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[music] Let's integrate x cubed from 0 to
2.

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In this case, the function x cubed is a
continuous function, therefore it's an

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integrable function.
So, all I need is just to start with a

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fine enough partition and it won't matter
how I pick those sample points inside the

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partition.
My Riemann sum will be close to the true

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value of the integral.
So, I'll cut the interval from 0 to 2 into

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n pieces.
Pictorially, I want to start with the

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interval between 0 and 2, and I want to
cut it up into n pieces.

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Now, the length of this whole interval is
two units long, so each is these pieces

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should have a went of 2 over n.
And that, let me write down the formula,

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the cut point x sub i will be 2 over n
times i, and that works out great .

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If i is 0, then I'm at the left hand
endpoint.

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And if i is n, I'm at the right hand
endpoint.

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And in between, I'm just marching from the
left hand side to the right hand side

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moving 2 over n each time.
Now, how should I choose my sample points?

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I'm just going to choose my sample points,
x sub i star, to be the right hand end

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points.
So, they'll be, in fact the same as x sub

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i.
Now, I just have to write down the Riemann

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sum for this particular equal size
partition and our chosen sample points.

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In general, the formula for Riemann sum is
the sum, I think it's the sum, i goes from

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

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i-th subinterval in my partition, right?
This is the area of the rectangle.

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And I'm adding up all those rectangles to
get an approximation of the area under the

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graph.
Well, in this specific case, what do I

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have?
X sub i star is the same thing.

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If I go back to this, x sub i star is the
same thing as x sub i.

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So, I can just plug in 2 over n times i,
and x sub i minus x sub i minus 1, that's

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the width of the i-th subinterval in my
partition.

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But, I rigged it so that all of the
subintervals have the same equal width.

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They've all got width 2 over n, okay?
So, that's the sum that I want to

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calculate.
Now, the function in this case is the

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cubing function.
So, this is the same as the sum i goes

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from 1 to n of 2 over n times i cubed
times 2 over n, which I could simplify

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somewhat.
This is the sum i goes from 1 to n of 16,

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2 cubed times 2, over n to the 4th, it's n
cubed over n times i cubed.

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So that's the particular Riemann sum in
this case.

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Quizzes are just a specific Riemann sum
for a specific value of n.

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To get the value of the intregal, I'm
going to take the limit an n approaches

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infinity.
Yeah, the integral from 0 to 2 of this

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function x cubed is the limit as n goes to
infinity of this Riemann sum.

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Now, strictly speaking, right?
This integral would be a limit over all

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the possible partitions.
But since the function's integrable, it

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doesn't matter what partition I choose as
long as it's fine enough.

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And n going to infinity makes finer and
finer partitions.

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So, what I really want to calculate, in
order to calculate this integral, is to

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calculate this limit.
Well, let's see if we can do that.

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The good news here is that I've got a
formula for the sum of cubes.

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So, we will pull out that formula in a
minute.

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And I've also got a constant here.
16 over n to the fourth is a constant.

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It doesn't depend on i at all, so I can
factor that out of this sum by

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distributivity.
So, this is the limit as n approaches

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infinity of 16 over n to the 4th times the
sum of i cubed, i goes from 1 to n.

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And now, I happen to know what the sum of
the first n cubed is.

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I can use that formula that we got from
before.

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That's the same as the sum of the first n
whole numbers squared.

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And in this case, that's the limit as n
goes to infinity of 16 over n to the 4th

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times, that's n times n plus 1 over 2.
That's the sum of the first n whole

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numbers and the sum of the first n perfect
cubes is that squared.

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Now, I've got to figure out what this is.
Well, I can factor out a over 2 squared

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and that'll cancel part of the 16.
So, this is the limit of n squared times n

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plus 1 squared times 4 over n to the 4th.
And now this is the limit as n goes to

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infinity.
This ends up being n squares time n

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squares plus lower order terms.
So, this is, the biggest term here is 4n

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to the 4th over n to the 4th, so this
limit ends up being equal to 4.

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And that is in fact the integral from 0 to
2 of x cubed dx.

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That integral, therefore, is equal to 4.
Now, what if I wanted to do this integral

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over some other interval?
Well, I could repeat this same kind of

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calculation to deduce that the integral
from 0 to 1 of x cubed dx is equal to a

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quarter, right?
But that would be another whole

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calculation that I'd have to work out.
Let's say, I wanted to integrate over the

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interval 1 to 2.
Well, there's a trick that you could use.

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So, the integral from 0 to 2 of x cubed
dx, we already calculated.

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But forget the exact answer.
What I want to note here is that this is

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related to two other integrals.
Integrating from 0 to 2x cubed dx is the

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same as integrating from 0 to 1 of x cubed
dx, and adding to that the integral from 1

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to 2.
Now geometrically, something like that

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hopefully makes some sense, right?
If I draw a graph, here's 0, here's 1,

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here's 2.
Here's the function y equals x cubed, say

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this first interval from 0 to 1 is
calculating the area here, right?

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That's what this integral's calculating.
The second integral from 1 to 2 is

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calculating this area here, right?
And if I add together these two areas, I

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end up getting the area from 0 to 2, which
is what this integral's calculating.

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Well, the nice thing here is that, since I
know this integral and I know this

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integral, that's enough information to
recover this integral.

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The integral from 0 to 2, we saw a moment
ago, was 4.

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The integral from 0 to 1, I'm claiming
here is a quarter, and that's enough

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information to tell me that the integral
from 1 to 2 of x cubed dx, well, that must

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be exactly what I need to add onto a
quarter to get 4, that must be 15 fourths,

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right?
So, this red area in here must be 15 4th

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square units.
What we're seeing here is that you can

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evaluate some integrals just by going back
to the definition and using our knowledge

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of sums, but the sums are ugly, right?
These are really terrible calculations.

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This is the exact same situation that
we've been in before.

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A long, long time ago, we were faced with
the definition of derivative and we

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started computing some derivatives just by
going back to the definition of

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derivative.
But then, over time, we learned a bunch of

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rules which made differentiation a whole
lot easier.

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We're going to see the same thing
happening here.

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Right now, we're just using these bare
hand arguments to calculate integrals.

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But over time, we're going to develop more
tools for evaluating integrals.