 In this video we provide the solution to question number seven for the practice exam number one for math 1220 in which case we have to set up the integral that'll compute the area of the region bounded by the curves y equals x squared and y equals x plus two. Notice that we set up the integral but we do not evaluate it. I'm going to first try to consider the picture in play here. So we have the parabola y equals x squared. That's going to give you something like the following. We have to also consider the line y equals x plus two. That's a line whose x as y intercept is two and whose slope is positive. So we'd expect something like this. There's some point of intersection here. There's some point of intersection here. We're looking for this region. We want to find the area of that region. So let's first figure out where these things intersect each other. So I set the two equal to each other. x squared is equal to x plus two. If you set the right hand side equal to zero, you'll get x squared minus x minus two equals zero. I am expecting that one of the answers will be positive, one of them will be negative. I can probably get away with factoring this thing. Factors of negative two, let's try x minus two and x plus one equals zero. That would give us the solutions of two and negative one. And sure enough, negative two times one is negative two. Negative two x plus one x is equal to negative x. That is the correct factorization. So these numbers have got to be two and negative one as I anticipated. Now during this interval, the line is always above the parabola for the entire time. So that would tell us that the area will be the integral from negative one to two. We're going to take the function on top, which is x plus two. And then we subtract from the function at the bottom, which is x squared. So we would get the integral from negative one to two of the function x plus two minus x squared dx. And do remember the dx, that is part of the integral. If you don't have it, you would be missing something there. The bounds are necessary and that then gives us the integral. We don't have to evaluate it. So that actually is all of the problem.