WEBVTT

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Let's say that I have
some function, s of t,

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which is positioned
as a function of time.

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And let me graph a potential
s of t right over here.

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We have a horizontal
axis as the time axis.

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Let me just graph something.

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I'll draw it kind
of parabola-looking.

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Although I could
have done it general,

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but just to make things a
little bit simpler for me.

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So I'll draw it kind
of parabola-looking.

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We call this the y-axis.

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We could even call this y equals
s of t as a reasonable way

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to graph our position as a
function of time function.

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And now let's think
about what happens

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if we want to think about the
change in position between two

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times, let's say between time
a-- let's say that's time

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a right over there-- and then
this right over here is time b.

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So time b is right over here.

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So what would be the change
in position between time a

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and between time b?

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Well, at time b, we
are at s of b position.

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And at time a we were
at s of a position.

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So the change in
position between time

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a and time b-- let me write this
down-- the change in position

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between-- and this
might be obvious to you,

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but I'll write it
down-- between times a

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and b is going to be equal
to s of b, this position,

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minus this position,
minus s of a.

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So nothing
earth-shattering so far.

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But now let's think
about what happens

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if we take the derivative of
this function right over here.

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So what happens when
we take the derivative

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of a position as a
function of time?

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So remember, the
derivative gives us

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the slope of the tangent
line at any point.

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So let's say we're looking
at a point right over there,

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the slope of the tangent line.

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It tells us for a very
small change in t--

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I'm exaggerating it visually--
for a very, very small change

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in t, how much are we
changing in position?

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So we write that as ds dt is
the derivative of our position

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function at any given time.

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So when we're talking about the
rate at which position changes

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with respect to
time, what is that?

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Well, that is equal to velocity.

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So this is equal to velocity.

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But let me write this
in different notations.

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So this itself is going
to be a function of time.

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So we could write this
is equal to s prime of t.

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These are just
two different ways

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of writing the derivative
of s with respect to t.

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This makes it a
little bit clearer

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that this itself is
a function of time.

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And we know that this is the
exact same thing as velocity

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as function of time, which
we will write as v of t.

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So let's graph what v of t
might look like down here.

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Let's graph it.

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So let me put another
axis down here that

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looks pretty close
to the original.

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I'll give myself
some real estate,

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so that looks pretty good.

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And then let me try
to graph v of t.

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So once again, if this is my
y-axis, this is my t-axis,

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and I'm going to graph
y is equal to v of t.

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And if this really
is a parabola,

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then the slope over here is
0, the rate of change is 0,

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and then it keeps increasing.

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The slope gets steeper
and steeper and steeper.

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And so v of t might look
something like this.

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So this is the graph of
y is equal to v of t.

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Now, using this
graph, let's think

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if we can conceptualize
the distance, or the change

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in position, between time
a and between time b.

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Well, let's go back
to our Riemann sums.

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Let's think about what an
area of a very small rectangle

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would represent.

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So let's divide this into
a bunch of rectangles.

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So I'll do it fairly
large rectangles

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just so we have some
space to work with.

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You can imagine
much smaller ones.

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And I'm going to do a
left Riemann sum here,

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just because we've
done those a bunch.

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But we could do it
right Riemann sum.

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We could do a trapezoidal sum.

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We could do anything we want.

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And then we could keep going
all the way-- actually,

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let me just do three right now.

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Let me just do three
right over here.

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And so this is actually a
very rough approximation,

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but you can imagine
it might get closer.

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But what is the area of
each of these rectangles

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trying-- what is it
an approximation for?

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Well, this one right over
here, you have f of a,

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or actually I should say v of a.

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So your velocity at time a is
the height right over here.

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And then this distance
right over here

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is a change in
time, times delta t.

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So the area for that
rectangle is your velocity

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at that moment times
your change in time.

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What is the velocity
at that moment

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times your change in time?

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Well, that's going to be
your change in position.

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So this will tell you--
this is an approximation

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of your change in
position over this time.

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Then the area of this rectangle
is another approximation

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for your change in position
over the next delta t.

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And then, you can imagine,
this right over here

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is an approximation
for your change

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in position for
the next delta t.

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So if you really wanted
to figure out your change

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in position between
a and b, you might

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want to just do a Riemann sum
if you wanted to approximate it.

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You would want to take
the sum from i equals 1

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to i equals n of v of-- and
I'll do a left Riemann sum,

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but once again, we
could use a midpoint.

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We could do trapezoids.

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We could do the
right Riemann sum.

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But I'll just do a left
one, because that's

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what I depicted right
here-- v of t of i minus 1.

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So this would be t0, would be a.

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So this is the first rectangle.

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So the first rectangle, you use
the function evaluated at t0.

