WEBVTT 00:00:01.500 --> 00:00:09.704 Topic 19. The Fundamental Theorem of Calculus and the Definite Integral. Let's read through the 00:00:09.706 --> 00:00:19.700 Fundamental Theorem of Calculus. For any continuous function f on the interval from a to b, notice that 00:00:19.702 --> 00:00:32.800 this is a closed interval, the integral from a to b of f of x dx is equal to, now the capital F of x 00:00:32.802 --> 00:00:46.703 means the antiderivative, evaluated from a to b, which means to substitute b into the antiderivative, 00:00:46.705 --> 00:00:55.505 a into the antiderivative and subtract. Once again, the derivative of capital F of x is f of x. 00:00:55.507 --> 00:01:01.604 Now at this point we're not going to discuss at all the meaning or applications of the fundamental theorem. 00:01:01.606 --> 00:01:10.700 We will save that for topic 20. Instead what we do here is practice the skill of evaluating using 00:01:10.702 --> 00:01:24.700 this part of the fundamental theorem. Let's take a look at many examples. Number one. 00:01:24.702 --> 00:01:41.099 The integral, and this is read from 1 to 3, so that's a and that's b, of x cubed dx is. 00:01:41.101 --> 00:01:47.200 Now most of the time the challenge here is still to find the antiderivative, and the substitution part, they 00:01:47.202 --> 00:01:53.704 are rather tedious sometimes, but they're not particularly difficult. So let's concentrate on finding 00:01:53.706 --> 00:02:06.300 the antiderivative first. We have here x to the n, and so the antiderivative would be x to the n plus 1 00:02:06.302 --> 00:02:19.600 over 4, but instead of putting plus c, you evaluate this function from 1 to 3. We substitute the 3 in first, 00:02:19.602 --> 00:02:34.500 into the antiderivative. This would be 3 to the fourth over 4, then minus, now this is the F of b, minus 00:02:34.502 --> 00:02:47.304 and then the 1 in, so that's 1 to the fourth over 4. That's minus F of a. That's this part, and then this is 00:02:47.306 --> 00:02:53.301 the tedious part. Obviously we're going to have to deal with a lot of fractions because this rule itself 00:02:53.303 --> 00:03:00.701 because of the division by the power tends to give us lots of fractions. 3 to the fourth would be 9 squared, 00:03:00.703 --> 00:03:16.098 so that would be 81 over 4, minus 1 over 4. So that's 80 over 4 which is 20. So what we've done is 00:03:16.100 --> 00:03:34.504 we've used the fundamental theorem to evaluate this definite integral. This is different because the 00:03:34.506 --> 00:03:43.301 result is going to be an actual constant rather than a function. When you work with an indefinite integral 00:03:43.303 --> 00:03:54.901 you don't have values to substitute in, and you end up with a function. Alright, let's look at example two. 00:03:54.903 --> 00:04:07.702 The integral from negative 1 to 1 of negative 6 dx. Alright, the antiderivative is negative 6x plus c, 00:04:07.704 --> 00:04:14.205 evaluated from negative 1 to 1. Now about this time I think we'd better stop and talk about, 00:04:14.207 --> 00:04:23.400 so what happened to this c? Well clearly it is not part of this, but why? Well, because when you sub in here 00:04:23.402 --> 00:04:33.604 and you get the antiderivative F of b, what you'd have is you'd have F of b plus c, then you'd have minus in parenthesis 00:04:33.606 --> 00:04:41.200 F of a plus c. Which means you have c minus c, which is 0. That explains why the c is just completely 00:04:41.202 --> 00:04:49.003 non-existent in this evaluation process. Ok, we substitute in this number first. This is always b. 00:04:49.005 --> 00:04:56.801 So that would be negative 6 times 1 is negative 6 minus, this becomes an exercise in keeping track 00:04:56.803 --> 00:05:08.800 of negative signs. When we substitute negative 1 into here, it's negative 6 times negative 1 which is 6. 00:05:08.802 --> 00:05:23.300 Let's do that again. This is F of b. This is F of a. F of b is 1 into here. F of b into here, negative 6, 00:05:23.302 --> 00:05:38.000 a into here would be 6. So I have negative 6 minus 6, and so the value of this definite integral is negative 12. 