area 1, we would want to add
them together.
[00:54:20.90] OK, so back to part b, trying to compute the total distance traveled by this particle
in the first 3 seconds. Which we said that what we want is the sum of these two areas. So the sum
of the distance it's traveled, regardless of direction. So we want area 1 plus area 2. Which means
if you think about the graph of the velocity, that instead of subtracting the portion that's below
the x-axis, we want to add it as if the functional were actually positive. Like this.
[00:55:10.47] To do that, it would be enough to take the absolute value of the function. Then all
negative portions like this would become positive. So what we can do is, we can take the integral
from 0 to 3 of absolute value of the velocity, with respect to time. That's what gives us the total
distance. We need to discard the negative portions. Not discard them entirely, but rather, count
them as positive.
[00:55:39.85] So how do we compute the integral of an absolute failure of a function? We need
to break the function into pieces, the pieces that are positive. Well, absolute value of V of t-absolute value of anything, really, is either the thing, so V of t-- and that happens when V of t is
greater or equal to 0. Which means on the interval from 0 to 2.
[00:56:11.65] And when the function is negative, we want to make it positive to get its absolute
value. One way to do that with a negative quantity is to take its negative. So we take negative V
of t when V of t is less or equal to 0, which means on the interval from 2 to 3. So what we need
here is the integral from 0 to 2 of the velocity. On this interval, V of t is positive, so its integral is
actually area 1. Plus the integral from 2 to 3 of the negative of the velocity with respect to time.
[00:56:56.70] Because the velocity itself was negative on this interval, so putting a negative sign
in front of it makes it positive. And now we're integrating the positive amount, and we're getting
exactly area 2, not its negative.
[00:57:13.29] All right, so let's do it. Integral from 0 to 2, the velocity was t squared minus 5t
plus 6. We're integrating with respect to t. And then we have to separately compute the integral
from 2 to 3 of the negative of the velocity. So negative t squared plus 5t minus 6.
[00:57:35.37] Now we need to compute each one of these two integrals separately. For the first
one, we get an antiderivative of t squared minus 5t plus 6 is 1/3 t cubed, minus 5/2 t squared plus
6t evaluated between the bounds of zero and 2. And then separately for the second one, we're
going to get negative t cubed over 3, plus 5 over 2 t squared, minus 6t evaluated between the
bounds of two and 3.
[00:58:16.78] First integral. Let's put some square brackets around everything because we're
going to have to subtract things. And then we have the second one in square brackets, too. So
replacing 2 in for t cubed, we get 8/3 minus 5/2 times 4, plus 6 times 2 is 12. And when we plug
in the lower bound, we see that every term cancels out. So we get just negative zero. So
continuing with the first one this is 8/3 plus 2. 2 times 3 is 6, bringing to common denominator, 6
plus 8 is 14. So this entire expression here is 14 over 3. That's our first area.
[00:59:12.47] Plus. Now integrating the second function, the negative of V of t gave us this
antiderivative. It's time to replace the bounds. So the first time we put in the upper bound, minus
3 cubed over 3, plus 5/2 times 3 squared, minus 6 times 3, which is 18. And then we subtract
from that this whole expression with two replaced in it. So minus 8/3, plus 5/2 times 4, minus 12.
[00:59:51.38] Now notice that this expression here is exactly what we computed up above,
because we plugged in 3 already when we were computing this. Except that we were plugging 3
into the velocity, not its negative. And we got 9/2, or 4.5 meters. So that helps us fill out our
algebra here. This would be negative 5/2. And then we're subtracting. Well, this expression
should also be familiar. It's exactly this expression here, but with a negative sign distributed
everywhere.
[01:00:24.76] So this expression would be negative 14 over 3. So first area is 14/3 plus-- and
here we have negative 9/2, plus 14 over 3. Bring to the common denominator. Multiply the first
one by 3 and the second one by 2 gives us negative 27 over 6, plus 28 over 6. So altogether we're
left with 1 over 6.
[01:01:02.39] Now combining these two together, we need to multiply this one by 2. We get 28
plus 1, 29 over 6 meters. So the total distance traveled in the first three seconds is 29/6 of a
meter. And we saw that these area 1, this is the one that corresponds to the first integral. So that's
14/3. And the second area is the one that correspond to the second integral, so it's 1/6.
[01:01:57.84] And really, the difference between displacement and total distance is the fact that,
for displacement we took this first 14/3 and subtracted the 1/6. And for the total distance, we
added them up. We could do the first without worrying about t is equal to 2 at all by just taking
the definite integral. Because that makes sure you subtract the negative parts.
[01:02:20.87] But for the total distance, we had add the two pieces. So we have to actually
compute each one of them separately, and then add them together by this breaking of the integral
into positive values or V of t and negative values of V of t.
[01:02:36.09] OK, I'm going to stop here. I will see you guys next time.
[01:02:39.78]