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PROFESSOR: Welcome back
to recitation.

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In this video, I want us to
compute the following limit,

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if the limit is n goes to
infinity of the sum for i

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equals 0 to n minus 1 of the
following, 2 over n times the

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quantity 2i over n quantity
squared minus 1.

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Now this might look a little
intimidating to try and take a

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limit of this, but what I'd like
you to do as a hint to

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you is that you should think
about this as being

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potentially a Riemann sum
of a certain function.

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So if you can figure out the
function, and you can figure

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out the appropriate interval
that you're taking a Riemann

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sum over, as n goes to infinity,
you should be able

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to write this as an integral.

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We know how to use the
fundamental theorem of

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calculus to determine
that a definite

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integral in many cases.

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Hopefully this is a function
for which we

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know a way to do that.

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So that's my hint to you.

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Think about it, it's a Riemann
sum approximating an integral,

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and I'll give you a while
to work on it, and

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then I'll be back.

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OK, welcome back.

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Well hopefully it's been fun
for you to look at this

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problem so far.

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Let me just remind you
what we were doing.

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We were trying to compute a
limit as n goes to infinity of

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the sum from i equals 0 to n
minus 1 of 2 over n times 2i

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over n squared minus 1.

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So I gave you the big hint that
this is probably going to

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be written as an integral.

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So let me show you some pieces
of this sum that should help

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us see what the integral is,
and then I'll make a guess

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about what this is, and then
I'll try to give an educated

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way to check my guess.

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So the first thing we noticed
is that there is one thing,

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this is a product of two
functions, and one of them,

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well of n I guess.

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But this is a product of two
things, one thing appears in

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every single term that
you have for i.

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So the sum has n terms, and
they're all going to be 2 over

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n times this, and the i
is going to change.

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But this does not change,
this 2 over n

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does not change, right?

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In fact, I could even pull
that out if I wanted.

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But, I don't want to pull it
out of the sum right now.

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I want us to look at
what's actually

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going on in this product.

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So if this thing is appearing
over and over again, and we

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know this is probably a Riemann
sum, then chances are

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this is our delta x.

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So delta x being equal to 2 over
n, we know delta x equals

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b minus a over n, where b and
a are our left endpoint, oh

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sorry, our right endpoint,
and our left endpoint.

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Right?

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We integrate from a to b.

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So b minus a is the length
of the interval.

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So this is really dividing up
whatever interval we're

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integrating over, into n
equals subintervals.

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So that's my first thought, is
that b minus a over n is equal

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to 2 over n.

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And now we want to try
and figure out

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what the heck is this.

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Well, when we take a Riemann
sum, remember when we take a

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Riemann sum what we get.

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We get the sum of delta x times
f of x sub i, and i is

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what's varying from
0 to n minus 1.

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Let me put a little curve in
here so we see those are two

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different things.

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So this is i equals
0 to n minus 1.

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I have this delta x here.

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I'm anticipating this is
some f of x sub i.

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And so the question
is, what f is it?

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Right?

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If I know what f it is, than I
know that this sum will be

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equal to something, the integral
from a to b of f of x

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dx, and a and b will
differ by 2.

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So that's where I'm heading.

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So now the question is what
is this a function of?

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What function is this,
I should say.

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Now first guess would be
something like, well, I'm

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taking some quantity,
I'm squaring it,

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and subtracting 1.

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So my first guess for this
function is x squared minus 1.

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I mean, that seems easy to me.

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Let's see if this would actually
even make sense just

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by looking at the subscripts,
or sorry, the index, the

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indices I have here.

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So what do I have?

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Well, when I put
in i equals 0--

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let's put down some of these
values-- when I put in i

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equals 0, I get 2 times 0
over n squared minus 1.

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When I put in i equals 1,
I get 2 times 1 over n

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squared minus 1.

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And I go all the way up, to
2 times n minus 1 over n

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squared minus 1.

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So it's kind of a long sum
there, but these are, this is

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what our sum of these things
looks like if I pull

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out the 2 over n.

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So here I get 0 squared
minus 1.

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That looks pretty good.

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Here I get 2 times 1 over
n squared minus 1.

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So it does look like I'm doing
something, taking something,

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squaring it, subtracting 1.

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Does it make sense that these
are the kind of x values I

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would expect to get if this
were the Riemann sum of x

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squared minus 1?

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It does, and let's
think about why.

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I'm starting at x equals 0
here it sure looks like.

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Let's look at what happens
when I go all

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the way over here.

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What happens when n gets really,
really big, is it that

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this ratio approaches 2.

