2026 AP Calculus AB FRQ #2 Answers

Added: 2026-05-14

[00:00:00]Welcome back guys. In the last video, I explained FRQ1 from the 2026 calc exam.

[00:00:06]And today, now I'm going to explain FRQ2. So in this FRQ, we're given the function f is defined by this. And the function g is defined by this. The graphs of f and g intersect at 1, 2 and a, comma b. We're told that for x is greater than or equal to zero, the equation y= g ofx can be rewritten as x= h of y equals this. We're told the graph of this is a horizontal line and y = 3.5 as shown in the figure. Then it asks us let r be the region bounded by the graph of g the x-axis and the y-axis and the vertical line x= z as shown in the figure. Find the area of the region R.

[00:00:44]And we need to show the setup for our calculation. So basically all I need to do here is the function g of x is this.

[00:00:52]Let's actually write it in a in a different color here. Function of g of x is this. And so if I want to find the area of region R, I literally just need to find the integral from 0 to 1 of G of X, right? So I'm just going to have to do so G of X is this. I'll write out the integral from 0 to 1 of G of X dx. I guess I could write out the integral from 0 to 1 of 14x + 12 over x + 2 or sorry 12 dx. So if I do that, the integral from 0 to 1 of 14x + 12 / x + 12 dx going to get this. So the answer there would just be 1.513.

[00:01:32]Now in the next question, we're told region R described in part A is the base of a solid for so region R as described in part A is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a rectangle whose height is 1/3 the length of its base in the region r. And then we're told that uh we need to write but not evaluate in integral. Right? So what we know is that the area of this rectangle is going to be base time height. Right? We also know though that the height is equal to 1/3* the base. So we could actually just write out that the area is equal to the base * 1/3 of the base which is just equal to 1/3 base squar. Now what is the base? Well the base is going to be since it says here each crosssection perpendicular to the x-axis is a rectangle. It says region r is the base of a solid. And so basically it's saying that g of x minus so this is g of x minus 0. So minus whatever this is like the distance between g of x and zero that is whatever my base is right and so this is going to mean then that the base is actually just g of x. So that means then that the area is equal to 1/3 * g of x^2. And so what we end up writing then is that we're going to have the integral from 0 to 1 because it's still the region r of just 1/3 * g ofx^2. And then we'll have our dx. And if you want, you could write out your 14x + 12 over x + 12^2 dx. I personally wouldn't care to do all that. But if you're really paranoid on the exam, I believe you would probably end up doing that. Now, in part C, we're then told the shaded region in figure one is bounded by the graphs of f and g on the interval x is 0 to x= a. So it says is bounded by the graphs of f of g. So the shaded region, okay, is just here and here. And then it says, um, find the area of the shaded region. Well, first we need to find a. How can we do that?

[00:03:39]Well, we would have to set f ofx equal to g ofx. But I'm I'm actually going to just use a regression. I'm going to do 1.43 to the x1 plus.57. And I mean, you can do this in your calculator as well if you have a TI 84 plus CE. So it's equal to 14 x1 + 12 over x1 + 12. It says that x is going to be 3.255.

[00:04:03]So that's what the a value is. If it told me that a x was equal to one, which it could be, I would just say like x is greater than two for example because it looks here like I need it can be greater than two, right? So anyway, we get that uh a is going to equal 3.25582, right? So find the area of the shaded region. Well, first we're going to go from 0 to one where the top integral it looks like is the f ofx, right? So that's going to be of f ofx minus g ofx.

[00:04:36]And then after when we're going from one to a, which is 3.2 whatever, so one to a, it looks like g of x is now on top, right? So that'll be g ofx minus f ofx, right? And so we could write out this whole thing like we could do integral of I'll actually just do this. So I'll say f ofx equals and then I'll delete this.

[00:04:58]I know on the exam you can't copy and paste but for the sake of saving time I'm going to just copy and paste. So we have that this is g ofx and then we can just write g ofx equals and then I can just write out uh the integral from 0 to 1 of f ofx - g ofx plus and this should be dx here plus the integral from 1 to a which was 3.25582 2 5582 of g ofx - f ofx dx right and it gives us this value here so equals 632.

[00:05:40]Going on to the final part here we're told that t be the region bounded by the graph of y= or sorry x= h of y the y- ais and the horizontal line y= 3.5 as shown in the figure. Right? But do not evaluate an integral expression that gives the volume of the solid. Well, for this what we're going to do is the disk method, right? So, we're revolving around the y ais. So, it's going to look like this, right? And so, it's just going to be the distance. So, I mean, the values for our bounds will be 1 to three and a half here. So, 1 to 3.5. And then we're going from h of y to zero.

[00:06:19]That is the distance between these two.

[00:06:21]So, it would be like h of y minus 0. And remember we do pi * our r^ 2, right? And so our r here is the h of i minus 0. And so it ends up just being pi * the integral from 1 to 3.5 of h of y^ 2 d y.

[00:06:40]And again, like in all the other ones, if you wanted to write out the whole h of y, you could, but I don't think that is necessary.