- [Instructor] So, we've got a Riemann sum.
We're gonna take the limit as N approaches infinity
and the goal of this video
is to see if we can rewrite this
as a definite integral.
I encourage you to pause the video
and see if you can work through it on your own.
So, let's remind ourselves
how a definite integral can relate to a Riemann sum.
So, if I have the definite integral from A to B
of F of X, F of X, DX,
we have seen in other videos
this is going to be the limit
as N approaches infinity of the sum, capital sigma,
going from I equals one to N
and so, essentially we're gonna sum the areas
of a bunch of rectangles
where the width of each of those rectangles
we can write as a delta X,
so your width is going to be delta X
of each of those rectangles
and then your height
is going to be the value of the function
evaluated some place in that delta X.
If we're doing a right Riemann sum
we would do the right end of that rectangle
or of that sub interval
and so, we would start at our lower bound A
and we would add as many delta Xs as our index specifies.
So, if I is equal to one,
we add one delta X,
so we would be at the right of the first rectangle.
If I is equal two, we add two delta Xs.
So, this is going to be delta X
times our index.
So, this is the general form
that we have seen before
and so, one possibility, you could even do a little bit
of pattern matching right here,
our function looks like the natural log function,
so that looks like our func F of X,
it's the natural log function,
so I could write that,
so F of X looks like the natural log of X.
What else do we see?
Well, A, that looks like two.
A is equal to two.
What would our delta X be?
Well, you can see this right over here,
this thing that we're multiplying
that just is divided by N
and it's not multiplying by an I,
this looks like our delta X
and this right over here looks like delta X times I.
So, it looks like our delta X is equal to five over N.
So, what can we tell so far?
Well, we could say that, okay, this thing up here,
up the original thing is going to be equal to
the definite integral,
we know our lower bound is going from two
to we haven't figured out our upper bound yet,
we haven't figured out our B yet
but our function is the natural log
of X and then I will just write a DX here.
So, in order to complete writing
this definite integral
I need to be able to write the upper bound
and the way to figure out the upper bound
is by looking at our delta X
because the way that we would figure out a delta X
for this Riemann sum here,
we would say that delta X is equal to the difference
between our bounds divided by how many sections
we want to divide it in, divided by N.
So, it's equals to B minus A,
B minus A over N, over N
and so, you can pattern match here.
If this is delta X is equal to B minus A over N.
Let me write this down.
So, this is going to be equal to B,
B minus our A which is two,
all of that over N,
so B minus two
is equal to five
which would make B equal to seven.
B is equal to seven.
So, there you have it.
We have our original limit, our Riemann limit
or our limit of our Riemann sum
being rewritten as a definite integral.
And once again, I want to emphasize
why this makes sense.
If we wanted to draw this
it would look something like this,
I'm gonna try to hand draw the natural log function,
it looks something like this
and this right over here would be one
and so, let's say this is two
and so going from two to seven,
this isn't exactly right
and so, our definite integral is concerned with the area
under the curve from two until seven
and so, this Riemann sum you can view
as an approximation when N isn't approaching infinity
but what you're saying is look,
when I is equal to one,
your first one is going to be of width five over N,
so this is essentially saying our difference
between two and seven,
we're taking that distance five,
dividing it into N rectangles,
and so, this first one is going to have a width of five
over N and then what's the height gonna be?
Well, it's a right Riemann sum,
so we're using the value of the function right over there,
write it two plus five over N.
So, this value right over here.
This is the natural log,
the natural log of two plus five over N,
and since this is the first rectangle
times one, times one.
Now we could keep going.
This one right over here
the width is the same, five over N
but what's the height?
Well, the height here, this height right over here
is going to be the natural log
of two plus five over N times two, times two.
This is for I is equal to two.
This is I is equal to one.
And so, hopefully you are seeing that this makes sense.
The area of this first rectangle
is going to be natural log of two
plus five over N times one
times five over N
and the second one over here,
natural log of two plus five over N times two
times five over N
and so, this is calculating the sum
of the areas of these rectangles
but then it's taking the limit as N approaches infinity
so we get better and better approximations
going all the way to the exact area.