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PROFESSOR: Hi.

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Welcome back to recitation.

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Today we're going to do a
couple of exercises on

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computing average values
and probabilities using

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integration.

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So for this problem, I'm going
to let R be the name of the

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region that's bounded above by
the curve, by the line I

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guess, y equals x and below by
the curve y equals x cubed.

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So I have a little picture of
r here, it, you know, lives

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all in the first
quadrant there.

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It's just this little sliver.

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So the first part of the
question is, if I look at all

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the points in r, what's
that average

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x-coordinate of those points?

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What's the average value
of the x-coordinate?

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And the second problem is, if
I choose a point at random

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somewhere in R, what's the
probability that its

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x-coordinate is larger
than 1/2?

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So what's the probability that
it lies on the right

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hand side of R?

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So why don't you pause the
video, take a couple minutes

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to work through these problems.
Come back and we can

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work on them together.

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Alright, welcome back.

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So hopefully you've had a chance
to get some good work

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done on these questions.

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So let's start to work
through them.

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So remember, to compute the
average value of a function

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over some region what you need
to do is you need to compute a

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weighted average.

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And the weighting here is that
you have to consider the fact

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that, you know, for x very near
zero, there aren't very

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many points in r in this
little corner.

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Right?

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And for x very near 1, there
aren't very many

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points there either.

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In the middle, this region is a
little higher, so there are

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more points there.

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So those points will sort of
weigh more when you take the

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average of all points, than
the points near the edges.

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So the way we account for that
is we have this weight

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function that is the, in our
case, since we're interested

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in the x-coordinate, the weight
at a given x-coordinate

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is the slice of the-- how much
of the region lies above that

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x-coordinate.

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What, you know, the area of a
little rectangle above that

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x-coordinate says, tells you
how many of the points have

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that x-coordinate.

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So then we want to average
the function x, right?

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Because we're interested in
the x-coordinate, so the

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function we're averaging is x.

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Over this region, with
that weighting.

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So, all right, so let's write
down what that means.

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So we have want average of the
function f of x, and the thing

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we're computing the average of
is just the x-coordinate, so

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it's just the function value x
over R. So we want the average

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of this function, f of x over R.
So what we need to compute

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is the, so we have two integrals
we need to compute.

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We need one integral that is
the numerator and so that

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numerator-- so I'm going to
just write average for the

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average that we want.

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So the numerator is
the integral--

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OK, and so we have to integrate,
we have to take all

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possible x values into
consideration.

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So x going, In this case that's
x going from 0 to 1 and

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now we want to multiply the
function that we're averaging,

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which in this case is x, by the

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appropriate weight function.

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And the weight function is
how much of the region is

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associated with that x value.

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And that's the height of this
little rectangle, which in

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this case is x minus
x cubed dx.

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OK.

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But then, this is an average, we
have to divide by the total

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weight of the region.

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The weight, in this case,
is just the area.

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So we have to divide by the
integral from 0 to 1 of just

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this x minus x cubed dx.

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OK, so we have to compute these
two integrals and then

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we have to take their ratio.

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So let's do them separately.

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So the first one is--

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well, the second one is actually
a little simpler.

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The one in the denomenator, so
let's handle that first. The

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integral from 0 to 1 of
x minus x cubed dx.

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Well, this is a pretty
easy integral.

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It's x squared over 2 minus
x to the fourth over 4

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between 0 and 1.

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So that's 1/2 minus 1/4 minus,
well when you put in 0 you

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just get 0.

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So that's 1/4.

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And the first one, the top, the
numerator is the integral

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from 0 to 1.

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OK, we can multiply through, so
that's x squared minus x to

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the fourth dx.

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So that integral from 0 to 1
again is x cubed over 3 minus

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x to the fifth over
5 between 0 and 1.

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When we put in 0 we get 0, put
int 1 we get 1/3 minus 1/5.

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So that's common
denominator 15.

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2/15.

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So the numerator of our average
value is 2/15, the

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denomenator is 1/4.

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So the average value we're
interested in is 2/15 divided

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by 1/4, which is 8/15.

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So the average x-coordinate is
just, so 8/15 is just a tiny

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bit larger than 1/2.

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So this is saying somehow the
average x-coordinate is just

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slightly shifted to
the right of 1/2.

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So this, I may have drawn this
region sort of symetrically,

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but in fact it's a little bit
shifted to the right there.

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So that's the the first
part of the problem.

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For the second part, we want to
compute the probability--

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so OK, so we choose a point at
random in this set R and we

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want to know what's the
probability that its

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x-coordinate is larger
than 1/2.

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Well, since all points are, you
know, equally likely, all

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regions the probability is just
has to do with the area

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of the region.

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What we really want to know is,
what's is area of R to the

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right of the line
x equals 1/2.

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So that's the good region, and
then we want to know how much

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of the entire region is that.

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So we want to know the area of
the good region to the right

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of 1/2 divided by
the total area.

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Now luckily we've already
computed the

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total area, right here.

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This was this denominator
that we had over here.

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So we just need the numerator
of that fraction.

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So let's go over here, let
me write that down.

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So the probability equals--

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I'm going to put good area
in quotation marks--

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so what I mean by good area is
the area of the set of points

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that satisfies our condition.

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And in our case, the
good area is just--

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so a point we're interested in,
if it's x-coordinate is

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bigger than 1/2, so we just want
to take the part of this

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region to the right of 1/2.

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So in order to compute that,
we just take this integral

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from 1/2 to 1 instead
of from 0 to 1.

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So we're only counting those
points to the right of 1/2 and

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then we want the, you know,
the area between these two

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curves on that region.

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OK, so, and again
this is a fairly

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simple integral to compute.

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So this gives me x squared over
2 minus x to the fourth

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over 4 between 1/2 and 1.

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All right, so this is maybe a
tiny bit hairy, so this gives

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me 1/2 minus 1/4 minus-- well we
put in 1/2 here we get 1/8

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minus 1/2 to the fourth is 1/16
divided by 4 is 1/64.

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So this is all going to
be in sxty-fourths.

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So OK.

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So let's see, 32/64 minus
16/64 minus 8/64 plus 1.

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If I just put it all over
that common denominator.

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All right, so I think that
looks like 9/64,

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if I did that right.

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So, OK.

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So that's the good area.

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And then the probability that
I want, I have to divide the

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good area by the total area.

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And we saw before that the
total area was 1/4.

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And that was what this
computational was.

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So the total area's 1/4,
so the probability--

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I'm just goint to write
pr for probability--

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that we're interested in
is 9/64 divided by

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1/4 which is 9/16.

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OK, so this says--

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also 9/16 is also a little
bit bigger than a half.

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So this is a different way of
saying, our region, there's a

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little bit more of it
to the right then

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there is to the left.

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You know, so it's slightly
more likely that a random

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point is to the right of the
line y equals 1/2 than it is

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to the left of that line.

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So just to sum up, we had these
two different problems

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that we did concerning this
region R. So first we computed

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the average value of the
x-coordinate of a point in

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this region.

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So that was part a.

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We computed the average value of
the x-coordinate, and that

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was over here.

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So we had to, you know, do this,
the weighted average of

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the x-coordinate divided
by the total area.

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And then second, we computed
the probability that a

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randomly chosen point has
x-coordinate larger than 1/2.

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And we did that over here.

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And for that we needed to
compute the area of the good

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region, which was the part to
the right of the line y equals

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1/2 and then we needed to divide
it by the total area.

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So I'll end there.

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