2015 AP Calculus AB 2a | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Let f and g be the functions defined by ( f(x) = 1 + x + e^{x^2 - 2x} ) and ( g(x) = x^4 - 6.5x^2 + 6x + 2 ). 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. The first thing they want us to figure out is to find the sum of the areas of region R and S.

So the sum of those areas, you could think about it, we're going to go from ( x = 0 ) right over here to ( x = 2 ). So we're going to take the integral from ( x = 0 ) to ( x = 2 ).

Let me write this down: the area of R + S is equal to... let me write that a little bit neater. So the area of R + S is going to be equal to...

Let's see, we could take the integral from ( x = 0 ) to ( x = 2 ). 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.

Remember, 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. At some point, ( g(x) ) is above ( f(x) ), and at other points, ( f(x) ) is above ( g(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 ( f(x) - g(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 0 to 2 of the absolute value ( f(x) = 1 + x + e^{x^2 - 2x} - g(x) , (-x^4 + 6.5x^2 - 6x - 2) ) take 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. And if you're wondering why did I say minus two instead of plus two, remember we're subtracting.

We're subtracting ( g(x) ); we're finding the difference between them.

So let's input this function into my calculator, and I'm going to do 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_1 ). So I am going to take the absolute value.

So let me see where the absolute value is; it's been a little while since I last used one of these. So maybe some math. Math. Number... Oh, there you go, absolute value.

So it's the absolute value of ( 1 + x + 2 e^{x^2 - 2x} - x^4 + 6.5x^2 - 6x - 2 ), and then we have to close the parentheses around the absolute value.

All right, so we've inputted ( y_1 ), and so now let's go over here and evaluate this definite integral. So we go to math and we scroll down to definite for function integral.

So click on that, and we're going to use ( y_1 ). So we go to variable, we go to the right to go to ( Y ) variables. It's a function variable that we just defined, and so we select ( y_1 ).

That's what we just inputted. Our variable of integration is ( x ), and our bounds of integration? Well, we're going to go from ( x = 0 ) to ( x = 2 ).

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... time to calculate. It's still munching on it.

Let's see, this is taking a good bit of time. There you go! All right, so it's approximately 2. If you want to get a little more precise, it's 2.4. So this is approximately 2.4.