WEBVTT

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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

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on computing average values
and probabilities using

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integration.

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So for this problem,
I'm going to let

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R be the name of
the region that's

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bounded above by the
curve, by the line I guess,

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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.

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It, you know, lives 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

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is, if I look at
all the points in R,

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what's that average
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

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that its x-coordinate
is larger than 1/2?

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So what's the
probability that it lies

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on the right-hand side of R?

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So why don't you
pause the video,

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take a couple minutes to
work through these problems,

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come back, and we can
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 done

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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

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is you need to compute
a 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,

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there aren't very 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

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very many points there either.

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In the middle, this
region is a little higher,

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so there are more points there.

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So those points
will sort of weigh

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more when you take the
average of all points,

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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,

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since we're interested in
the x-coordinate, the weight

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at a given x-coordinate
is the slice

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

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have 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,

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so the 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,

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and f of x, the thing we're
computing the average of

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is, just the x-coordinate, so
it's just the function value x.

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Over R. So we want the
average of this function, f

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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,

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and so that numerator--
so I'm going

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to just write average for
the average that we want.

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So the numerator is
the integral-- OK,

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and so we have to
integrate, we have

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to take all possible
x-values into consideration.

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

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

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which in this case is x, by the
appropriate weight function.

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And the weight function
is how much of the region

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is associated with that x-value.

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And that's the height of
this little rectangle,

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which in 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

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to divide by the total
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

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from 0 to 1 of just
this x minus x cubed dx.

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OK, so we have to compute
these two integrals

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and then 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--
well, the second one

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is actually a little simpler.

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

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The 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

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you just get 0.

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So that's 1/4.

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And the first one,
the top, the numerator

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is the integral from 0 to 1.

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OK, we can multiply
through, so that's

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x squared minus x
to the fourth dx.

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So that integral
from 0 to 1 again

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is x cubed over 3 minus x to the
fifth over 5 between 0 and 1.

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When we put in 0 we get 0,
put in 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,

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the denominator is 1/4.

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So the average value
we're interested in

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is 2/15 divided by
1/4, which is 8/15.

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So the average
x-coordinate is just,

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so 8/15 is just a tiny
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 symmetrically,

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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-- so

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

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and we want to know
what's the probability

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that its x-coordinate
is larger than 1/2.

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

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all regions, the
probability just

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has to do with the
area of the region.

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

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to the right of the
line x equals 1/2.

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So that's the good
region, and then we

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want to know how much of
the entire region is that.

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So we want to know the
area of the good region

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to the right of 1/2
divided by the total area.

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Now luckily we've
already computed

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the 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-- I'm going

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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-- so

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a point we're interested in,
if its x-coordinate is bigger

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

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to the right of 1/2.

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So in order to
compute that, we just

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take this integral 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

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and then we want the,
you know, the area

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between these two
curves on that region.

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OK, so, and again this
is a fairly simple

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integral to compute.

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So this gives me x squared over
2 minus x to the fourth over 4

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between 1/2 and 1.

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

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so this gives me 1/2 minus 1/4
minus-- well we put in 1/2 here

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we get 1/8 minus-- 1/2 to the
fourth is 1/16, divided by 4

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is 1/64.

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So this is all going
to be in sixty-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, 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,

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I have to divide the 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 computation was.

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

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I'm just going to write pr
for probability-- that we're

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interested in is 9/64
divided by 1/4 which is 9/16.

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OK, so this says-- also 9/16
is also a little bit bigger

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than a half.

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

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a little bit more
of it to the right

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then there is to the left.

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You know, so it's
slightly more likely

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that a random point is
to the right of the line

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y equals 1/2 than it is
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.

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So first we computed the average
value of the x-coordinate

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of a point in this region.

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So that was part a.

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We computed the average
value of the x-coordinate,

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and that was over here.

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

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of the x-coordinate
divided by the total area.

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And then second, we
computed the probability

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that a 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

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the area of the
good region, which

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was the part to the right
of the line y equals 1/2

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

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So I'll end there.