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PROFESSOR: Hi.

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Welcome back to recitation.

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We've been talking
in lecture about

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various different applications
of definite integrals.

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And one of them has been that
we can use definite integrals

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to find areas of regions
that previously we

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wouldn't have been able to.

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So, one simple
example of that is

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that we can use a
definite integral

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to find the area of a region
bounded between two curves.

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So I have an example of
such a question right here.

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So the question is to compute
the area of the region that's

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bounded between the curves
y equals x cubed and y

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equals 3x minus 2.

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So one thing you'll notice
is I haven't given you

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endpoints for this region.

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So I've just given you
the bound and curves.

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So one thing you're
going to have to do right

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at the beginning is to figure
out what this looks like

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and what region you're going
to be integrating over.

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What interval you're going
to be integrating over.

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So, this is a kind
of tricky one.

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Why don't you pause the
video, think about it

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for a few minutes, try
and work it out yourself,

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and you can come back and we can
try and work it out together.

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All right, welcome back.

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So hopefully you've
had some luck

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figuring out what this
situation looks like

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and then computing the integral.

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So let's talk about
it for a minute.

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So we have these two curves.

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And so somewhere-- I'm asking
about the region bounded

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by them-- so what I'm
telling you is somewhere

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these curves intercept and they
surround some bounded region.

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And so I'm asking for what
the area of that region is.

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So in order to
figure that out, we

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should figure out where
these curves intercept.

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So to find the,
in other words, we

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need to find the endpoints of
the interval over which we're

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going to integrate.

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So in order to that,
we have to solve

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for where we have the
intersection between y equals x

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cubed and y equals 3x minus 2.

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So we have to
solve the equation.

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So we need to solve the equation
x cubed equals 3x minus 2.

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Or x cubed minus
3x plus 2 equals 0.

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So this a polynomial equation.

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It's not quadratic, right?

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It's a cubic equation.

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So that means it's hard
to solve in general.

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Luckily in this case,
it's-- there are some--

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there's a root we can
recognize fairly easily.

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Which is, it's
pretty-- you know,

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one thing you can always do is
check small positive integers

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or use the rational
root theorem,

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if you remember that
from high school math.

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So in this case, if
you check x equals 1,

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it's fairly easy to
see that x equals

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1 is a root of this equation.

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So in other words, we can
factor the left-hand side.

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We can divide out a
factor of x minus 1.

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OK, and so now we have
to do long division

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or synthetic division, whatever
kind of division you want,

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to divide through here.

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So we get x squared minus--
so we've got a-- nope, I lied.

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Plus x minus 2 after you divide.

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How's that look?

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We can just double check.

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You know, we get an x cubed
minus x squared plus x squared

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minus 2x minus x plus 2.

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OK.

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So that adds up to
the right thing.

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So we either have
that 1 is a root

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or that x squared you
know, rather for the places

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where these intersect,
we have either x equals 1

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or x squared plus
x minus 2 equals 0.

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And now here you can,
you know, again factor

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or use the quadratic
equation or what have you.

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And you can see so we have that
this actually fully factors

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as x minus 1 squared
times x plus 2 equals 0.

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Which means that we
have intersections

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when either x minus 1 is
0 or when x plus 2 is 0.

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So intersections at x equals
1 and x equals minus 2.

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All right.

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So you can take that information
and you can, you know,

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put it together and you can
make a nice picture like this.

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Or, I suppose, you could
have made the nice picture

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before you had that information.

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And so we see that we've got
this line, y equals 3x minus 2

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here.

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And we have the curve
y equals x cubed.

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And they have two
intersection points.

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They intersect once down
at x equals minus 2,

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y equals minus 8.

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And they intersect again
up at the point (1, 1).

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And they're actually
tangent there.

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So x cubed, the curve y equals
x cubed stays above the line

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at this point.

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So one way you can read
that off is here you

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had a double root at
minus 1, if you like.

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But, OK, so then, the region in
question is this region here.

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So it's a region bounded
between those two curves

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that we're trying to
compute the area of.

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So now that we
have the endpoints

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this isn't such a
tricky problem at all.

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Right?

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so we've got the endpoints
and we know-- now

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that we've drawn this picture--
we know which curve is on top

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and which curve is on bottom.

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Right?

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So the height, when we imagine
cutting this region into lots

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of little rectangles or
approximate rectangles,

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the height is going to be
x cubed minus 3x minus 2.

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Minus the quantity 3x minus 2.

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Right?

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The x cubed is on top and
3x minus 2 is on the bottom.

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So the area is equal
to the integral.

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OK, and now we know where we
have to integrate from and to.

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So we're integrating from
x equals minus 2 to 1

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to get this whole region of
x cubed minus the quantity

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3x minus 2 dx.

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OK?

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Because the y equals x
cubed is the top curve.

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And y equals 3x minus
2 is the bottom curve.

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So this is the height of those
rectangles, which is positive.

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OK, and so now, OK, well now
this is pretty straightforward.

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We're integrating a
polynomial at this point.

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So this is the integral.

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Well, OK, so
integrating x cubed,

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that gives me x to
the fourth over 4.

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Integrating minus 3x gives
me minus 3x squared over 2.

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And integrating plus
2 gives me plus 2x.

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Between x equals minus 2 and 1.

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So OK, so now I just
do this difference.

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So this is equal to 1/4
minus 3/2 plus 2 minus-- OK,

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now I put in minus 2 I get 2
to the fourth-- sorry-- minus 2

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to the fourth is 16, over
4, so that's minus 4.

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OK, minus 3 times 4 over 2 is 6.

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And then plus 2 times
minus 2 is minus 4 again.

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OK, and now we've just
got some arithmetic.

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So this all becomes a plus 6.

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And I have to add
everything together.

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So that's something like, well
I've got a denominator of 4,

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so it's, yeah.

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All right, so there's a
secret I should tell you.

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Which is that mathematicians
are not actually very

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good at arithmetic, usually.

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So this is minus 6/4
plus 2 plus 6, so plus 8.

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So this is 8 minus 5/4,
which is 6 and 3/4.

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OK, that's 6 plus 3/4.

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So 6 and 3/4.

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This was just arithmetic.

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Here was the calculus part.

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And, in fact, you know, one
feature of a problem like this

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is that you have a fair
amount of work sometimes

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to see the picture of
what you're working with.

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Then the integration here
was pretty straightforward.

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Right?

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This was a fairly easy
integral to compute.

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If you're, say, better at
fraction arithmetic than I am,

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at least.

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So, there you go.

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So, we had this region.

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So we found, actually the
endpoints of the interval.

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We found which was the top curve
and which was the bottom curve.

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And that led us right
down this integral.

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And then that
integral was fairly

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easy to compute
from that point out.

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And then we ended up with
this as our final area.

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So I'll end there.