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

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

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

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points as long as the partition is fine
enough.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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n, and the function being the squaring
function.

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

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

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

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

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this particular partition as my particular
choice of sample point.

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

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

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

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

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

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

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

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

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

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

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

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before.
And this is calculating the integral from

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0 to 1 of x squared.
All right.

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

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

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

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

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

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

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right?
That formula tells me that this is the

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

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

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over 6.
Now, I could expand that out.

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

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

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matters, right?
What these highest powers are.

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

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

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3rd.
Let's summarize this with an official

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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times, right?
And these are the same.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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pieces and fit that thing entirely inside
my curved region.

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

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

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

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right?
This whole story isn't actually about

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

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

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

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

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

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I like them to be.