WEBVTT
00:00:00.280 --> 00:00:04.040
The fundamental theorem of calculus: evaluating definite integrals.
00:00:04.340 --> 00:00:09.090
In this video, we will learn how to evaluate definite integrals using the fundamental theorem of calculus.
00:00:09.360 --> 00:00:11.770
The theorem is usually stated in two parts.
00:00:11.950 --> 00:00:15.030
For the purpose of this video, weβll be focusing on the second part.
00:00:15.400 --> 00:00:24.740
This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ is any antiderivative of lowercase π.
00:00:24.980 --> 00:00:29.730
Here expressed as capital πΉ prime of π₯ is equal to lowercase π of π₯.
00:00:30.120 --> 00:00:39.700
Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to uppercase πΉ of π minus uppercase πΉ of π.
00:00:40.190 --> 00:00:45.830
Now here, an interesting point to note is that weβve said capital πΉ is any antiderivative of lowercase π.
00:00:46.050 --> 00:00:50.570
This means that if there are many antiderivatives, it does not matter which one we pick to use our theorem.
00:00:50.880 --> 00:00:54.480
To make sense of this, let us think back to what we mean by an antiderivative.
00:00:54.720 --> 00:01:03.080
By now, we should be familiar with the fact that the first part of the fundamental theorem of calculus essentially tells us that differentiation and integration are inverse processes.
00:01:03.210 --> 00:01:10.870
This means that we can find the general form for the antiderivative of a function, lowercase π of π₯, by evaluating the indefinite integral of that function.
00:01:11.110 --> 00:01:12.800
Letβs consider an example function.
00:01:12.830 --> 00:01:15.110
Lowercase π one of π₯ is equal to two π₯.
00:01:15.400 --> 00:01:21.800
Its antiderivative, uppercase πΉ one of π₯, would be equal to the indefinite integral of two π₯ with respect to π₯.
00:01:22.020 --> 00:01:27.220
To solve this, we use the power rule of integration, raising the power of π₯ by one and dividing by the new power.
00:01:27.540 --> 00:01:30.740
Of course, here, we mustnβt forget to add on our constant of integration, π.
00:01:31.050 --> 00:01:33.720
More simply, our answer is written as π₯ squared plus π.
00:01:33.970 --> 00:01:37.600
Now, this constant of integration π can take any value we like.
00:01:37.730 --> 00:01:43.010
And our expression would still be one of an infinite number of antiderivatives of our original function πΉ one of π₯.
00:01:43.220 --> 00:01:47.180
If we took the case of π equals zero, πΉ one of π₯ would simply be π₯ squared.
00:01:47.550 --> 00:01:50.980
π equals five would give us an antiderivative of π₯ squared plus five.
00:01:51.270 --> 00:01:55.690
π could even be negative π, which would mean our antiderivative would be π₯ squared minus π.
00:01:56.050 --> 00:01:58.630
All of these are antiderivatives of two π₯.
00:01:58.950 --> 00:02:04.930
Given that the fundamental theorem of calculus allows us to use any of these antiderivatives to evaluate a definite integral, what do we do.
00:02:05.220 --> 00:02:08.590
Letβs continue to use our example function π one of π₯ equals two π₯.
00:02:08.760 --> 00:02:12.330
That now, weβre looking to the definite integral between π and π of this function.
00:02:12.540 --> 00:02:18.720
The theorem tells us that this is equal to the antiderivative capital πΉ one of π minus capital πΉ one of π.
00:02:19.010 --> 00:02:23.170
We can arbitrarily pick one of our antiderivatives, for example, π₯ squared plus five.
00:02:23.330 --> 00:02:27.840
If πΉ one of π₯ equals π₯ squared plus five, then πΉ one of π equals π squared plus five.
00:02:27.910 --> 00:02:29.680
Similar reasoning follows for πΉ one of π.
00:02:29.900 --> 00:02:34.340
Simplifying this, we see that there is a plus five and a minus five term, which cancel each other out.
00:02:34.440 --> 00:02:36.790
And weβre therefore left with π squared minus π squared.
00:02:37.130 --> 00:02:42.910
In fact, if we had used the general form which would involve any constant π, the same thing wouldβve happened.
00:02:43.090 --> 00:02:46.260
So regardless of the constant we use, we arrive at the same result.
00:02:46.420 --> 00:02:51.790
Given this fact, we can simply choose to ignore the constant of integration altogether when evaluating a definite integral.
00:02:51.960 --> 00:02:55.460
And this is essentially equivalent to taking the case where π is equal to zero.