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So it's 2 times n minus 1 over
n. n minus 1 over n, as n gets

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arbitrarily large, as n gets
really big, then this

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approaches 2.

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So this is approaching
2 squared minus 1.

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So it's giving me more evidence
that this is probably

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the function x squared
minus 1.

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And now I'm starting
to believe the

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interval is 0 to 2.

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I know it's length 2 interval,
and it's looking like the

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interval is 0 to 2.

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Let's come back and talk
about one more thing.

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The one other thing that you
should notice is that how does

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this value differ from this
value, and the next, and the

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next, and the next?

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They differ by 2 over n.

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So each time whatever input I
had previously, I'm now adding

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2 over n to the next input.

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And that should make sense of
what we know about Riemann

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sums, because what I do, is I
divide my interval into these

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subintervals of length
2 over n.

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I'm evaluating it at one x value
that I'm starting, in

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this case, at 0.

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Then the next interval
is 2 over n more.

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Then I evaluate at
that x value.

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The next one it 2 over
n more, and I

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evaluate at that x value.

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So this is looking like--

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I'm going to write it here,
this is my guess--

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integral from 0 to 2 of
x squared minus 1 dx.

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And now to make myself feel
good about this--

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I'm pretty sure it's that.

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To make you feel good about
this, I'm going to divide this

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into four subintervals, and I'm
going to show you what the

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Riemann sum with four intervals
looks like, and then

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we can talk about how it relates
to this one over here.

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OK, so let me draw a graph.

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Actually, I'll use just
white chalk again.

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Let me draw a graph of x squared
minus 1 from 0 to 2.

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So 0, 1, 2, minus 1.

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OK, so at 0, x squared minus
1 is negative 1.

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At x equals 1, x squared
minus 1 is 0.

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And at 2, x squared
minus 1 is 3.

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So hopefully is all going to go
into the video and in the

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video screen I mean.

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And there we go, something
like that.

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So this is, you know, it
continues over here, but I'm

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really only interested
in this part.

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So now let's look at what
the subintervals are.

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And now I'm going to get
some colored chalk.

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So what are the subintervals?

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I'm taking 1 over 4, OK?

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And so delta x, in this
case, is 2 over 4,

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which is equal to 1/2.

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Right?

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And so what are my, what
are, what is my sum

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going to look like?

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Well, I am going to tell you
that I'm also going to use

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left-handed endpoints.

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And I mentioned earlier why
that is, I believe.

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Maybe I didn't.

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But, I started off at i
equals 0, and my first

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input value was 0.

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My last input value had an n
minus 1 in it instead of an n.

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So I was doing, somehow, the
second to last place that I

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was interested in here.

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So it's definitely more of a
left-hand endpoint thing.

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So I'm going to do this with
left-hand endpoints.

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And I'm not going to
simplify as I go.

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I'm going to write it out in
sort of an expanded form.

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OK, so let's write it out
in expanded form.

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So the Riemann sum, this is y
equals x squared minus 1.

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The Riemann sum is, the first
term is 1/2 times what?

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It's the value, this x value,
which is 0, evaluated on this

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curve, so 0 squared minus 1.

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The next term--

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I'll just write them right
below each other--

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is 1/2.

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'Cause again, let's draw
a picture of what the

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first one is, sorry.

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It's this rectangle.

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00:09:11,345 --> 00:09:16,540

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Right?

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It's the value, length 1/2,
and it's the function

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evaluated at 0.

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The next one is length 1/2,
and it's going to be the

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function evaluated
at whatever this

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left-hand endpoint is here.

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So it's going to be this area.

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So it's going to be length 1/2,
and then the height is

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going to be at x equals
1/2, so 1/2 quantity

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squared minus 1.

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00:09:40,590 --> 00:10:11,590

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The next one is going
to be this interval.

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Well, there's no rectangle to
draw because it's just, the

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output is zero at the
left endpoint there.

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So it's just going to be, it's
going to have an output equal

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to zero at length 1/2
and height 0.

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But, we'll write
it out anyway.

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It's going to be 1/2
times the quantity.

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Now, I went up 1/2 more, so it's
going to be two 1/2's ,

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two times 1/2 squared minus 1.

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Let me just show you
why I did this.

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OK, if we look at the picture,
here I'd gone up 1/2's for my

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initial value.

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Here I'd gone up another 1/2
for my initial value.

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So that's two 1/2 from my
initial value of 0.

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The next one is going to be
three 1/2's, so this is three

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1/2's away, or commonly
known as 3/2.

