WEBVTT

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What I want to do in this video
is find the area of this region

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that I'm shading in yellow.

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And what might seem challenging
is that throughout this region,

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I have the same lower function.

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Or I guess the
lower boundary is y

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is equal to x squared
over 4 minus 1.

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But I have a different
upper boundary.

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And the way that
we can tackle this

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is by dividing this
area into two sections,

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or dividing this region into two
regions, the region on the left

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and the region on
the right, where

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for this first region,
which I'll do--

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I'll color even more in
yellow-- for this first region,

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over that entire interval in x.

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And it looks like x is
going between 0 and 1.

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y equals-- when x is equal to
1, this function is equal to 1.

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When x is equal to 1, this
function is also equal to 1.

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So this is the point 1 comma 1.

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That's where they intersect.

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So for this section, this
subregion right over here,

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y equals square root of x is the
upper function the entire time.

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And then we can have
a-- we can set up

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a different-- we can
separately tackle figuring out

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the area of this region.

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From x is equal to 1
to x is equal to 2,

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where y equals 2 minus
x, is the upper function.

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So let's do it.

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So let's first think
about this first region.

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Well, that's going to be
the definite integral from x

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is equal to 0 to
x is equal to 1.

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And our upper function is square
root of x, so square root of x.

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And then from that, we want to
subtract our lower function--

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square root of x minus x
squared over 4 minus 1.

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And then of course,
we have our dx.

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So this right over here, this is
describing the area in yellow.

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And you could imagine it, that
this part right over here,

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the difference between
these two functions

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is essentially this height.

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Let me do it in a
different color.

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And then you
multiply it times dx.

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You get a little
rectangle with width dx.

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And then you do that for each x.

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Each x you get a
different rectangle.

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And then you sum them all up.

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And you take the limit as
your change in x approaches 0.

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So as you get ultra,
ultra thin rectangles,

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and you have an
infinite number of them.

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And that's our definition,
or the Riemann definition

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of what a definite integral is.

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And so this is the area
of the left region.

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And by the exact same
logic, we could figure out

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the area of the right region.

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The right region--
and then we could just

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sum the two things together.

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The right region, we're
going from x is equal to 0

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to x-- sorry, x is equal to
1 to x is equal to 2, 1 to 2.

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The upper function is 2 minus x.

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And from that, we're going to
subtract the lower function,

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which is x squared
over 4 minus 1.

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And now we just
have to evaluate.

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So let's first simplify
this right over here.

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This is equal to the
definite integral

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from 0 to 1 of square root of x
minus x squared over 4 plus 1,

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dx-- I'm going to write
it all in one color now--

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plus the definite integral
from 1 to 2 of 2 minus x,

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minus x squared over 4.

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Then subtracting a negative is
a positive 3-- or a positive 1.

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We could just add it to this 2.

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And so this 2 just becomes a 3.

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I said 2 minus
negative 1 is 3, dx.

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And now we just have to
take the antiderivative

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and evaluate it at 1 and 0.

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So the antiderivative
of this is-- well,

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this is x to the 1/2.

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Increment it by 1.

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Increment the power by
1, you get x to the 3/2,

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and then multiply
by the reciprocal

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of the new exponent-- so
it's 2/3 x to the 3/2.

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Minus-- the antiderivative
of x squared over 4

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is x to the third, divided by 3,
divided by 4, so divided by 12,

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plus x.

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That's the antiderivative of 1.

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We're going to
evaluate it at 1 and 0.

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And then here the
antiderivative is

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going to be 3x minus x
squared over 2 minus x

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to the third over 12.

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Once again, evaluate it
at-- or not once again.

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Now we're going to
evaluate at 2 and 1.

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So over here, you evaluate
all of this stuff at 1.

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You get 2/3 minus 1/12 plus 1.

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And then from that, you
subtract this evaluated at 0.

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But this is just all
0, so you get nothing.

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So this is what the yellow
stuff simplified to.

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And then this purple stuff,
or this magenta stuff,

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or mauve, or whatever color this
is, first you evaluate it at 2.

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You get 6 minus-- let's see,
2 squared over 2 is 2, minus 8

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over 12.

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And then from that,
you're going to subtract

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this evaluated at 1.

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So it's going to be 3 times 1--
that's 3-- minus 1/2 minus 1

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over 12.

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And now what we're
essentially left with

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is adding a bunch of fractions.

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So let's see if we can do that.

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It looks like 12 would
be the most obvious

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common denominator.

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So here you have 8/12
minus 1/12 plus 12/12.

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So this simplifies
to-- what's this?

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This is 19/12, the part
that we have in yellow.

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And then this business,
let me do it in this color.

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So 6 minus 2, this is
just going to be 4.

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So we can write this as
48/12-- that's 4-- minus 8/12.

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And then we're going to have to
subtract a 3, which is 36/12.

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Then we're going to add to
1/2, which is just plus 6/12,

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and then we're
going to add a 1/12.

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So this is all going to simplify
to-- let's see, 48 minus 8

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is 40, minus 36 is 4, plus
6 is 10, plus 1 is 11.

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So this becomes plus 11/12.

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Let me make sure
I did that right.

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48 minus 8 is 40,
minus 36 is 4, 10, 11.

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So that looks right.

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And then we're ready
to add these two.

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19 plus 11 is equal to 30/12.

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Or if we want to simplify
this a little bit,

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we can divide the numerator
and the denominator by 6.

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This is equal to
5/2, or 2 and 1/2.

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And we're done.

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We figured out the area
of this entire region.

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It is 2 and 1/2.