00:02:55.650 --> 00:03:00.220
If we had considered this case to begin with, we wouldβve reached the same answer, but with fewer steps of working.
00:03:00.320 --> 00:03:02.560
Let us now see an example of our theorem in practice.
00:03:02.670 --> 00:03:05.350
Let π of π₯ equal six π₯ squared plus one.
00:03:05.670 --> 00:03:09.690
Evaluate the definite integral of π from π₯ equals two to π₯ equals three.
00:03:09.950 --> 00:03:14.990
For this question, weβve been asked to evaluate a definite integral, which using standard notation would look like this.
00:03:15.120 --> 00:03:17.440
We see that the integrand is our function π of π₯.
00:03:17.530 --> 00:03:21.010
And the limits of integration are two and three, as specified by the question.
00:03:21.110 --> 00:03:25.740
To evaluate this definite integral, weβre gonna be using the second part of the fundamental theorem of calculus.
00:03:25.770 --> 00:03:33.830
This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ prime of π₯ is equal to lowercase π of π₯.
00:03:34.030 --> 00:03:37.410
In other words, capital πΉ is an antiderivative of lowercase π.
00:03:37.680 --> 00:03:45.270
Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
00:03:45.440 --> 00:03:47.520
Letβs now apply this theorem to solve our problem.
00:03:47.770 --> 00:03:51.390
Our function, lowercase π of π₯, is equal to six π₯ squared plus one.
00:03:51.500 --> 00:03:55.930
To find this antiderivative, capital πΉ of π₯, we can integrate lowercase π of π₯.
00:03:56.060 --> 00:04:05.260
If we apply the power rule of integration, raising the power of π₯ in each of our terms and then dividing by the new power, we get an answer of six π₯ cubed over three plus π₯.
00:04:05.360 --> 00:04:07.570
And we also add our constant of integration π.
00:04:07.800 --> 00:04:14.110
At this point, we remember that the fundamental theorem of calculus allows us to use any antiderivative, which means π can take any value.
00:04:14.370 --> 00:04:18.180
It makes sense for us to choose the simplest case possible, which is when π is equal to zero.
00:04:18.470 --> 00:04:20.560
Essentially, this means we can ignore this constant.
00:04:20.860 --> 00:04:24.830
Simplifying, the antiderivative that weβll then use is two π₯ cubed plus π₯.
00:04:25.050 --> 00:04:25.330
Great.
00:04:25.360 --> 00:04:28.840
Letβs set this antiderivative to one side and go back to our original calculation.
00:04:28.890 --> 00:04:31.460
The question has asked us to evaluate this definite integral.
00:04:31.720 --> 00:04:33.500
Our upper limit of integration is three.
00:04:33.620 --> 00:04:34.980
And our lower limit is two.
00:04:35.350 --> 00:04:42.220
The fundamental theorem of calculus then tells us that this integral is equal to capital πΉ of three minus capital πΉ of two.
00:04:42.530 --> 00:04:49.050
Now, since we have just found capital πΉ of π₯, our antiderivative, we can substitute in the values of three and two into this function.
00:04:49.200 --> 00:04:53.090
After we have input these values, we can perform a few simplifications on our new expression.
00:04:53.310 --> 00:04:56.720
After working through these simplifications, we reach an answer of 39.
00:04:56.910 --> 00:04:58.710
With this step, weβve completed our question.
00:04:58.930 --> 00:05:01.920
The definite integral given in the question evaluates to 39.
00:05:02.150 --> 00:05:06.330
We evaluated our integral using the antiderivative of the given function πΉ of π₯.
00:05:06.460 --> 00:05:10.390
And the tool that we used to help us was the second part of the fundamental theorem of calculus.
00:05:10.850 --> 00:05:13.240
Okay, before moving on, a quick word on notation.
00:05:13.450 --> 00:05:20.670
Given the definite integral between π and π of a function π of π₯ with respect to π₯, very often, youβll see the next step written in the following form.
00:05:20.800 --> 00:05:27.440
With the antiderivative of the given function written in brackets like so and the limits of integration carried over to the right-hand bracket.
00:05:27.660 --> 00:05:28.970
And this is simply a shorthand.
00:05:29.080 --> 00:05:35.110
Itβs a way of expressing the antiderivative as a function before we substitute in our limits of integration and evaluate.
00:05:35.290 --> 00:05:39.670
It is equivalent to what we should now be familiar with, which is capital πΉ of π minus capital πΉ of π.
