Writing Riemann sum limit as definite integral

- [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.