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Studying for a test? Prepare with these 3 lessons on AP Calculus AB solved exams.

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# 2015 AP Calculus AB 2a

Video transskription

- [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.