00:05:39.910 --> 00:05:44.370
Youβll almost always see this step, since itβs a very useful shorthand, which helps to tidy up our working.
00:05:44.460 --> 00:05:47.980
Looking back at our previous example function, π one of π₯ is equal to two π₯.
00:05:48.170 --> 00:05:57.820
If we were to consider the definite integral between one and three of this function with respect to π₯, our next step would look something like this, with the antiderivative of π one of π₯ in brackets, as shown.
00:05:58.020 --> 00:06:02.700
We would then proceed to continue our evaluation by substituting in three and one, the limits of integration.
00:06:02.800 --> 00:06:04.430
And weβd eventually reach an answer of eight.
00:06:04.850 --> 00:06:07.260
Let us look at an example question using this notation.
00:06:07.600 --> 00:06:13.640
Evaluate the integral between zero and two of two sin π₯ minus three π to the π₯ with respect to π₯.
00:06:14.090 --> 00:06:18.170
To answer this question, weβre going to be using the second part of the fundamental theorem of calculus.
00:06:18.440 --> 00:06:26.670
This tells us that if lowercase π is a continuous function on the closed interval between π and π and capital πΉ is any antiderivative of lowercase π.
00:06:26.880 --> 00:06:34.580
Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
00:06:35.040 --> 00:06:41.890
Looking back at our question, the first thing we might notice is that the function lowercase π, which is our integrand, consists of two different terms.
00:06:42.060 --> 00:06:44.580
The first term involves the trigonometric function sine.
00:06:44.740 --> 00:06:46.720
And the second involves the exponential π.
00:06:46.990 --> 00:06:53.500
Now, we should be familiar with the fact that both trigonometric and exponential functions of this form are continuous over the entire set of real numbers.
00:06:53.530 --> 00:07:01.670
We have therefore fulfilled the criteria that our function π must be continuous over the closed interval between π and π, which in our case is the closed interval between zero and two.
00:07:01.890 --> 00:07:07.070
Now, given that we do have two terms, we might find that our working is clearer if we split these up into separate integrals.
00:07:07.100 --> 00:07:11.330
We would do this like so, remembering to keep the limits of integration the same across both terms.
00:07:11.510 --> 00:07:13.700
We can now evaluate each of these integrals separately.
00:07:13.850 --> 00:07:17.250
The antiderivative of two sin π₯ is negative two cos π₯.
00:07:17.320 --> 00:07:21.760
And the antiderivative of negative three π to the π₯ is negative three π to the π₯.
00:07:22.120 --> 00:07:27.200
Of course, remember, we can ignore the constant of integration in both cases since weβre working with definite integrals.
00:07:27.390 --> 00:07:33.710
Here, we note that weβve expressed our antiderivative in brackets, with the limits of integration being carried over to the right-hand bracket in both cases.
00:07:33.820 --> 00:07:37.650
Given that both of these brackets have the same limits of integration, we can simply combine them.
00:07:37.910 --> 00:07:43.830
Now, you might notice that we couldβve moved directly from our original integral to this step, by treating each of the terms individually.
00:07:44.070 --> 00:07:49.720
Instead of splitting our integral into two and then recombining, we would simply have found the antiderivative of each of our terms.
00:07:50.130 --> 00:07:52.890
If youβre not sure, however, thereβs no harm in writing out the method in full.
00:07:53.250 --> 00:07:57.550
To move forward with our question, we now substitute in the limits of our integration, which are zero and two.
00:07:57.750 --> 00:07:59.300
We then reach the following expression.
00:07:59.540 --> 00:08:02.630
With our first set of parentheses, thereβre no simplifications necessary.
00:08:02.630 --> 00:08:03.720
So we can just leave this.
00:08:03.780 --> 00:08:07.400
For the second set of parentheses, we might recall that cos of zero is equal to one.
00:08:07.430 --> 00:08:10.190
So negative two cos zero is equal to negative two.
00:08:10.380 --> 00:08:12.800
Alongside this, π to the power of zero is also one.
00:08:12.930 --> 00:08:15.490
So negative three π to the power of zero is negative three.
00:08:15.680 --> 00:08:19.930
Our second set of parentheses therefore becomes negative two minus three, which is negative five.
00:08:20.050 --> 00:08:21.930
But for our final answer, weβre subtracting this.
00:08:21.960 --> 00:08:23.430
So weβre left with a positive five.
00:08:23.550 --> 00:08:25.100
And weβve now reached our final answer.
00:08:25.360 --> 00:08:31.940
The definite integral given in the question evaluates to negative two cos of two minus three π squared plus five.
