 In this video, we provide a solution to question number one from exam practice exam one for math 1220, in which case we have to evaluate the integral from negative two to two of the function three x plus one squared with respect to x. My first thoughts when I look at this one is probably just foil the thing out the three x plus one squared. If you do that, you would end up with a nine x squared plus six x plus one. I then they're going to choose to integrate that thing find an anti derivative anti derivative of nine x squared will become a three x cubed. The anti derivative of six x to be three x squared and the anti derivative of one would then be an x at that point. We can plug in negative two and positive two. So then we're going to just work through that calculation there. So we get three times two, two cube, which is an eight. Then we're going to get three times two squared, which is four. And then we get a two. Then we subtract from that when we plug in the negatives there, you're going to get three times negative eight plus three times negative four actually positive four square there. And then you're going to get a negative two like so. So some things to note here, three times eight is 24. And here because there's a double negative negative, you're actually going to get 24 twice. So that gives us a 48. In this situation of three times four, which is 12 over here, you also get a 12, which is a minus. So actually cancel each other out. And then lastly, you're going to get two minus a negative two, which will actually make it a positive four. So you get 48 plus four, which then simplifies to be a 52. So we see that the correct answer is in fact D. And that's I think the simplest way to approach this one. If you wanted to, you could have also made this work using a U substitution. You could have made U equal three X plus one D U equals in that case three for which you could have a three right here and a one third right here. If you're going to do that, you probably should change the bounds as well. When you plug two inside of this, you end up with a seven. So you could rewrite your integral as one third, the integral from seven on top. When you plug a negative two in the bottom, you'll get a negative five. And then your function looks like U squared D U for which then you can integrate it from there. You're going to get a one ninth U cubed as you go from negative five to seven. And so you get one ninth, you're going to get seven cubed. And then you'll get a negative negative five cubed. And I'll let you finish this out and show that's equal to 52. The U substitution is a viable option, but I think with this one just foiling out and working out turned out to be much simpler. And so that's the recommendation that the recommended approach I would take on this one.