- [Voiceover] Let F and G be the functions defined by
F of X is equal to one plus X plus E to the X squared
minus two X, and G of X equal X to the fourth
minus 6.5X squared plus six X plus two.
Let R and S be the two regions enclosed by the graphs
of F and G shown in the figure above.
So here I have the graphs of the two functions
and they enclose regions R and S.
So the first thing they want us to figure out
is find the sum of the areas of region R and S.
So the sum of those areas,
you can think about it.
We're going to go from X equals zero right over here
to X equals two, so we're going to take the integral
from X equals, I'm going to write this down
area of R plus S is equal to,
let me just write that a little bit neater.
So the area of R plus S is going to be equal to,
let's see we can take the integral from X equals zero
to X equals two,
X equals zero to X equals two,
and what are we going to integrate?
Well we're going to integrate the difference
between the two functions
and really the absolute value of the difference
where we want the sums,
we don't want to have negative area here,
we want the sum of the areas of regions R and S
and at some point you have G of X is above F of X
and at other points F of X is above G of X,
but if we take the absolute value,
it doesn't matter which one we're subtracting from the other
we're just getting the absolute value of the difference.
So let's take the absolute value of,
the absolute value of F of X minus,
minus G of X DX.
So that's going to be the sum of the areas,
or we could say this is going to be the integral
from zero to two of the absolute value
F of X is one plus X plus E to the X squared
minus two X, minus G of X,
so minus X to the fourth, plus 6.5X squared,
minus six X minus two.
Do the absolute value.
DX.
Now this would be pretty hairy to solve
if we did not have access to a calculator
but lucky for us on this part of the AP exam
we can use a graphing calculator,
so let's do that to evaluate this definite integral here.
If you're wondering why I'm seeing minus two
instead of plus two remember we're substracting,
we're subtracting G of X, we're finding
the difference between them.
So let's input this function into my calculator
and I'm going to do it with the same thing
that I did in part one.
Where I'm just going to define,
let me turn it on.
All right.
So I'm going to actually clear that out
and I'm going to define this whole expression as Y one.
So I am going to take the absolute value,
so let me see where the absolute value
it's been a little while since I last used one of these.
So should give me some math,
math number, oh there you go.
Absolute value, so it's the absolute value
of one plus X, plus 2nd E to the
and then we have X squared
minus two X and then close that parenthesis
and then you have minus X to the fourth power
plus 6.5 times X squared minus six times X
minus two and then we have to close the parenthesis
around the absolute value.
All right, so we inputted Y one
and so now let's go over here
and also evaluate this definite integral.
So we go to math, and then we scroll down
to definite for function integral.
So click on that and we're going to use Y one,
so we go to variables,
we go to the right to go to Y variables,
which it's a function variable that we just defined,
and so we select Y one,
that's what we just inputted.
Our variable of integration is X
and our bounds of integration,
we'll we're going to go from X equals zero
to X equals two.
So we go from zero to two
and then we let the calculator munch on it a little bit
and we get, it's taking some time to calculate,
it's still, it's still munching on it.
Let's see this is taking a good bit of time.
There you ago, all right, so it's approximately 2.00,
if you want to get a little bit more precise,
2.004.
So this is approximately 2.004.