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# AP Calculus BC exams: 2008 1 b&c

Parts b and c of problem 1 (free response). Created by Sal Khan.

## Want to join the conversation?

• How would I use a ti-84 to find the number approximating the x values of the points of intersection?
• [2nd] → [Calc](above [Trace]) → intersect
Select curve 1
Select curve 2
Move your thing to approximately where the intersection is. If you're sure it only intersects at one point, you don't need to guess you can just press enter.
Answer will be at the bottom of the screen.
• Is there a cubic equation like the quadratic equation that we could use to solve third degree polynomials? And for that matter, also fourth, fifth and so on?
• At , Sal drew the x and y axes slanted. I don't understand why he did this.
• He's drawing a three-dimensional solid. If you were to draw the x- and y-axes on the ground, that's what the function would look like from a location off to the side. As for the shape of it, each cross-section of the solid is the same shape as the functions he's been talking about for the last couple of videos. The basic shape is outlined at or so, but Sal never (not even in the next few videos) draws the actual solid in its entirety.
• Should / could those limits of integration be found with (2 or 3 iterations of) newton's method instead of a calculator?
• why is the height -2- (x^3-4x) looking at the graph? I understand the -2 because it's below the y=2 line. But why do we subtract the cubic equation from that?
(1 vote)
• For the first problem, wouldn't you need to take the absolute value of that integral because the answer you would get would be negative (seeing as areas cannot be negative)?
(1 vote)
• Nope; if set up correctly, that integral should be positive.
(1 vote)
• Very dumb question: Why is the integral set up at -2 - (x^3 - 4x) instead of (x^3-4x) - (-2)?
(1 vote)
• This is because the area below the curve is negative. Taking the definite integral of f(x) where f(x) < 0 for a given interval results in an negative area which you can't have. It's all down to preference but It's clearer to an examiner that you understand the area is below the x axis.
(1 vote)
• Would it have been acceptable to the graders to recognize that x^3-4x below the y=-2 line is the same as x^3x-4x+2 below the x axis (y = 0 line). This is just a simple translation (which preserves everything). The integral (using the same points of integration) would result in -x^4/4 +2x^2 -2x. The same result as in the video if Sal calculated the antiderivative (which I know was not required). But the set up would be slightly different (letting S stand for the integral symbol): S (0-(x^3-4x+2))dx (the range would be the same). I don't see any loss of information or lack of understanding in this answer. For me, this would have saved me time on the test.
(1 vote)
• I don't know about the graders, since I don't know what they grade upon. But you are correct, if you move the function up by 2 units and then flip it over so the integral gives you a positive value (by being above the x-axis), then you get the exact same result.
(1 vote)
• can't you do the integral with respect with dy so that the bounds are integers?
(1 vote)
• Rasheed,

You could integrate with respect to y. However, you would have to find the length of the slices in terms of y. Considering that the endpoints of the slices would land on the same function, this method would be more difficult.
(1 vote)