00:08:32.560 --> 00:08:40.800
Okay, if we look back to the fundamental theorem of calculus, another important condition to consider is the continuity of the function lowercase π for which weβre integrating.
00:08:41.070 --> 00:08:45.700
Remember that the theorem states that our function must be continuous on the closed interval between π and π.
00:08:45.910 --> 00:08:48.020
This π and π form the limits of our integral.
00:08:48.230 --> 00:08:53.600
If lowercase π is not continuous over this interval, then we cannot confidently say that this relationship is true.
00:08:54.020 --> 00:08:57.140
To illustrate this, let us consider the function one over the square root of π₯.
00:08:57.320 --> 00:09:00.090
If we were to draw the graph of this function, it might look something like this.
00:09:00.220 --> 00:09:04.670
Now letβs consider the definite integral of our function between one and two with respect to π₯.
00:09:04.870 --> 00:09:09.490
This could be interpreted as the area under the curve between one and two, as shown on our graph.
00:09:09.680 --> 00:09:15.090
To evaluate this, we might want to reexpress one over the square root of π₯, so that the power of π₯ is in a more manageable form.
00:09:15.290 --> 00:09:17.460
We then use the familiar power rule of integration.
00:09:17.710 --> 00:09:18.640
And we would simplify.
00:09:18.870 --> 00:09:20.570
If we continued, weβd find no problems.
00:09:20.570 --> 00:09:21.910
And weβd reach numerical answer.
00:09:22.340 --> 00:09:27.470
Okay, what would happen instead if we were asked to evaluate the definite integral between negative one and one?
00:09:27.680 --> 00:09:29.450
Here, we might start to run into some problems.
00:09:29.570 --> 00:09:33.940
It should be clear from our graph that π of π₯ is undefined when π₯ is less than or equal to zero.
00:09:34.090 --> 00:09:37.680
Trying to imagine the area under our curve between these bounds would be meaningless.
00:09:37.810 --> 00:09:43.170
Since our function is undefined over part of the closed interval between negative one and one, it cannot be said to be continuous.
00:09:43.260 --> 00:09:48.830
Given this fact, thereβs no sense in continuing, since we cannot use the fundamental theorem of calculus to evaluate this integral.
00:09:49.210 --> 00:09:51.260
Letβs take a look at an example to illustrate this.
00:09:51.510 --> 00:09:57.080
Evaluate the integral between four and nine of negative two times the square root of π₯ with respect to π₯.
00:09:57.330 --> 00:09:59.800
For this question, weβve been asked to evaluate a definite integral.
00:10:00.030 --> 00:10:07.200
With questions of this type, it can sometimes be useful to move constant factors, such as the negative two, from the inside of the integrand to the outside.
00:10:07.580 --> 00:10:14.080
Next, we might also find it useful to reexpress our square root of π₯ as π₯ to the power of a half or π₯ to the power of 0.5.
00:10:14.110 --> 00:10:15.370
And weβll see why in a moment.
00:10:15.630 --> 00:10:19.950
To move forward with this question, weβre gonna be using the second part of the fundamental theorem of calculus.
00:10:20.190 --> 00:10:25.810
This gives us a way to evaluate definite integrals using the antiderivative of the function which forms the integrand.
00:10:26.350 --> 00:10:32.770
At this point, we note that the theorem states the function, lowercase π, must be continuous on the closed interval between π and π.
00:10:33.230 --> 00:10:36.750
π and π are the limits of integration, which in our case are four and nine.
00:10:37.050 --> 00:10:43.340
Now, the function that weβre now working with, lowercase π, is the square root of π₯, which weβve just expressed as π₯ to the power of a half.
00:10:43.550 --> 00:10:49.670
This function is not continuous over the entire set of real numbers, but rather is only continuous when π₯ is greater than or equal to zero.
00:10:49.920 --> 00:10:54.220
Luckily, both of the limits of our definite integral, four and nine, are greater than or equal to zero.
00:10:54.390 --> 00:10:59.580
And so we can therefore say that the square root of π₯ is continuous on the closed interval between four and nine.
00:10:59.850 --> 00:11:01.870
This means that itβs perfectly fine to use our theorem.
00:11:02.150 --> 00:11:07.520
To proceed with our evaluation, we use the power rule of integration, raising the power of π₯ by one and dividing by the new power.
00:11:07.740 --> 00:11:12.840
The antiderivative of π₯ to the power of half is therefore two over three times π₯ to the power of three over two.