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OK?

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So that one is going to be 1/2
is the base length again,

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times the quantity 3 times
1/2 squared minus 1.

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And that is in the picture,
this rectangle.

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00:10:47,265 --> 00:10:52,170

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Great.

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So what are we see here?

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If we look at this, these four
pieces, what do we have?

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We have a 1/2 in front
each time.

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Which, what was the 1/2?

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It was b minus a over n.

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So b minus a was 2, n was 4.

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So maybe I should have kept
that as 2 over 4.

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But, it's a little easier to
write it as 1/2 because of

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what I'm doing next.

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I square something,
I subtract 1.

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I go up by the value that this
is from the initial one here.

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And so now I'm taking the output
of what was in here.

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I now take the output at 1/2
more than what was here.

246
00:11:27,410 --> 00:11:30,640
Now I take it at two 1/2's more
than what was here, or

247
00:11:30,640 --> 00:11:34,260
1/2 more than what was there,
and then three 1/2's more than

248
00:11:34,260 --> 00:11:36,430
what was here, or one more
than what was there.

249
00:11:36,430 --> 00:11:38,320
That's kind of confusing, but
let's go back to the picture

250
00:11:38,320 --> 00:11:41,090
and see what it is.

251
00:11:41,090 --> 00:11:43,480
My delta x was 1/2, right?

252
00:11:43,480 --> 00:11:46,580
So I evaluate at the first
place, and I evaluate one more

253
00:11:46,580 --> 00:11:50,440
up, and then I evaluate one more
up, and I evaluate one

254
00:11:50,440 --> 00:11:52,630
more up, which gives
me outputs here,

255
00:11:52,630 --> 00:11:53,750
there, there, and there.

256
00:11:53,750 --> 00:11:55,370
Right?

257
00:11:55,370 --> 00:11:58,820
So really if you go back and you
look at the formulation of

258
00:11:58,820 --> 00:12:05,760
the, of the sum, this was 2 over
n times quantity 2i over

259
00:12:05,760 --> 00:12:13,460
n squared minus 1, you can see
the 2 over n is my 1/2, and

260
00:12:13,460 --> 00:12:16,530
then this is maybe the hardest
part to see, but that's the 2

261
00:12:16,530 --> 00:12:20,470
over n is my 1/2 again, and
the i is this thing that's

262
00:12:20,470 --> 00:12:24,400
coming in as 1, 2, 3.

263
00:12:24,400 --> 00:12:27,660
And so that i was going
from 0 to n minus 1--

264
00:12:27,660 --> 00:12:31,880
so I should have said
0, 1, 2, 3.

265
00:12:31,880 --> 00:12:31,980
Right?

266
00:12:31,980 --> 00:12:35,520
So that i is the 0 to n minus
1, and then I'm evaluating

267
00:12:35,520 --> 00:12:37,740
that, and then I add
them all up.

268
00:12:37,740 --> 00:12:44,620
So when I take the sum, I get,
for n equals 4, I get this.

269
00:12:44,620 --> 00:12:47,722
So in fact, this is just a
guess, still maybe you should,

270
00:12:47,722 --> 00:12:49,480
maybe you should convince
yourself more.

271
00:12:49,480 --> 00:12:53,110
I'm actually convinced at this
point, based on not just this

272
00:12:53,110 --> 00:12:55,380
evidence, but the evidence I
understood before about how

273
00:12:55,380 --> 00:12:57,030
the function works.

274
00:12:57,030 --> 00:13:00,620
Maybe you want to compare
it when n equals 6, or

275
00:13:00,620 --> 00:13:01,330
something like that.

276
00:13:01,330 --> 00:13:03,010
You may need a little more
evidence to make you

277
00:13:03,010 --> 00:13:06,150
understand this particular
piece.

278
00:13:06,150 --> 00:13:07,770
But, hopefully that makes sense
to you that this is

279
00:13:07,770 --> 00:13:11,780
really just i times delta x, and
then evaluated somewhere.

280
00:13:11,780 --> 00:13:15,900
That's the main idea
of this component.

281
00:13:15,900 --> 00:13:18,250
OK, well hopefully this
is informative to you.

282
00:13:18,250 --> 00:13:20,410
If you want to know the exact
answer of how to compute the

283
00:13:20,410 --> 00:13:22,720
sum, obviously you just
take the integral.

284
00:13:22,720 --> 00:13:24,570
So I know you can do that.

285
00:13:24,570 --> 00:13:26,400
So that's where I'll stop.

286
00:13:26,400 --> 00:13:26,606