If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:6:54

AP Calc: CHA‑5 (EU), CHA‑5.A (LO), CHA‑5.A.1 (EK)

What I want to do in this video
is find the area of this region that I'm shading in yellow. And what might seem challenging
is that throughout this region, I have the same lower function. Or I guess the
lower boundary is y is equal to x squared
over 4 minus 1. But I have a different
upper boundary. And the way that
we can tackle this is by dividing this
area into two sections, or dividing this region into two
regions, the region on the left and the region on
the right, where for this first region,
which I'll do-- I'll color even more in
yellow-- for this first region, over that entire interval in x. And it looks like x is
going between 0 and 1. y equals-- when x is equal to
1, this function is equal to 1. When x is equal to 1, this
function is also equal to 1. So this is the point 1 comma 1. That's where they intersect. So for this section, this
subregion right over here, y equals square root of x is the
upper function the entire time. And then we can have
a-- we can set up a different-- we can
separately tackle figuring out the area of this region. From x is equal to 1
to x is equal to 2, where y equals 2 minus
x, is the upper function. So let's do it. So let's first think
about this first region. Well, that's going to be
the definite integral from x is equal to 0 to
x is equal to 1. And our upper function is square
root of x, so square root of x. And then from that, we want to
subtract our lower function-- square root of x minus x
squared over 4 minus 1. And then of course,
we have our dx. So this right over here, this is
describing the area in yellow. And you could imagine it, that
this part right over here, the difference between
these two functions is essentially this height. Let me do it in a
different color. And then you
multiply it times dx. You get a little
rectangle with width dx. And then you do that for each x. Each x you get a
different rectangle. And then you sum them all up. And you take the limit as
your change in x approaches 0. So as you get ultra,
ultra thin rectangles, and you have an
infinite number of them. And that's our definition,
or the Riemann definition of what a definite integral is. And so this is the area
of the left region. And by the exact same
logic, we could figure out the area of the right region. The right region--
and then we could just sum the two things together. The right region, we're
going from x is equal to 0 to x-- sorry, x is equal to
1 to x is equal to 2, 1 to 2. The upper function is 2 minus x. And from that, we're going to
subtract the lower function, which is x squared
over 4 minus 1. And now we just
have to evaluate. So let's first simplify
this right over here. This is equal to the
definite integral from 0 to 1 of square root of x
minus x squared over 4 plus 1, dx-- I'm going to write
it all in one color now-- plus the definite integral
from 1 to 2 of 2 minus x, minus x squared over 4. Then subtracting a negative is
a positive 3-- or a positive 1. We could just add it to this 2. And so this 2 just becomes a 3. I said 2 minus
negative 1 is 3, dx. And now we just have to
take the antiderivative and evaluate it at 1 and 0. So the antiderivative
of this is-- well, this is x to the 1/2. Increment it by 1. Increment the power by
1, you get x to the 3/2, and then multiply
by the reciprocal of the new exponent-- so
it's 2/3 x to the 3/2. Minus-- the antiderivative
of x squared over 4 is x to the third, divided by 3,
divided by 4, so divided by 12, plus x. That's the antiderivative of 1. We're going to
evaluate it at 1 and 0. And then here the
antiderivative is going to be 3x minus x
squared over 2 minus x to the third over 12. Once again, evaluate it
at-- or not once again. Now we're going to
evaluate at 2 and 1. So over here, you evaluate
all of this stuff at 1. You get 2/3 minus 1/12 plus 1. And then from that, you
subtract this evaluated at 0. But this is just all
0, so you get nothing. So this is what the yellow
stuff simplified to. And then this purple stuff,
or this magenta stuff, or mauve, or whatever color this
is, first you evaluate it at 2. You get 6 minus-- let's see,
2 squared over 2 is 2, minus 8 over 12. And then from that,
you're going to subtract this evaluated at 1. So it's going to be 3 times 1--
that's 3-- minus 1/2 minus 1 over 12. And now what we're
essentially left with is adding a bunch of fractions. So let's see if we can do that. It looks like 12 would
be the most obvious common denominator. So here you have 8/12
minus 1/12 plus 12/12. So this simplifies
to-- what's this? This is 19/12, the part
that we have in yellow. And then this business,
let me do it in this color. So 6 minus 2, this is
just going to be 4. So we can write this as
48/12-- that's 4-- minus 8/12. And then we're going to have to
subtract a 3, which is 36/12. Then we're going to add to
1/2, which is just plus 6/12, and then we're
going to add a 1/12. So this is all going to simplify
to-- let's see, 48 minus 8 is 40, minus 36 is 4, plus
6 is 10, plus 1 is 11. So this becomes plus 11/12. Let me make sure
I did that right. 48 minus 8 is 40,
minus 36 is 4, 10, 11. So that looks right. And then we're ready
to add these two. 19 plus 11 is equal to 30/12. Or if we want to simplify
this a little bit, we can divide the numerator
and the denominator by 6. This is equal to
5/2, or 2 and 1/2. And we're done. We figured out the area
of this entire region. It is 2 and 1/2.

AP® is a registered trademark of the College Board, which has not reviewed this resource.