00:11:13.160 --> 00:11:17.070
Again, weβre gonna shift this constant factor outside of our brackets to make our calculations easier.
00:11:17.420 --> 00:11:19.670
Next ,we substitute in our limits of integration.
00:11:19.790 --> 00:11:25.330
Now, at this stage, it might be more useful to see our power of three over two expressed as the cube of a square root.
00:11:25.630 --> 00:11:28.250
Conveniently, nine and four are both square numbers.
00:11:28.250 --> 00:11:31.560
And we can simplify our parentheses to be three cubed minus two cubed.
00:11:31.760 --> 00:11:33.830
We now move forward with a few more simplifications.
00:11:33.860 --> 00:11:36.930
And eventually, we reach an answer of negative 76 over three.
00:11:37.130 --> 00:11:38.980
This is the final answer to our question.
00:11:39.200 --> 00:11:44.160
We evaluated the given definite integral using the second part of the fundamental theorem of calculus to help us.
00:11:44.540 --> 00:11:51.020
Along the way, we were sure to confirm that our function lowercase π was continuous over the closed interval between the limits of integration.
00:11:51.250 --> 00:11:53.490
One final point that we didnβt really go into earlier.
00:11:53.650 --> 00:12:01.820
Weβre able to say that the square root of π₯ is not continuous when π₯ is less than zero, because, in fact, the square root of π₯ is undefined over the real numbers when π₯ is less than zero.
00:12:02.210 --> 00:12:05.640
And of course, a function cannot be continuous at points where it is not defined.
00:12:05.900 --> 00:12:06.710
Letβs now move on.
00:12:06.970 --> 00:12:12.530
Another thing worth considering are the cases of piecewise functions or cases involving the absolute value of a function.
00:12:12.710 --> 00:12:19.250
The reason we might need to think about these functions carefully is that they can be thought of as having different behaviors over different regions of their domain.
00:12:19.520 --> 00:12:21.760
Let us see how to deal with this in the following example.
00:12:22.040 --> 00:12:28.360
Evaluate the definite integral between negative four and five of the absolute value of π₯ minus two with respect to π₯.
00:12:28.580 --> 00:12:33.070
For this question, weβve been asked to evaluate the definite integral of a function, which weβll call lowercase π.
00:12:33.320 --> 00:12:36.890
This function is the absolute value or the modulus of π₯ minus two.
00:12:37.140 --> 00:12:41.450
Now, for any real number, we can express an absolute value function as a piecewise function.
00:12:41.680 --> 00:12:49.730
We can do this by recalling that if π₯ minus two evaluates to a negative number or absolute value, weβll multiply this by negative one to turn it into a positive number.
00:12:49.990 --> 00:12:54.140
Okay, so when π₯ minus two is greater than or equal to zero, our function is simply π₯ minus two.
00:12:54.390 --> 00:12:58.060
But when π₯ minus two is less than zero, our function is multiplied by negative one.
00:12:58.090 --> 00:13:00.460
So it is negative π₯ minus two.
00:13:01.050 --> 00:13:05.400
Of course, itβs probably more useful to us to isolate π₯ on one side of these inequalities.
00:13:05.630 --> 00:13:07.410
We do so by adding two to both sides.
00:13:07.720 --> 00:13:11.690
Now, it might also be useful for us to simplify this as negative π₯ plus two.
00:13:12.340 --> 00:13:12.660
Okay.
00:13:12.850 --> 00:13:17.020
Now that we have reexpressed our function piecewise, we can think about how it might look graphically.
00:13:17.290 --> 00:13:18.340
Here we see the graph.
00:13:18.600 --> 00:13:26.450
Although the scale is not exact, both the graph and the piecewise definition should show us the difference in behavior of our function either side of π₯ equals two.
00:13:26.970 --> 00:13:30.070
We see a sharp corner at the point two, zero on our graph.
00:13:30.240 --> 00:13:33.890
In fact, we would say that our function is not differentiable when π₯ equals two.
00:13:34.120 --> 00:13:36.300
But it is continuous when π₯ equals two.
00:13:36.460 --> 00:13:42.210
This is important, because in order to evaluate our definite integral, weβll be using the second part of the fundamental theorem of calculus.
00:13:42.300 --> 00:13:49.570
This allows us to evaluate a definite integral using the antiderivative, uppercase πΉ, of the function which forms our integrand, lowercase π.
00:13:49.820 --> 00:13:56.220
The condition for doing so is that lowercase π must be continuous on the closed interval between π and π, which are the limits of the integration.
