00:01
Okay, in this problem we need to evaluate the integral from negative one to one of three x squared minus eight x dx.
00:10
All right, so what we need to realize when we're taking the integral like this is what's kind of, i forget what it's called, but it's like the power rule.
00:19
And what that means is we're going to take our exponent here and we're going to raise it by one.
00:24
And then we're going to divide by that number, right? so we're going to have x cubed and then we're going to divide by three.
00:33
So we'll have one third, that's dividing by three, times three x and then we'll have raised our x squared to x cubed.
00:41
Now we're going to say minus, we're going to take our x here and we're going to raise that from x to the first to x squared.
00:49
And that means that we'll have eight and then we're going to divide by two.
00:53
So i'm going to say instead of divide by two, i'm going to say multiplying by one half.
00:58
All right? so then we're going to take that and we're going to substitute negative one and one into them and subtract the difference.
01:05
So one third of three, that's easy, that's one.
01:08
So we'll have x cubed and then half of eight is four.
01:12
So we'll have x cubed minus four x squared evaluated from negative one to one.
01:18
All right? so i'm going to take one and substitute it in and then i'm going to subtract that value from what i get when i substitute in negative one...