Calculus, all content (2017 edition)
- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c
2015 AP Calculus AB 2a
Areas between curves.
Want to join the conversation?
- Hmm. I usually just split the functions up into multiple intervals. In this case it was from 0 to 1.033 for R, and 1.033 to 2 for S. I guess with a calculator this way is much easier, but I was taught to separate the integrals.(8 votes)
- Yeah, separating them into parts is pretty standard as well. Really, it's accomplishing the same thing because when you evaluate the integral of an absolute value function, at every point the inner function would have switched signs you would have to split the integral and evaluate the positive or negative version of whatever is inside the absolute value signs. That amounts to being the same thing you would do if you split the integral right away to solve this problem.
Still, it's important to know how to split integrals in case you have a shape that's defined by three or more functions and the interaction isn't as simple as a single intersection point.(5 votes)
- Let's say I want to complicate my life a little and find the areas as the sum of two separate areas R and S.
1) How do solve the equation in order to find the point of intersection? (x^4 - 6.5 * x^2 + 5 * x + 1 - e^(x^2 - 2 * x) = 0) It appears to be unsolvable.
2) Assuming problem #1 has been resolved, what is the effective way to find which function is greater? Or should I simply take an absolute value and that should do the job?
3) How do you solve the integral of (e^(x^2 - 2 * x)dx)?
4) If I do not have a calculator on the exam, does that mean I am doomed?(5 votes)
- From my experience, yes. There is no easy way to solve an equation where: a^f(x) = g(x). The only way to do this simply is with the equation solver on a TI-84 or other calculator with similar abilities. If you plan on taking the AP Calculus exam, you should invest in a calculator. Additionally, on problem 1 in this series, you have to take the definite integral of sin(x^2) or on this question the definite integral of e^(x^2 - 2x)dx. This isn't possible without a graphing calculator or memorization of complex irregular integrals that the test designers don't expect you to know.
As for your question 2, I look at the graphs above. Perhaps this is something I can only see, but the graph of g(x) is clearly the top border of R and the bottom border of S. This may have something to do with the fact that g(x) is a degree 4 polynomial and f(x) is an exponentially based function.
Another way to explain is to take the derivative of g(x) and see that it should have 2 roots between 0 and 2 - horizontal tangents, while the exponential function has at maximum one.(5 votes)
- For the question 2a at1:44, why don't we have to split up the two portions of the graph to integrate them. In section R, g(x) is above the f(x) line however in section S the reverse is true. Wouldn't this affect the answer?(3 votes)
- I never used the way he uses here to solve this kind of problem because I always have been splitting the functions into two intervals, but the way he uses is much easier and more efficient with the aid of calculator.(3 votes)
- Nice solution! Much easier than what I was trying to do , which as Joe pointed out was trying to figure out where f and g crossed in the interval and separate them out. I thought it kinda looked like where the inflection point was in the polynomial but that was 1.041 and actual value was 1.032832 (close). Can someone PLEASE tell me what the indefinite integral is on the exponential function , I was thinking x + x^2/2 + e ^(x^2-6) but I'm guessing there must be something more on the e term given the complex exponent. Doesnt look like a simple substitution cause if u= x^2 -6 then I cant get rid of the x. Soooo, I set u equal to the whole e term, then when integrating I get u^2/2 and when resubstituting back in, the 2 can come down from the exponent and cancel with the denominator and your left with just the original e term in the function as your integral on that term. Is that correct??(3 votes)
- I still don't get how he shrunk the area of both regions into the integral of | f(x)-g(x) |dx. Would it work it he still kept the same limits of integration 0 and 2 and wrote it as the integral of [g(x)-f(x)] + [f(x)-g(x)] for R + S?(2 votes)
- At1:24how does he know which function f(x) is and which one g(x) is? how does he know which one to subtract from which in the integral? (as in, how does he know to do | f(x) - g(x) | as opposed to | g(x) - f(x) |?)(3 votes)
- He doesn't have to know which function is above or below on what interval because he is taking absolute value of the difference between the two functions.(4 votes)
- You are not allowed to use absolute value; my AP calculus teacher told us that a point was awarded for doing the correct sequence in the integral of top-bottom(0 votes)
- [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.