Area Between Curves Example 3

Uploaded by TheIntegralCALC on 14.06.2011

Hi, everyone! Welcome back to today, we’re going to be doing another area
between curves problem. And in this one we’ve been given the two curves f(x) = 12x – 3x^2
and g(x) = 6x – 24. And we’ve been asked to find the area between those curves on the
range x = 1 to x = 7. So let’s go ahead first and take a look
at the graphs of these two functions. We have the purple graph here which represents f(x)
so let’s go ahead and indicate this one here as f(x) and we have the orange graph
which represents g(x). And if you have access to a graphing calculator, you can go ahead
and graph these on that, otherwise these two functions are pretty simple to sketch the
graphs of without the calculator. Either way, we’ve got f(x) and g(x) and we’ve been
asked to find the area on the range 1 to 7. SO this is x= 1 right here and x = 7 is right
here through g(x). So we have to find the area between the curves between these two
lines here. So you can tell that the graph, these two functions cross each other. So g(x)
is above f(x) from 4 to 7. From 1 to 4 though, f(x) is above g(x). They cross each other
at 4 right here on the center. So what we need to do is go ahead and confirm
if the intersection is in fact at 4. It looks like it is based on these graphs but if you
couldn’t visually, you’d want to find where they intersected anyway and the way
that we do that is just by setting them equal to each other. So we’ll say 12x – 3x^2
= 6x – 24. And we move everything to the same side, we’ll get 3x^2 + 6x – 12x – 24
and when we simplify, we’;ll get 3x^2 – 6x – 24. If we divide by 3, we’ll get x^2
-2x – 8. And then you can factor this to get (x – 4)(x + 2). So to know that the
graphs cross each other, x has to be equal to 4 and -2. So that’s where the graphs
intersect. Luckily for us, we just have to concern ourselves with one point because -2
here is outside our range. Remember that we only care about the range 1 to 7 and -2 is
outside that range so we can go ahead and ignore it. 4 is the only intersection point
that we care about and we’ve confirmed that in fact our intersection point is at 4 as
we saw on the graph right here. So what that means for us is that we need
to go ahead and write our integral. First, we need an integral from 1 to 4. So the left-hand
end point to the intersection point at 4 and then from the intersection point at 4 to the
right-hand endpoint at 7. And from 1 to 4, remember that f(x) is above g(x). It’s greater
than g(x). You can see here on our graph, f(x) is higher, it’s above, it’s greater
than g(x). So we will go ahead and write our integral from 1 to 4, the left endpoint to
the point of intersection. And because f(x) is above g(x), we’ll write f(x) first 12x
– 3x^2. And then subtract the lower function which is g(x) 6x – 24 dx. Then we need to
add to that the area from 4 to 7. From the intersection point 4 to the right endpoint
of our range 7. And because g(x) is above f(x), the graphs have crossed each other,
we’ll go ahead and write g(x) first so 6x – 24 and then we’ll subtract from that
the lower function so we’ll get 12x – 3x^2 dx.
So we’ve now set up our integrals to find the area and all we have to do is evaluate
these definite integrals. The first thing we want to do is simplify these integrals
as much as possible so we’ll get 12x – 3x^2 – 6x + 24 dx and then plus the integral
from 4 to 7 of 6x – 24 – 12x + 3x^2 dx. Then we can simplify further. Let’s order
our terms from the highest degree to the lowest degree. We’ll put –3x^2 first since it’s
the highest degree term. 12x – 6x = 6x, plus 24 dx and then we add to that the integral
from 4 to 7 of 3x^2 first. 6x – 12x = -6x, minus 24 dx.
So now, we can go ahead and integrate because we’ve simplified as much as possible and
remember that when we’re integrating, we will add one to the exponent of each of these
terms and then divide the coefficient on the term by the new exponent. So if we take 3x^2
here, we’ll add one to the exponent so instead of x^2, we get x^3. We added 1 to 2 to get
3 and then we will divide the coefficient which is -3 by the new exponent which is currently
3. So we divide -3/3 and of course we just get -1 so the coefficient is -1. So –x^3.
6/2 = 3x^2 and then plus 24x and of course we’re going to be evaluating that on the
range 1 to 4. Then we add to that this other integral so we’ll get x^3 – 3x^2 – 24x
which we’re evaluating on the range 4 to 7.
So now that we’ve done that we will go ahead and plug in our limits of integration. So first, remember that when
you’re evaluating these definite integrals, you always plug in the top number first and
then subtract and then plug in the bottom number. So we’ll plug in 4 first. So we’ll
get -4^3 + 3(4^2) + 24(4) and then we will subtract from that what we get when we plug
in 1 which is the lower limit of integration. So -1^3 + 3(1^2) + 24(1) and then we will
add to that what we get when we plug in the limits of integration from 4 to 7. So again
we’ll plug in 7 first so we’ll get 7^3 – 3(1^2) – 24(x) and then subtract from
that what we get when we plug in 4, so we get 4^3 – 3(4^2) – 24(4). So simple as
that. It’s just arithmetic at this point. Everything’s set up.
So when we simplify this, you can see we get -64 + 48 + 96 and then we will subtract everything that we get here. So we’ll get
-1 + 3 + 24 then we’ll add to that 343 – 147 – 168 and
then subtract from that 64 – 48 – 96. So when we simplify these all out, we’ll
get -64 + 48 + 96 + 1 – 3 – 24 + 345 – 147 – 168 – 64 + 48 + 96. And if we do that
arithmetic on our calculator, we will get a final answer of Area = 162. And that’s
our final answer. That is the total area between the curves on the range 1 to 7. Visually,
that’s everything here right between those two curves from 1 to 4 and then between the
two curves here from 4 to 7. It’s this whole area right here.
So that’s the area between the curves on the range 1 to 7 when the graphs intersect
at 4. I really hope that video helped you guys and I will see you in the next one. Bye!