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For the second
rectangle, you use

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the function evaluated at t1.

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We've done this in
multiple videos already.

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And then we multiply it times
each of the changes in time.

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This will be an
approximation for our total--

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and let me make it
clear-- where delta t is

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equal to b minus a over the
number of intervals we have.

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We already know, from
many, many videos

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when we looked at
Riemann sums, that this

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will be an approximation
for two things.

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We just talked about it'll be
an approximation for our change

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in position, but it's also an
approximation for our area.

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So this right over here.

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So we're trying to approximate
change in position.

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And this is also approximate
of the area under the curve.

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So hopefully this
satisfies you that if you

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are able to calculate the area
under the curve-- and actually,

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this one's pretty easy,
because it's a trapezoid.

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But even if this was a function,
if it was a wacky function,

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it would still apply
that when you're

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calculating the area under the
curve of the velocity function,

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you are actually figuring
out the change in position.

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These are the two things.

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Well, we already
know, what could we

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do to get the exact
area under the curve,

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or to get the exact
change in position?

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Well, we just have
a ton of rectangles.

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We take the limit as
the number of rectangles

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we have approaches infinity.

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We take the limit as
n approaches infinity.

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And as n approaches
infinity, because delta t is

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b minus a divided
by n, delta t is

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going to become
infinitely small.

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It's going to turn into dt,
is one way to think about it.

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And we already have
notation for this.

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This is one way to think
about a Riemann integral.

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We just use the
left Riemann sum.

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Once again, we could use the
right Riemann sum, et cetera,

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et cetera.

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We could have used a
more general Riemann sum,

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but this one will work.

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So this will be equal
to the definite integral

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from a to b of v of t dt.

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So this right over here is
one way of saying, look,

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if we want the exact area under
the curve, of the velocity

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curve, which is going to be
the exact change in position

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between a and b, we
can denote it this way.

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It's the limit of this Riemann
sum as n approaches infinity,

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or the definite integral
from a to b of v of t dt.

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But what did we just figure out?

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So remember, this is
the-- we could call this

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the exact change in position
between times a and b.

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But we already figured
out what the exact change

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in position between
times a and b are.

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It's this thing right over here.

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And so this gets interesting.

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We now have a way of evaluating
this definite integral.

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Conceptually, we
knew that this was

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the exact change in
position between a and b.

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But we already figured
out a way to figure out

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the exact change of
position between a and b.

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So let me write all this down.

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We have that the definite
integral between a and b

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of v of t dt is equal
to s of b minus s of a

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where-- let me write this in
a new color-- where s of t

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is the-- we know v of t is
the derivative of s of t,

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so we can say where s of t is
the antiderivative of v of t.

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And this notion,
although I've written it

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in a very nontraditional--
I've used position velocity--

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this is the second fundamental
theorem of calculus.

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And you're probably
wondering about the first.

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We'll talk about that
in another video.

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But this is a super
useful way of evaluating

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definite integrals
and finding the area

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under a curve, second
fundamental theorem

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of calculus, very closely
tied to the first fundamental

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theorem, which we
won't talk about now.

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So why is this such a big deal?

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Well, let me write it in
the more general notation,

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the way that you might
be used to seeing it

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in your calculus book.

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It's telling us that
if we want the area

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under the curve between
two x points a and b

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of f of x-- and so this
is how we would denote

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the area under the curve
between those two intervals.

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So let me draw that
just to make it clear

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what I'm talking about
in general terms.

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So this right over
here could be f of x.

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And we care about the area
under the curve between a and b.

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If we want to find the
exact area under the curve,

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we can figure it out by taking
the antiderivative of f.

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And let's just say that capital
F of x is the antiderivative--

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or is an antiderivative, because
you can have multiple that

00:11:36.095 --> 00:11:43.916
are shifted by constants--
is an antiderivative of f.

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Then you just have to take--
evaluate-- the antiderivative

00:11:47.330 --> 00:11:49.440
at the endpoints and
take the difference.

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So you take the endpoint first.

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I guess you subtract the
antiderivative evaluated

00:11:56.230 --> 00:12:00.200
at the starting point from
the antiderivative evaluated

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at the end point.

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So you get capital F of
b minus capital F of a.

00:12:07.871 --> 00:12:10.370
So if you want to figure out
the exact area under the curve,

00:12:10.370 --> 00:12:13.030
you take the
antiderivative of it

00:12:13.030 --> 00:12:16.050
and evaluate that
at the endpoint,

00:12:16.050 --> 00:12:18.840
and from that, you subtract
the starting point.

00:12:18.840 --> 00:12:20.340
So hopefully, that makes sense.

00:12:20.340 --> 00:12:23.530
In the next few videos,
we'll actually apply it.