00:05:38.002 --> 00:05:46.903 Example three. Notice as we go through this we're doing a really good review of finding antiderivatives, and 00:05:46.905 --> 00:06:02.903 here we're still using the simplest rules. Example three. The integral from 1 to 4 of 2 x to the negative 3/2 dx. 00:06:02.905 --> 00:06:21.202 The antiderivative would be 2 x to the, add 1, negative 1/2 divided by negative 1/2, so that's the same thing 00:06:21.204 --> 00:06:31.904 as times. This is divided by negative 1/2. That's the same thing as times negative 2, evaluated from 1 to 4. 00:06:31.906 --> 00:06:39.602 Now why did I do this? Just to save steps, make it neater, easier to work with. Now I'm going to simplify 00:06:39.604 --> 00:06:50.103 this expression before I substitute in. The numerator would be negative 4. The denominator would be the 00:06:50.105 --> 00:07:10.804 square root of x. Now evaluate at the endpoints. I substitute 4 in. That's F of b minus, and then 00:07:10.806 --> 00:07:25.801 I substitute a in, 1 into here. The square root of 1 would be 1. So I have negative 4 over 2, 00:07:25.803 --> 00:07:43.104 which is negative 2, and then plus 4, which is 2. Go back and check your work. The antiderivative. 00:07:43.106 --> 00:07:53.202 Add 1, divide by negative 1/2, so that's times negative 2. Simplify. This would be the square root 00:07:53.204 --> 00:08:10.302 in the denominator. Sub in the 4, the 1, check your terms, 2. Let's look at quite a few more. Example four. 00:08:10.304 --> 00:08:28.908 The integral from 1 to 2 of e to the x dx. Alright, the antiderivative is e to the x. We evaluate from 1 to 2. 00:08:28.910 --> 00:08:46.802 So we get e squared minus e to the 1, which is just e, and unless we have some reason why we want a decimal 00:08:46.804 --> 00:08:56.403 representation, that's perfectly sufficient. Example five. The integral from 1 to 2, and you notice I use 00:08:56.405 --> 00:09:01.500 a lot of 1's and 2's. This is a skill practice here and those are easy numbers to work with so we 00:09:01.502 --> 00:09:13.905 practice with those numbers. The integral from 1 to 2 of 3 over x dx. So this is 3 times 1 over x. 00:09:13.907 --> 00:09:30.200 The antiderivative would be 3 ln absolute value of x evaluated from 1 to 2. 2 into here would be 3 ln 2, 00:09:30.202 --> 00:09:44.807 and 1 into here would be 3 ln 1. What's ln 1? Well, remember that this is the power you raise e to to get 1. 00:09:44.809 --> 00:09:52.104 So what power do you raise e to to get 1? Zero. Once you realize this term is 0, you can substitute 00:09:52.106 --> 00:10:04.201 0 into here. 0 times negative 3 is 0, and so we get 3 times the natural log of 2, which is a constant. 00:10:04.203 --> 00:10:12.301 Let's try another example where we have to use some more complicated techniques or at least something different. 00:10:12.303 --> 00:10:32.902 Example six. We have the integral from 0 to 1 of x minus 2 times x plus 1 dx. Aha. We have a product. 00:10:32.904 --> 00:10:41.000 So we have a couple of choices here. We could multiply, or we could maybe think that this would be 00:10:41.002 --> 00:10:46.901 u to a power and maybe this would be the derivative of u. Well that's obviously not the case because 00:10:46.903 --> 00:10:53.904 the derivative of either one of these is 1. So the simplest thing to do in this case is to multiply first 00:10:53.906 --> 00:11:09.801 before we do anything else. We would get x squared minus x minus 2 dx, and now we can use the difference rule 00:11:09.803 --> 00:11:26.404 to get the antiderivative. It would be x cubed over 3 minus x squared over 2 minus 2x. We want to evaluate 00:11:26.406 --> 00:11:37.401 the entire thing, so we need some parentheses here, from 0 to 1. So we're going to have to find F of b, 00:11:37.403 --> 00:11:55.400 this part, by subbing the 1 into here. So I get 1/3 minus 1/2 minus 2 times 1 is 2. That's F of b, and 00:11:55.402 --> 00:12:04.800 F of a would be 0 minus 0 minus 0. So that's very convenient. Alright, let's see what we have here. 00:12:04.802 --> 00:12:18.801 This would be, we need a common denominator here, let's go with 6. So this would be 2/6 minus 3/6 minus 12/6. 