If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

AP Calculus BC exams: 2008 1 c&d

parts c and d of problem 1 in the 2008 AP Calculus BC free response. Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user Daniel Aaron
    At , I've put the exact same definite integral into my calculator (Casio fx-9860 GII) and it just doesn't seem to give me the same answer. I get approx. 10.227. I'm positive that I have the integral entered correctly and I'm positive that I understand the concept. Is there anything I can check to determine whether or not I have it right?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Aditya Baradwaj
      I encountered the same problem while using my TI-84 plus. Turns out the calculator was in degree mode. That means that the input for the trigonometric functions (sin, cos, tan, etc.) would be in degrees. But when we do problems like these, we want the input to be in radians, the more fundamental unit. That way, the zeroes and critical points of the function 'sin(pi*x)' will align with rational values of 'x'. Just change your calculator to radian mode, and you will get the right answer. In the TI-84, you can do this by pressing (MODE) and selecting (RADIAN). Hope I helped!
      (2 votes)
  • blobby green style avatar for user khill
    how do you ad fraction with decimals
    (0 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Drew Culpepper
      I'm not sure I understand your question, but you would have to convert both of them to the same representation. That most likely means converting the fraction to a decimal, since the decimal is probably irrational. Then you can add them as normal numbers. That's more a question for the arithmetic playlist, though.
      (4 votes)
  • blobby green style avatar for user hendrix.coli55
    bad web connection how do I fix this
    (0 votes)
    Default Khan Academy avatar avatar for user

Video transcript

So we were doing part c of the first problem on the Calculus BC exam, the free response part, and so what, I'll re-read it, it says the region r, this is the region r, is the base of a solid. I redrew the region r here, but with a little perspective, so we can hopefully visualize it in three dimensions. So it says for this solid, each cross section perpendicular to the x-axis, right? There's two ways you can do a cross-section, you could do them, you could cut this way, but that would be parallel to the x-axis. We wanted to cut perpendicular to the x-axis, or parallel to the y-axis, right? So if we cut it along the line like that. So they say, each cross-section perpendicular to the x-axis is a square. So I drew a couple of squares here. That's one. So here, this would be the base, and since we know the cross-section is the square, the height has to be the same length as the base. Same thing here. So here, the height will be really high, because this is kind of maybe our maximum point in terms of the base width. But it's still pretty wide, then it gets narrow again. So how do we figure out the volume of this solid? Which is kind of hard to visualize. And it's in some ways you easier to visualize it than it is to draw, so you have to give me some credit. But anyway, what we do is, we take the area of each of these squares, and I drew one of them here. You take you each of the area of each of these squares, multiply them by a super small change in x, and that we know, from everything we've learned in calculus hopefully, that that super small change in x, I'm trying to draw a little perspective, is dx. So if we multiply dx times the area of this square, that is the volume of this kind of part the entire solid. And if we were to sum up all of these infinitely thin solids, we would get the volume for the whole thing. So how do we do that? Well let's write our integral expression. So what is the area of each of these squares? Each of these cross-sections, what is the area? Well the base is going to be the difference between our two functions, right? This top function right here, that was sin of pi x. And this bottom function, right here, that is y is equal to x to the third minus 4x. So the base of the functions, the base of this distance right here is going to be the difference between the top function and the bottom function. So each base is going to be sin of pi x, minus this function, so minus x to the third, plus 4x, right switch the sign. Minus x to the third, plus 4x. So far this might look pretty similar to part a, but what's the twist here? We want the area. The area of each of the squares, not just this distance. So what's the area? It's going to be this distance squared, so we have to square this entire thing. So that's the area of each of these squares, and then we have to multiply them times a little bit of dx. And that gives us the volume for each of these, I guess, parts of the entire solid. And then what are our boundaries of integration? Well it's the same thing as in part a. This is 0, this is 2. So the boundaries are pretty straightforward. So we essentially just have to evaluate this now. And just like in part b, I've first tried to evaluate this analytically, and you end up getting a very, very, very hairy integral. Which you can do analytically, you but you have to know some power reduction formulas in trigonometry, you have to do some integration by parts, it would probably use up all of the time on the AP Exam. So since they said that a graphing calculator is required for some parts of the problem, I say why not use our graphing calculator? Because the graphing calculator is very good at numerically evaluating definite integrals like this. So let's get out my TI-85 emulator again. Here we go. I want you see the key strokes so that you, ok turned on, exit out of this. So we're going to use the calculus function here, so second, calculus. And this function right here, this is definite integral, a very useful function to use. Press F5, definite integral. And then we just type in the expression. So the expression, let me move it down a little bit. So it's open parentheses sin of where is pi? Pi is -- I haven't used one of these calculators in a long, long time -- oh there it is. Second pi x sin of pi x minus x to the third power, plus 4x. All of that squared. And then this definite integral function, you have to tell it which is the independent variable, or kind of, you know, what variable are we integrating across, and that's the variable x. And then you just tell it the boundaries of integration, and we're done. So integrate from 0, from x is equal to 0, to x is equal to 2. If I haven't made a mistake, I can hit enter, and let the calculator do the rest of the work. Let's see what it ends up with. OK, 9. -- that's the answer, that is the volume of this solid -- it's 9.9783 So you could write that this is equal to 9.9783. And I'm pretty sure they want you to use a calculator, because frankly, computing the integral, that's kind of just chug math, you know, very mechanical math, although it's pretty sophisticated, but it would take you forever. But this is kind of, I think, what they wanted you to do, set up the integral, recognize that each of the squares, the area of each of the squares is just going to be this distance, the distance between the functions squared. And then you integrate that from 0 to 2. Let's see how much time I have left. I have a couple of minutes. So let's do part d. The region r models, let me, paste, oh, I didn't want to do that, edit, undo, edit, paste, there you go. OK, so I wanted to make this a little smaller. So what does part d say? The region r models the surface of a small pond, so that's the surface now. At all points r at a distance x from the y-axis, the depth of the water is given by h of x is equal to 3-x. So essentially 3-x is the depth, right? So at this point of this pond, the depth is just 3, right 3-0. And at this point, the depth is 3-2 which is 1. So essentially the pond is going to get shallower and shallower or as we go further to the right. You could almost imagine it, let me see if I can draw it again. So this is the sin function with some perspective. This is the polynomial function below it. This is, it's probably drawn in, that's the x-axis. This is the y-axis. And so here, the depth of the pond is given by the function h of x is equal to 3-x. So over here, the depth is 3, so if I were to draw, so if I were to go straight down the depth is, you know maybe it's 3. And the pond essentially gets shallower and shallower as we go to the right. So how do we figure out the volume of this? Over here, what is the depth? It's going to be 1, right, 3-2. This is x is equal to 2. So here the depth is going to be 1. So if we take the cross-section just along the x-axis, the depth is going to look something like this, but then this is the top of it. I know that's kind of hard to visualize. But anyway, how do we figure out what the volume of this lake is? But actually, I realize I'm pushing nine minutes, so I will continue this in the next video. See you soon.