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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.