00:12:18.803 --> 00:12:32.601 2/6 minus 3/6 minus 12/6. You always have to check the fraction part. So this would be negative 13/6. 00:12:32.603 --> 00:12:39.502 And what did we do here? We multiplied first in order to find the antiderivative. Let's take an example 00:12:39.504 --> 00:12:55.100 where we have to divide. Example seven. Let's evaluate the integral from 1 to 2 of x squared plus x plus 1 00:12:55.102 --> 00:13:07.400 all over x squared, dx. We have a quotient. The simplest thing to do is to divide. So we will divide x squared 00:13:07.402 --> 00:13:15.004 into each of these terms. We rewrite the function as the integral, or the expression as the integral from 1 to 2 00:13:15.006 --> 00:13:33.700 of 1 plus 1 over x plus 1 over x squared dx. Notice again a lot of the work here is finding the antiderivative. 00:13:33.702 --> 00:13:45.800 This is going to be a natural log. This one is going to be a power rule. So the antiderivative of 1 is x. 00:13:45.802 --> 00:13:57.604 The antiderivative of 1 over x is ln absolute value of x, and the antiderivative of x to the negative 2 is going 00:13:57.606 --> 00:14:14.402 to be add 1, over that, evaluated from 1 to 2. Let's clean this up a little bit. This would be x plus 00:14:14.404 --> 00:14:28.802 ln absolute value of x minus 1 over x evaluated from 1 to 2. Ok, we have the antiderivative. We've got this 00:14:28.804 --> 00:14:44.201 step. We're on to this step. 2 subbed in would be 2 plus ln 2 minus 1/2 minus. See how I use the parentheses? 00:14:44.203 --> 00:14:51.600 I would strongly encourage this. This is F of b minus F of a. These take a long time, but your chances of getting 00:14:51.602 --> 00:14:57.702 them right the first time are much higher if you set this up clearly with the parentheses and the negative. 00:14:57.704 --> 00:15:07.203 Ok, where were we? We were at putting in a 1, so we have 1 plus ln 1, which we know is 0, minus 1. So we 00:15:07.205 --> 00:15:22.101 have in these parentheses, 1 minus 1 is 0, and ln 1 is 0. This term is 0. So we have 2 plus ln 2 minus 1/2. 00:15:22.103 --> 00:15:36.500 This is 4/2 minus 1/2 is 3/2 plus ln 2. Alright, let's move on now to some examples where we have to use some 00:15:36.502 --> 00:15:59.900 type of substitution. Example eight. The integral from 1 to 5 of 2x minus 1 raised to the 5/2 power dx. 00:15:59.902 --> 00:16:17.201 So this is going to be a u to the n problem where u is 2x minus 1. The correct du/dx is 2, but 2 is not there, 00:16:17.203 --> 00:16:24.802 so I correct or adjust the constant by putting in a 2 and a 1/2 on the outside. Now the most common error 00:16:24.804 --> 00:16:37.500 on this problem is to lose the 1/2 so be careful. The antiderivative would be 1/2 times 2x minus 1. When 00:16:37.502 --> 00:16:57.504 we add 1 it's in the form of 2/2, so we have 7/2 divided by 7/2 which is times 2/7, evaluated from 1 to 5. 00:16:57.506 --> 00:17:11.105 Anybody lose track of where we are? Let's go back and check it. u to the n, we need a 2 for the du/dx, 00:17:11.107 --> 00:17:22.105 so we have a 1/2 constant out in front. There's the 1/2. This is u to the n plus 1 divided by 7/2 which is 2/7. 00:17:22.107 --> 00:17:34.400 Now we apply this part of the theorem. The 2x - 1 to the 7/2. Well, let's see. I think maybe we need 00:17:34.402 --> 00:17:44.604 another step here. 2 over 2 is 1, so let's bring the 1/7 out to the front. Then let's rewrite this as the 00:17:44.606 --> 00:17:59.101 square root of 2x minus 1 to the 7th. So that's the square root of 2x minus 1 raised to the 7th, 00:17:59.103 --> 00:18:07.203 evaluated from 1 to 5. Now you may wonder why I would choose a problem with numbers like this. Well, 00:18:07.205 --> 00:18:12.403 because these are the kinds of problems that come up a lot in applications, and you just have to take your time 00:18:12.405 --> 00:18:24.700 here when you evaluate. First of all, F of 5 would be 1/7 times the square root of 5 times 2 is 10 00:18:24.702 --> 00:18:37.403 minus 1 is 9. The square root of 9 is 3, so that's 3 to the 7th, minus 1/7. 