Evaluating definite integral of function with removable and jump discontinuities

0:00- [Instructor] We're told to evaluate the following integral, and f of x is defined in this way, and here is a graph of f of x, so pause the video and have a go at that. Okay, so this integral goes from x equals negative 6 to x equals 5. So x equals negative 6 is over here, all the way to x equals 5, which would be over here.

0:24So what we're gonna do is, we could think about all of this value. Now, it's not gonna be a positive value because our function is negative over this entire interval. And then it's going to go to this section right over here, when x is equal to 2, the function looks more like, or it is becomes this, and so we'll have this negative value.

0:49So another way to think about it's this area plus that area, but it would be negative. Now you might be saying, "Hey, you know, I have these little removable discontinuities here, won't that affect my area?" And the simple answer is no, because this is a point discontinuity.

1:04So you could say, "Okay, right below it you have an infinitesimally narrow section of no area," but it's infinitesimally small, so it doesn't actually take away from any of the area. So what we need to do is just break this up. So for this first negative area right over here, that is going to be the integral from negative 6 to 2, negative 6 to 2, and over that interval, this is what f of x looks like.

1:33It is x squared minus 9, over x plus 3, dx. And then to that, we are going to add... Maybe I'll do it in this color. We're going to do the integral over that interval, so that is going to be from x equals 2 to 5 of this business, negative 5 minus square root of x minus 1, and then we have dx.

2:02So now, if we evaluate this, you might say, "Hey, I still have... x can't be equal to negative 3," but I said it's equivalent to the area even if this discontinuity wasn't there. So this red part, we can say that this is going to be equal to the integral from negative 6 to 2.

2:18This numerator over here is x plus 3 over x minus 3. And for x not equaling negative 3, that would simplify to just x minus 3. So x minus 3, dx. And I can do this for the integral once again, because I'm visually looking at this area here, and I said, "Okay, that seems like this."

2:36Or I know that this little discontinuity isn't going to actually take anything out of my area. And then I'll have that other part right over there, plus the integral from 2 to 5 of negative 5 minus the square root of x minus 1. Actually, let me just evaluate this second integral over here.

2:57So the antiderivative of negative 5 is negative 5x, negative 5x, let me do some brackets here, and then the antiderivative of negative square root of x minus 1, what we have inside, you could do u-substitution here, or you can essentially, do a little bit of a shorthand of u-substitution, which is really the reverse chain rule and say, "Okay, I have the derivative of x minus one."

3:19Derivative of x minus 1 is 1. So I'm already multiplying that out here, so I can really just treat x minus 1 like an x." So what does that mean? This is x minus 1. Let me put this minus out here. This is x minus 1 to the 1/2 power. So if we increase that by one more power, this would be x minus 1 to the 3/2 power.

3:42And then I'm going to multiply that times the reciprocal of this. So times 2/3, and I'm going to evaluate that from 2 to 5. And actually, I'll leave you there. The important thing is because this is starting to look like a complex integral. But you have the tools in your toolkit to solve this integral...

4:06Actually, you know what? Let me not cop out. Let me just solve it for you. Okay, so let me move this a little bit down. So this red stuff, the antiderivative of x minus 3, is x squared over 2 minus 3x. And I'm gonna go from negative 6 to 2. And then I'm gonna step ahead on this part.

4:28Let me evaluate this. So if I evaluated it 5, it's negative, 5 times 5 is 25, minus 2/3, times 5 minus 1, is 4 to the 3/2. The square to 4 is 2, to the third is 8. So this is 8. And so, it's still not super clean. And then from that, I am going to subtract, so plus that, minus, evaluating all of that, at 2.

5:00So I'm gonna have negative 10, minus 2/3, 2 minus 1 is 2, to the 3/2, is just 1. So at least that came out a little bit simpler. So all of that business. And so, let's see, in red here, this is quite hairy, actually. So if I evaluate it at 2, I get 2 squared is 4 over 2, minus, 3 times 2 is 6, and then I'm gonna have minus, negative 6 squared is 36, divided by 2 is 18, negative 6 times negative 3 is plus 18.

5:41So all of the stuff that we have in red, is going to be, this is 2 minus 6, which is negative 4, and then we're gonna subtract 36 from that. All of that is going to be negative 40, in red. And then in green, let's see what I have here. If I were to rewrite all of that, it is going to be negative 25, minus 16 over 3, plus 10, plus 2 over 3.

6:17And so, this is going to be, if I take that part and that part, that gets us a negative 15. So I have a negative 15. And then if I take... Let's just use another color. If I take the negative 16 over 3 and 2 over three, if I sum that, that's negative 14 over 3.

6:36So minus 14 over 3, and so, this is going to be equivalent to... Let's see, this part right over here is negative 55, negative 55, and then this over here is, what's this, negative 4 and 2/3, negative 4 and 2/3. So minus 4 and 2/3, so this is equal to negative 59 and 2/3, very hairy math.

7:06But the main thing is to realize how you can breakdown that integral.