Uploaded by MIT on 07.01.2011

Transcript:

PROFESSOR: Hi.

Welcome back to recitation.

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

computing average values and probabilities using

integration.

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

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

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

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

all in the first quadrant there.

It's just this little sliver.

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

the points in r, what's that average

x-coordinate of those points?

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

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

somewhere in R, what's the probability that its

x-coordinate is larger than 1/2?

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

hand side of R?

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

to work through these problems. Come back and we can

work on them together.

Alright, welcome back.

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

done on these questions.

So let's start to work through them.

So remember, to compute the average value of a function

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

weighted average.

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

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

many points in r in this little corner.

Right?

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

points there either.

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

more points there.

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

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

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

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

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

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

x-coordinate.

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

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

that x-coordinate.

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

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

function we're averaging is x.

Over this region, with that weighting.

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

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

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

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

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

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

We need one integral that is the numerator and so that

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

average that we want.

So the numerator is the integral--

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

possible x values into consideration.

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

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

which in this case is x, by the

appropriate weight function.

And the weight function is how much of the region is

associated with that x value.

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

this case is x minus x cubed dx.

OK.

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

weight of the region.

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

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

this x minus x cubed dx.

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

we have to take their ratio.

So let's do them separately.

So the first one is--

well, the second one is actually a little simpler.

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

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

Well, this is a pretty easy integral.

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

between 0 and 1.

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

just get 0.

So that's 1/4.

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

from 0 to 1.

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

the fourth dx.

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

x to the fifth over 5 between 0 and 1.

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

So that's common denominator 15.

2/15.

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

denomenator is 1/4.

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

by 1/4, which is 8/15.

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

bit larger than 1/2.

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

slightly shifted to the right of 1/2.

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

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

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

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

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

want to know what's the probability that its

x-coordinate is larger than 1/2.

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

regions the probability is just has to do with the area

of the region.

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

right of the line x equals 1/2.

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

of the entire region is that.

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

of 1/2 divided by the total area.

Now luckily we've already computed the

total area, right here.

This was this denominator that we had over here.

So we just need the numerator of that fraction.

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

So the probability equals--

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

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

that satisfies our condition.

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

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

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

region to the right of 1/2.

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

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

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

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

curves on that region.

OK, so, and again this is a fairly

simple integral to compute.

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

over 4 between 1/2 and 1.

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

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

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

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

So OK.

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

If I just put it all over that common denominator.

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

if I did that right.

So, OK.

So that's the good area.

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

good area by the total area.

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

And that was what this computational was.

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

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

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

1/4 which is 9/16.

OK, so this says--

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

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

little bit more of it to the right then

there is to the left.

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

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

to the left of that line.

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

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

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

this region.

So that was part a.

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

was over here.

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

the x-coordinate divided by the total area.

And then second, we computed the probability that a

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

And we did that over here.

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

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

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

So I'll end there.