## Video transcript

Let's keep doing the first problem from the 2008 Calculus BC exam. We're on Part b. I think that's too thick. OK it says, the horizontal line y equals negative 2 splits the region r-- this is r-- into two parts. Write, but do not evaluate, an integral expression for the area of the part of r that is below the horizontal lines. Let's draw y is equal to negative 2. So y equals negative 2 would look like this. y is just a constant. That's not thick enough. I don't know if you can see that. Let me do it as a thick line and in a darker color maybe. y is equal to negative 2 will look something like that. And what they're saying is if it splits this region r into two parts, this part and this part. And what they want to know is an integral expression for the area of the part of r that is below the horizontal line. So they care about the area of this part of r. And remember they just want us to write the expression, not evaluate it. So that'll hopefully save us time. So how do we figure this out? Well the easy part is actually to figure out what the expression we're going to take the definite integral of. And it's going to be a little bit harder to figure out the boundary points. So what is the expression within the definite integral that we will use? Well, just like we did in Part a, think about we're going to take a sum of a bunch of rectangles. And the height of the rectangles is going to be the difference between the two functions. And this is y is equal to negative 2, and then this function right here is y is equal to-- and we had written it down in Part 3-- but that's x cubed minus 4x. That's this curve right here. So the height of each of these little rectangles is going to be minus 2 minus x to the third minus 4x. That's the height of each of these rectangles. And then the width of each of those rectangles we know from-- well, just learning calculus or learning integrals-- the width is dx. So we're going to multiply that times dx. And then we're going to take all of the sums from x is equal to whatever this point is to whatever this point is. So we need to figure out this point, this value of f, which is going to be here, and then this value of x, which is going to be there. And so these are really just two of the points where these two functions intersect each other. So how do we figure out those points? Well what we could do is we can set them equal to each other, so we could say at what x values does x to the third minus 4x equal minus 2. At what x values are the y values the same? So we just set them equal to each other. If we wanted to write this is a proper polynomial expression, we would get x to the third minus 4x plus 2 is equal to 0. And I actually just tried to record a video where I was doing this on the fly, and I kept staring at this and I was like, boy, this is a hard polynomial to factor. I kept trying to guess numbers, or figuring out-- I even tried to do Newton's method-- and I kept getting weird numbers, and I became suspicious of myself. And then I looked at the actual test-- actually I can show it to you. It says right there, a graphing calculator is required for some problems or parts of problems. And I realized that they probably want us to use a graphing calculator to figure out the roots of this polynomial. So let's do that. It's been a long time since I actually took AP Calculus, and now I remember that a graphing calculator was a big deal. I just actually downloaded this TI-85 emulator. So let's use this to figure out the roots of this polynomial. Let's turn it on. If we want to figure out the roots we use the poly function, so second poly. What's the order of this polynomial? What's the third-degree polynomial? f of x to the third. So order is 3, enter. And what are the coefficients? Well the coefficient on x to the third term is 1. Go down. What's the coefficient on the x squared term? Well there is no x squared term, right? So that coefficient is 0. Go down. What's the coefficient on the x term? It's minus 4. So minus 4. Go down again. And then the coefficient or the constant term. Well that's just going to be 2. And now we can just hit solve. And we get three crazy numbers, and this shows you that this would have been very hard to solve analytically if you have a normal brain. So let's see. There are three places where y equals negative 2 intersects y is equal to x to the third minus 4x. It intersects at minus 2.21. Well that's off of this graph, that's not even here. That's somewhere off to the left. This curve probably comes back down and intersects over there at minus 2. But it also intersects at 1.675 which is probably right here. Right, that looks like 1.675. It also intersects at 0.539, which is right there. So we can use those values that our graphing calculator gave us and put it into our definite integral. So this point right here our polynomial solver told us is x is equal to 0.539. So we'll put here 0.539. And then this point right here-- so this is our limits of integration for our definite integral, right? We're going to sum up these little rectangles from x is equal to 0.539 to x is equal to 1.675. And they told us that they do not want us to evaluate it, so we are done with Part b. We could just write this and we should hopefully get full credit. Maybe they'd want you to simplify this a little bit, but I'd be surprised if they were mark off for that. Anyway, let's do Part c. If we have time. Part c. So they say the region r is the base of a solid. For this solid each cross-section perpendicular to the x-axis is a square. Find the volume of the solid. OK, so this is interesting. Let me see if I can draw it. So that curve, I'm going to draw it kind of with a perspective so you can see the solid they're talking about. So it's the same region r. So we had a sin function on the top. Looks something like that. And then we had that polynomial function on the bottom that looks something like that. I'm trying to draw it at an angle. So just to show you the x-axis. This is going to be the x-axis. Let me draw the y-axis. The way I drew it now it looks something like this. Trying to do a little bit of perspective so that we can visualize what they're talking about. So that's the y-axis. So that's x y. And what they're saying, this is the region r again, just like in the previous two parts of the problem. They say the region r is the base of a solid. So this is the base of a solid. For this solid each cross-section perpendicular to the x-axis. So a cross-section perpendicular to the x-axis. Let's see if we can draw that. So this would be a cross-section. So it's like we took a knife and we cut like this, we cut parallel to the y-axis. Say we took this cross-section of this solid right now. They say that it is a square. So that means that the base has to be the same height, has to be the same disk size as the height. So if we took the cross-section of the solid here it would be like that. Here it would be a smaller square like that. If we took the cross-section there, it would be a small square as well. So what they want us to do is figure out the volume of the solid. You can kind of imagine what it looks like. It's small squares and the squares get really big and then they get small again. So how do we do that? Well we do the same thing. We take the area of each of these squares-- we know they're squares-- times each of the dx's-- the small differential-- and we sum them up over from 0 to 2. On the first diagram, I think that was 2. Oh, and I'm about to be out of time. So I will continue this problem in the next video. See you soon.