00:13:56.380 --> 00:14:03.710
Given that our function lowercase π is continuous when π₯ is equal to two, we are able to conclude that it is continuous over the entire set of real numbers.
00:14:03.850 --> 00:14:06.190
And hence, the continuity condition is satisfied.
00:14:06.640 --> 00:14:08.970
Okay, onto evaluating the definite integral.
00:14:09.430 --> 00:14:13.360
Now weβve already said that our function behaves differently either side of the line π₯ equals two.
00:14:13.600 --> 00:14:17.570
A useful first step for us then is to split up our integral into two parts.
00:14:17.740 --> 00:14:23.370
The first going from the lowest bound, negative four to two, and the second going from two to five.
00:14:23.610 --> 00:14:30.740
Since the upper limit of our first integral is the same as the lower limit of our second integral, the sum of these two will be the same as our original integral.
00:14:31.000 --> 00:14:39.900
Now that weβve split our integral into two parts, weβre able to substitute in the two different subfunctions that we defined using the piecewise definition of the absolute value of π₯ minus two.
00:14:40.170 --> 00:14:43.730
We can understand this by considering our integrals as the area under these lines.
00:14:43.830 --> 00:14:47.400
From negative four to two, our function behaves like minus π₯ plus two.
00:14:47.580 --> 00:14:50.400
And from two to five, our function behaves like π₯ minus two.
00:14:50.700 --> 00:14:54.460
We can interpret the sum of these two areas as being the same as our original integral.
00:14:54.730 --> 00:14:58.060
From this point, we can now move forward using the familiar power rule of integration.
00:14:58.200 --> 00:15:01.490
We raise the power of π₯ for each of our terms and divide by the new power.
00:15:01.750 --> 00:15:03.850
Letβs tidy up to make some room for the next steps.
00:15:04.070 --> 00:15:06.090
Here, weβve input the limits of both integrals.
00:15:06.120 --> 00:15:08.630
And some color has been added to help follow the calculation.
00:15:08.840 --> 00:15:11.230
Weβll need to go through a few more simplification steps.
00:15:12.670 --> 00:15:14.040
Again, weβll clear some space.
00:15:14.280 --> 00:15:15.690
And weβll continue to simplify.
00:15:15.940 --> 00:15:19.170
Eventually, we reach a point where weβll express everything in terms of halves.
00:15:19.550 --> 00:15:23.380
And we reach a final answer of forty-five halves or 45 over two.
00:15:23.700 --> 00:15:25.400
With this, weβve completed our question.
00:15:25.800 --> 00:15:30.480
We did this by first expressing the absolute value of π₯ minus two as a piecewise function.
00:15:30.910 --> 00:15:38.940
Then, by splitting our original integral into two parts and using the second part of the fundamental theorem of calculus to help us evaluate each individual part.
00:15:39.910 --> 00:15:40.260
Okay.
00:15:40.300 --> 00:15:42.160
To finish off, letβs go through some key points.
00:15:42.350 --> 00:15:57.420
The second part of the fundamental theorem of calculus tells us if lowercase π is a continuous function on the closed interval between π and π and if capital πΉ is any antiderivative of lowercase π, which we can express as capital πΉ prime of π₯ is equal to lowercase π of π₯.
00:15:57.580 --> 00:16:05.190
Then the integral between π and π of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital πΉ of π.
00:16:05.420 --> 00:16:09.230
Remember, we can use any antiderivative of the function lowercase π.
00:16:09.260 --> 00:16:12.930
This means we can choose to take the case, which makes our calculations as simple as possible.
00:16:13.110 --> 00:16:17.990
Which is when π, our constant of integration, is equal to zero, essentially allowing us to ignore this.
00:16:18.380 --> 00:16:26.070
Often when given a definite integral, our next step will be to write the antiderivative in brackets with the limits of integration carried over to the right-hand bracket.
00:16:26.300 --> 00:16:31.000
This is a way to express the antiderivative as a function before inputting the limits of integration.
00:16:31.180 --> 00:16:36.070
But this is usually an intermediate step on your way to capital πΉ of π minus capital πΉ of π.
00:16:36.370 --> 00:16:42.920
In order to use this theorem, remember to check that lowercase π is continuous and indeed defined on the closed interval between π and π.
00:16:43.280 --> 00:16:45.290
If it is not, then the integration may fail.
00:16:45.810 --> 00:16:56.560
Finally, piecewise functions or functions involving absolute values may require you to split the integral into multiple parts, since these types of functions behave differently over different regions of their domains.