1 into here would be, 00:18:37.405 --> 00:18:49.605 2 minus 1 is 1. Square root of 1 is 1. 1 to the 7th is 1. Let's go back and check this. 00:18:49.607 --> 00:18:59.100 5 into here. 10 minus 1 is 9. The square root of 9 is 3. So that's 3 to the 7th. 1 into here is 1. 00:18:59.102 --> 00:19:07.804 Square root of 1 is 1, and so we get, and we're not going to worry about 3 to the 7th. We'll leave this 00:19:07.806 --> 00:19:23.002 as 3 to the 7th minus 1/7. Ok, let's look at number nine. Remember here that we're practicing a skill, and 00:19:23.004 --> 00:19:37.501 this skill takes lots of practice. Example nine. The integral from 0 to 1 of e to the negative x dx. 00:19:37.503 --> 00:19:44.100 Now this one's very simple and innocent looking, but it's actually e to the u, and we're going to have to 00:19:44.102 --> 00:19:56.705 make an adjustment here. u is negative x. du/dx is negative 1, and we don't have that in here, so we need a 00:19:56.707 --> 00:20:04.803 negative here and a negative here. Negative 1 times negative 1 is positive. The antiderivative is going 00:20:04.805 --> 00:20:17.002 to be e to the u, but don't forget this out in the front. Then we're going to evaluate from 0 to 1. 00:20:17.004 --> 00:20:33.800 Subbing in the 1 and then the 0. Let's check it. There's a negative here. This is the negative that's 00:20:33.802 --> 00:20:45.502 part of the theorem and then negative here. Now, e to the 0 is 1. So this part is going to be plus 1, 00:20:45.504 --> 00:20:55.001 and this is negative 1 over e, and that's where we're going to leave this one. 00:20:55.003 --> 00:21:15.602 Another one with e. Example ten. The integral from 0 to 4 of e to the square root of x divided by 00:21:15.604 --> 00:21:26.104 the square root of x dx. I always like to do this one because when you back up and look at it it's a 00:21:26.106 --> 00:21:34.101 little intimidating. So we tear it apart, and we look at what we have. We have the integral from 0 to 4 00:21:34.103 --> 00:21:55.903 of what? This is e to the x to the 1/2 power, and this is 1 over x to the 1/2, and then we have dx. 00:21:55.905 --> 00:22:17.602 Let's see if this is the correct du/dx. u is x to the 1/2. du/dx is, very careful, 1/2 x to the negative 1/2. 00:22:17.604 --> 00:22:23.102 Alright, we're good on this part because x to the negative 1/2 is the same thing as 1 over x to the 1/2. 00:22:23.104 --> 00:22:35.400 So what do we need? We need the 1/2. So we put in a 1/2, and we multiply by a 2 out in the front. 00:22:35.402 --> 00:22:43.200 So now we have 2 times the antiderivative which is e, and I'm going to go back. Instead of writing 00:22:43.202 --> 00:22:52.204 x to the 1/2, let's switch back to the square root. Why? It's just easier to evaluate that way. If you would 00:22:52.206 --> 00:23:03.702 rather x to the 1/2, that's fine. We're going to evaluate from 0 to 4. We sub in the 4, so we have 2 e, 00:23:03.704 --> 00:23:24.904 square root of 4 is 2, minus 2 e to the 0. e to the 0 is 1, so this is 2 e squared minus 1 times 2, 00:23:24.906 --> 00:23:34.401 which is 2. You may notice on this one that I didn't use as many sets of parentheses. That's because 00:23:34.403 --> 00:23:40.304 everything in this term was positive. Both of these numbers were positive, and so I let this negative sign 00:23:40.306 --> 00:23:52.004 just stand out here because it's the only one in the problem that comes from the theorem. Number eleven. 