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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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It's 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

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the quantity 2i over n
quantity squared minus 1.

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Now this might look
a little intimidating

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to try and take a
limit of this, but what

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I'd like you to do,
as a hint to you,

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is that you should think about
this as being potentially

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a Riemann sum of a
certain function.

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So if you can figure
out the function,

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and you can figure out
the appropriate interval

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that you're taking a Riemann
sum over, as n goes to infinity,

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you should be able to
write this as an integral.

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We know how to use the
fundamental theorem of calculus

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to determine that a definite
integral in many cases.

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Hopefully this is a function for
which we 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,

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and 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 problem

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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

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as n goes to infinity of the
sum from i equals 0 to n minus 1

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of 2 over n times 2i
over n squared minus 1.

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

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to be written as an integral.

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So let me show you some
pieces of this sum that

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should help us see what the
integral is, and then I'll

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make a guess about
what this is, and then

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I'll try to give an educated
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.

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One thing appears
in every single term

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that you have for i.

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So the sum has n
terms, and they're all

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going to be 2 over n times this,
and the i is going to change.

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But this does not change, this
2 over n 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,

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and we know this is
probably a Riemann sum,

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then chances are
this is our delta x.

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So delta x being
equal to 2 over n,

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we know delta x equals b minus
a over n, where b and a are

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our left endpoint-- oh,
sorry, our right endpoint--

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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

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interval we're integrating
over, into n equal subintervals.

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

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is equal to 2 over n.

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And now we want to try
and figure out, well,

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what the heck is this.

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Well, when we take
a Riemann sum,

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remember when we take a
Riemann sum what we get.

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

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and i is what's varying
from 0 to n minus 1.

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Let me put a little
curve in here

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so we see those are
two 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

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be equal to something,
the integral from a

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to b of f of x 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

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like, well, I'm 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 indices

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I have here.

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So what do I have?

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Well, when I put in
i equals 0-- let's

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put down some of these values--
when I put in i equals 0,

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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

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over n squared minus 1.

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

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minus 1 over n squared minus 1.

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

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this is what our
sum of these things

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looks like if I pull
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

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the kind of x values
I would expect

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to get if this were the Riemann
sum of x 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 the way over here.

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

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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,

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as n gets arbitrarily
large, as n gets really big,

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then this 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 interval is 0 to 2.

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I know it's a length 2
interval, and it's looking

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like the 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

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is that how does
this value differ

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from this value, and the next,
and the 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,

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I'm now adding 2 over
n to the next input.

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And that should
make sense of what

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we know about Riemann
sums, because what I do,

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is I divide my interval
into these subintervals

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of length 2 over n.

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I'm evaluating it
at one x-value,

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that-- I'm starting,
in 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 is 2 over n more,
and I evaluate at that x-value.

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So this is looking like--
I'm going to write it here,

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this is my guess-- 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,

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I'm going to divide this
into four subintervals,

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and I'm going to show you
what the Riemann sum with four

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intervals looks
like, and then we

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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, this is all
going to go into the video,

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and-- in the 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,

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but I'm 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-- so what

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is my sum going to look like?

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

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to use 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,

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and my first 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

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that I 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

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is 0, evaluated on this
curve, so 0 squared minus 1.

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The next term-- 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

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the first one is, sorry.

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It's this rectangle.

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Right?

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It's evaluated--
It's length 1/2,

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and it's the function
evaluated at 0.

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The next one is
length 1/2, and it's

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going to be the
function evaluated

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at whatever this 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

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the height is going
to be at x equals 1/2,

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so 1/2 quantity squared minus 1.

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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,

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the output is zero at
the left endpoint here.

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So it's just going
to be-- it's going

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to have output equal to 0,
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-- now,

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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

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gone up 1/2 from
my initial value.

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

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

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

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is three 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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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.

227
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So b minus a was 2, n was 4.

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00:11:07,230 --> 00:11:10,127
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

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because of 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.

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Now I take it at two 1/2's more
than what was here, or 1/2 more

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than what was there,
and then three 1/2's

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more than what was here, or
one more than what was there.

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That's kind of confusing, but
let's go back to the picture

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00:11:38,846 --> 00:11:41,090
and see what it is.

240
00:11:41,090 --> 00:11:43,480
My delta x was 1/2, right?

241
00:11:43,480 --> 00:11:44,930
So I evaluate at
the first place,

242
00:11:44,930 --> 00:11:49,600
and I evaluate one more up, and
then I evaluate one more up,

243
00:11:49,600 --> 00:11:52,630
and I evaluate one more up,
which gives me outputs here,

244
00:11:52,630 --> 00:11:53,750
there, there, and there.

245
00:11:53,750 --> 00:11:55,370
Right?

246
00:11:55,370 --> 00:11:57,710
So really if you go
back and you look

247
00:11:57,710 --> 00:12:04,090
at the formulation of the
sum, this was 2 over n times

248
00:12:04,090 --> 00:12:09,250
quantity 2i over
n squared minus 1,

249
00:12:09,250 --> 00:12:13,210
you can see the 2
over n is my 1/2,

250
00:12:13,210 --> 00:12:15,850
and then this is maybe
the hardest part to see,

251
00:12:15,850 --> 00:12:18,200
but that's the 2 over
n is my 1/2 again,

252
00:12:18,200 --> 00:12:24,400
and the i is this thing
that's coming in as 1, 2, 3.

253
00:12:24,400 --> 00:12:27,810
And so that i was going
from 0 to n minus 1-- so I

254
00:12:27,810 --> 00:12:31,480
should have said 0, 1, 2, 3.

255
00:12:31,480 --> 00:12:31,980
Right?

256
00:12:31,980 --> 00:12:35,705
So that i is the 0 to n minus
1, and then I'm evaluating that,

257
00:12:35,705 --> 00:12:37,740
and then I add them all up.

258
00:12:37,740 --> 00:12:44,620
So when I take the sum, I get,
for n equals 4, I get this.

259
00:12:44,620 --> 00:12:46,512
So in fact, this
is just a guess,

260
00:12:46,512 --> 00:12:47,886
still maybe you
should, maybe you

261
00:12:47,886 --> 00:12:49,480
should convince yourself more.

262
00:12:49,480 --> 00:12:52,150
I'm actually convinced
at this point,

263
00:12:52,150 --> 00:12:54,280
based on not just this
evidence, but the evidence I

264
00:12:54,280 --> 00:12:57,030
understood before about
how the function works.

265
00:12:57,030 --> 00:13:00,140
Maybe you want to
compare it when n equals

266
00:13:00,140 --> 00:13:01,330
6, or something like that.

267
00:13:01,330 --> 00:13:02,788
You may need a
little more evidence

268
00:13:02,788 --> 00:13:05,797
to make you understand
this particular piece.

269
00:13:05,797 --> 00:13:07,380
But, hopefully that
makes sense to you

270
00:13:07,380 --> 00:13:10,150
that this is really just
i times delta x, and then

271
00:13:10,150 --> 00:13:11,780
evaluated somewhere.

272
00:13:11,780 --> 00:13:15,900
That's the main idea
of this component.

273
00:13:15,900 --> 00:13:17,989
OK, well hopefully this
is informative to you.

274
00:13:17,989 --> 00:13:19,780
If you want to know
the exact answer of how

275
00:13:19,780 --> 00:13:22,720
to compute the sum, obviously
you just take the integral.

276
00:13:22,720 --> 00:13:24,570
So I know you can do that.

277
00:13:24,570 --> 00:13:26,106
So that's where I'll stop.