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[music] Let's integrate x squared from 0 to 1 by using the definition of integral.

Now, since the function x squared is continuous, it's integrable.

So, it won't matter exactly how we choose our partition or how we choose our sample

points as long as the partition is fine enough.

So, I'm going to choose a partition in which each of the subintervals has the

same width, 1 over n. I'm going to divide the interval into n

pieces each with the same size. Now, I've got to figure out exactly where

those cut points are being made. And I'm labeling these points x sub i, so

my left hand n point is x sub 0 and my right hand n point is x sub n, and my cut

points in between where I'm cutting up to partition are i over n.

So, the first one here, or the one after the 0 of 1 anyway, is 1 over n.

The next one is 2 over n, the next one is 3 over n, and each of those subintervals

has the same width, 1 over n. Now, I also have to choose points to

sample the function at, and I'm just going to choose the right hand end point here.

So, my x sub 1 sample point will be 1 over n, my x sub 2 sample point will be 2 over

n, and so on. And I can write down the Riemann sum,

right? The Riemann sum is the sum, i goes from 1

to n, the function evaluated at the sample point times the width of this i-th

interval, right? Which I could write as x sub i minus x sub

i minus 1. Now, in this case some of this is a little

bit easier. I know I can write this a little bit more

nicely, right? F of x of i star, x of i star being i over

n, and the function being the squaring function.

I can rewrite this Riemann sum as i over n squared, and then x of i minus x of i

minus 1. That's just the width of the i-th

subinterval in my partition. And that's, you know, how far apart these

numbers are, and that's 1 over n. So, there is my Riemann sum associated to

this particular partition as my particular choice of sample point.

I can evaluate that exactly by using some of the facts we've learned about sums.

In fact, I want to do a little bit more than just evaluate this, right?

I want to take the limit of this as n goes to infinity.

And by choosing finer and finer partitions, that's going to give me better

and better approximations to the true area under the graph of this function, right?

That's the area that I'm trying to calculate by evaluating this particular

Riemann sum and taking the limit as the number of pieces in my equal length

partition goes to infinity, okay? So, this is really what I want to

calculate, alright? I want to calculate the limit as the

number of things in my partition, each of equal size, goes to infinity, right?

That's a finer and finer partition. And this is the Riemann sum that we had

before. And this is calculating the integral from

0 to 1 of x squared. All right.

Well, how do I do this now? Well, this is the limit as n approaches

infinity of the sum, i goes from 1 to n, and this is i squared over n squared times

1 over n. Now, n is a constant, so I can pull this

out of a sum by using distributivity. This is the limit as n approaches infinity

of 1 over n cubed times the sum of i squared, i goes from 1 to n.

Now we think back, I've got a formula for the sum of the first n perfect squares,

right? That formula tells me that this is the

limit as n goes to infinity of 1 over n cubed times, and what's the sum of the

first n perfect squares? It's n times n plus 1 times 2n plus 1 all

over 6. Now, I could expand that out.

This is the limit as n goes to infinity of 1 over n cubed times 2n cubed plus 3n

squared plus n all over 6. And this limit, well, it really just

matters, right? What these highest powers are.

There's an n cubed in the denominator, or a 6n cubed in the denominator, and a 2n

cubed in the numerator. And as n approaches infinity, this is 1

3rd. Let's summarize this with an official

looking statement. So, the integral from 0 to 1 of x squared

dx, right? We've just calculated this as 1 3rd.

And what that's really saying is that if I've got say, this graph here, say this is

the graph of x equals x squared, this integral is calculating this area, right?

The area between 0 and 1 underneath the graph of x squared, right?

This red area here is 1 3rd square unit as a result of the integral calculation.

Now, I can play around with this a little bit.

So, to say that the integral from 0 to 1 of x squared dx is 1 3rd, well, one thing

I could do with this is I could multiply this by 3, right?

What's 3 times the integral from 0 to 1 of x squared?

Well, that's 3 times a 3rd, that's equal to 1.

But 3 times an integral, that's the same as integrating from 0 to 1 3 times the

function, right? And this is really worth thinking about.

I mean, here I'm taking this area and multiplying it by 3.

Here, I'm calculating the area of the graph stretched in the y direction by 3

times, right? And these are the same.

And this is saying that the integral from 0 to 1 of 3x squared dx is equal to 1.

So, what does that mean? So here, I've graphed the function y

equals x squared from 0 to 1. And the area in there is 1 3rd of a square

unit. That's the first calculation that we did.

And, now what I'm claiming is that if I took this same function here but

multiplied it by 3, right? Which has the effect of stretching it in

the y direction by 3 times as much. Well, that triples the area.

So, the original area here is a 3rd, but now there's new area 1 square unit, right?

Which is what I've written here, okay? So, what does it mean to say that the area

under the graph is one square unit? What I don't mean is that I can take a

square of side length 1 and cut it up and exactly cover that region.

That's really not what I'm talking about when I claim that, that region has area 1.

If I claim that, that region is area one, I really mean two different things.

I mean that if I had a square with just a little tiny corner nicked off of it, I

could take that thing that has area just under one unit and chop it up into little

pieces and fit that thing entirely inside my curved region.

Conversely, I mean that I could take a square of side length 1 and just a little

bit of extra area, and I could cut that thing up into little tiny pieces and cover

up my curved region, right? But it's fundamentally all about limits,

right? This whole story isn't actually about

achieving anything. It's just saying that I can get as close

as I want, right? It's saying that I can take this almost 1

square unit entirely inside, or I could take a little bit more than 1 square unit

and cover it up. But I'm never actually asserting that

things are equal. I'm just saying that they are as close as

I like them to be.