00:23:52.006 --> 00:23:58.804 Can you guess what kind of rule number eleven is going to be? We've done a lot of x to the n. 00:23:58.804 --> 00:24:07.603 Then we did e to x. Then we did ln x. Then we did u to the n, and then we did e to the u. I guess 1 over u 00:24:07.605 --> 00:24:27.703 is probably next. So let's look at example eleven. The integral from 0 to 1 of x over 1 plus 2 x squared, dx. 00:24:27.705 --> 00:24:39.502 So why is this 1 over u? Because the term in the denominator is u raised to the first power. So let's rewrite 00:24:39.504 --> 00:24:53.901 it as 1 over 1 plus 2 x squared times x, and as your skill at this gets better you start leaving spaces in 00:24:53.903 --> 00:25:03.209 places where you know things are going to need to go. For example, u equals 1 plus 2 x squared. 00:25:03.211 --> 00:25:12.205 So this is 1 over u. So the derivative of u with respect to x is 4x, and you could see that as you begin to 00:25:12.207 --> 00:25:21.103 do the problem. I need a 4 here. I didn't leave enough space here though. I need a 1/4 here. Antiderivative. 00:25:21.105 --> 00:25:44.005 1/4 natural log of u, absolute value, evaluated from 0 to 1. The 1 in here would be 1 squared is 1, 00:25:44.007 --> 00:25:57.000 times 2 is 2, plus 1 is 3. 1/4 ln 3. Keep going. Minus the 0 into here. Now sometimes you're tempted to 00:25:57.002 --> 00:26:04.701 think when this is 0 that just means the term is going to be 0. No, that's the x that's 0. So this is x squared, 00:26:04.703 --> 00:26:15.301 so that's 0 squared, times 2 is 0, plus 1 is 1. So I end up with 1/4 ln 1. But what actually happens is 00:26:15.303 --> 00:26:29.305 once again ln 1 is 0. 0 times a fourth is 0. This whole term is 0, and my final answer is 1/4 natural log 3. 00:26:29.307 --> 00:26:46.910 Alright, we have two more cases to look at. Example twelve. We have the integral from 1 to 2 of ln x 00:26:46.912 --> 00:26:57.304 over x, dx, and you've seen this before, and you found the antiderivative before. You have to recognize that 00:26:57.306 --> 00:27:04.401 even though this is natural log it is not going to be the rule that ends up with the antiderivative 00:27:04.403 --> 00:27:19.503 being natural log. This is u to the 1. The correct du/dx is what? It's 1 over x, and it's there. 00:27:19.505 --> 00:27:32.706 So the antiderivative is the natural log of x, squared, over 2, evaluated from 1 to 2. Once again using 00:27:32.708 --> 00:27:46.701 the fundamental theorem, we sub in 2, so this would be ln 2 quantity squared divided by 2 minus 00:27:46.703 --> 00:28:00.103 ln 1 quantity squared divided by 2. What's ln 1 again? For about the third or fourth time, it's 0. 00:28:00.105 --> 00:28:14.104 Final answer, natural log of 2 squared divided by 2, and we're done. One last example. 00:28:14.106 --> 00:28:34.006 The integral from 1 to 2 of e to the x over e to the x plus 5, dx. This is u to the 1. 00:28:34.008 --> 00:28:44.805 It's not e to the u. It's u to the 1 in the denominator. u is e to the x plus 5, so the correct du/dx 00:28:44.807 --> 00:28:55.899 is e to the x, and yes, we are excited. It is correct, and so the antiderivative, no adjustment necessary, 00:28:55.901 --> 00:29:10.499 is the natural log of e to the x plus 5, evaluated from 1 to 2, which is, the natural log of the 00:29:10.501 --> 00:29:20.103 absolute value of e squared plus 5, which is a positive number, but we have the bars. We'll just leave them. 00:29:20.105 --> 00:29:30.704 And then the natural log of, we put in a 1. This would be e to the 1 plus 5. Let's check this one. 00:29:30.706 --> 00:29:41.601 Here's u. The correct du/dx is here, so the antiderivative is the natural log of the absolute value of u, 00:29:41.603 --> 00:29:56.104 evaluated at 2, minus the evaluation at 1. 2 into here is e squared plus 5, e plus 5. At that point we will stop. 00:29:56.106 --> 00:30:01.905 I would say that at this point we have reviewed most of the types of integration that we've covered. 00:30:01.907 --> 00:30:06.103 We now know how to use the fundamental theorem to evaluate. 00:30:06.105 --> 00:30:12.900 We move on to the next topic where you finally find out what all of this means.