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PROFESSOR: Welcome
back to recitation.

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Today what I'd
like to do is work

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on using Riemann sums to
approximate definite integrals.

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So, what we're
going to do first is

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I'm going to let you
work on the problem.

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So, let me state the problem.

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I'll give you a little
time to work on it.

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And then I'll come back.

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So the problem is the following.

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We want to use four
subintervals and left endpoints

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to approximate the
following definite integral:

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the integral from minus 1 to
3 of the function x cubed.

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And I've drawn a rough
sketch of what x cubed,

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y equals x cubed looks
like to help you out

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as the starting point.

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So why don't you use the four
subintervals and left endpoints

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to approximate the
definite integral, then

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I'll come back and
show you how I do it.

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OK, welcome back.

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So what we want to
do, again, is use

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Riemann sums to approximate
this definite integral.

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And I've given you
the specifications

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of four subintervals
and left endpoints.

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So I'm going to
draw what these four

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subintervals and
their left endpoints

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are going to give
us on the graph.

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Then we're going to calculate
what the actual value is

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of this estimate.

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And then I'll show you
a way you can write it

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in some different notation.

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So let me actually
find some blue chalk.

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Excuse me.

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So, using blue chalk
I'm going to show you

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where the four subintervals
and the left endpoints.

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So it's minus 1 to 3,
so I've conveniently

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done it so we have
every unit length is

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one of the subintervals.

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They're all unit length.

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And if I want left endpoints,
then my rectangle's heights,

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if you remember, are going to
be the output at the left side

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of the interval.

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So I'm going to designate
the left interval as output.

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And what do we have here?

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I'm going to now draw
rectangles of those, using those

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with the one side on
x-axis and one side

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at the height of the
left endpoint output.

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So there's one rectangle.

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Oh, I almost did
something wrong here.

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This left endpoint's
rectangle has no height.

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Because the left
endpoint's value is 0.

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The output is 0.

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There's another rectangle.

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And there's the last rectangle.

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So, to designate
everything carefully,

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I will call this value y_0,
the output value here y_0.

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This value will be y_1.

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The output value
here will be y_2.

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And to show exactly what
do I mean by output value,

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here I have enough room
to write the height

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is really what it is, is y_3.

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So y_0, y_1, y_2 and y_3
designate the heights

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of these four rectangles.

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And the length of each
rectangle, if you remember,

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is really what we
designate as delta x.

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So this right here is delta x.

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And in this case delta x
is each time equal to 1.

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So, remember, formally,
what we want to do

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is, if we want to find
this definite integral,

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we want to find the area under
the curve between the x-axis

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and the function x cubed.

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And remember this
is signed area.

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So anything below the
x-axis has a negative sign

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associated to it.

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Anything above the x-axis has a
positive sign associated to it.

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So, when I'm trying
to determine this,

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when I'm trying
to do an estimate,

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I'm finding the areas
of these rectangles.

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So my first approximation
that I've given here,

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is we want the four subintervals
and the left endpoints.

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What we really need
is this delta x,

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which is the base,
times the height, which

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is y_0, 1, 2 and 3, of
each of these rectangles.

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So, to write it
carefully-- well I

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guess this won't be the most
careful way we write it--

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but I have the base, which
is delta x in each case,

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times y_0 plus y_1
plus y_2 plus y_3.

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And the only thing I
have to be careful of

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is that the area
of the rectangle

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here will be negative.

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But how do I pick that up?

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I pick that up because this
y_0 value, as an actual output,

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is negative.

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So that actually will be
picked up automatically

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by the value of the
output on the function.

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So let's evaluate
these things. y_0

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should be the value of the
function at x equal negative 1.

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When x is negative 1-- the
function, remember, is x cubed,

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so I get negative 1 cubed,
which is just negative 1.

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So delta x is 1.

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y_0 is negative 1.

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y_1 is the value the
function at x equals 0,

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because my left endpoints
are minus 1, 0, 1 and 2.

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So the second one
is 0 is the input.

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At x equals 0 I
get 0 as an output.

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The y_2 is going to be
this third left endpoint.

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That's at x equal 1.

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1 cubed is 1.

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So I get a 1 there.

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And then, the third left
endpoint, I had minus 1,

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0, 1, 2.

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So the third left endpoint is 2.

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2 cubed is 8.

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And so y_3 is 8.

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Hopefully the subscripts
here didn't confuse you

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because I actually was very
close to having the subscripts

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represent the x-value.

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But that's not in fact what
the subscripts are doing.

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The subscripts are
just representing what

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interval we are looking at.

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So the four intervals are
designated by 0, 1, 2 and 3.

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So I get, when I put this all
together I get 1 times-- well,

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minus 1 plus 1 is 0-- 1 times 8.

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So I get 8.

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And if I wanted to
look at the picture,

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how is that represented?

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Well this rectangle has
base 1 and height 8.

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So that has area 8.

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This rectangle is
actually a square.

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It has base 1, height 1.

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And this one is also a square.

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It has base 1, height 1.

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But because it's below,
it has a minus sign

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associated to it, when I think
about it in terms of area.

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So I get minus 1 plus 1 plus 8.

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And that's where
the 8 comes from.

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Now if I wanted to write this
in terms of a Riemann sum,

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you see them sometimes
written in this way.

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You can think about it in
a formal sum in this way.

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You've probably seen this in the
lecture videos more like this.

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The sum from i equals 0 to
3 of f of x sub i delta x.

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And now, in this
case, we could even

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be more deliberate
if we wanted to.

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And we could let x sub i be
designated by some value.

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We could say,
starting at negative 1

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and adding some
number to it each time

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based on the interval.

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But this is sort
of the easiest way

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to write out what
we're interested in.

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And then the x sub i's are going
to be designated separately.

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We can write x sub 0 is equal to
minus 1, x sub 1 is equal to 0,

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x sub 2 is equal to-- sorry, not
minus 1-- is equal to plus 1.

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And x sub 3 is equal to 2.

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So in that case, that's
what we would get there.

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And we could write
it in that way

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and then get the same values.

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So I think I will stop there.

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