WEBVTT
00:00:01.350 --> 00:00:07.490
In this video, weβll learn how to apply integration to find the area bounded by the curves of two or more functions.
00:00:07.900 --> 00:00:14.670
By this stage, you should feel confident in applying processes for integration to evaluate definite and indefinite integrals.
00:00:14.970 --> 00:00:21.530
Weβll now look at how integration can help us to find the area of the regions that lie between the graphs of two or more functions.
00:00:22.330 --> 00:00:32.330
Consider the region that lies between the curve with equation π¦ equals π of π₯ and the π₯-axis which is bounded by the vertical lines π₯ equals π and π₯ equals π.
00:00:32.590 --> 00:00:34.090
Iβve shaded this region pink.
00:00:34.560 --> 00:00:44.820
If π is a continuous function, we know that we can evaluate the area of this region by integrating the function π of π₯ with respect to π₯ between the limits of π and π.
00:00:45.400 --> 00:00:47.730
Now, letβs add another curve to our diagram.
00:00:47.960 --> 00:00:57.130
This time, the curve has the equation π¦ equals π of π₯, where π is continuous and π of π₯ is less than or equal to π of π₯ in the closed interval π to π.
00:00:57.480 --> 00:01:09.790
Once again, we can find the area of the region between the curve π¦ equals π of π₯ and the π₯-axis and these two vertical lines, now Iβve shaded this region in yellow, by evaluating the integral of π of π₯ between the limits of π and π.
00:01:10.380 --> 00:01:20.790
We can now see that if we subtract the area between the curve π of π₯ and the π₯-axis from the area between the curve of π of π₯ and the π₯-axis, that weβll be left with this region π΄ three.
00:01:21.200 --> 00:01:25.790
This is the region between the two curves π¦ equals π of π₯ and π¦ equals π of π₯.
00:01:25.920 --> 00:01:38.090
We can, therefore, say that the area π΄ three of the region bounded by π¦ equals π of π₯ and π¦ equals π of π₯ is the integral of π of π₯ evaluated between π and π minus the integral of π of π₯ evaluated between π and π.
00:01:38.500 --> 00:01:45.700
But we also know that the sum or difference of the integral of two functions is equal to the integral of the sum or difference of these functions.
00:01:45.800 --> 00:01:55.240
So, we can say that the area is equal to the integral of π of π₯ minus π of π₯ with respect to π₯ evaluated between π₯ equals π and π₯ equals π.
00:01:55.710 --> 00:01:57.680
This brings us to our first definition.
00:01:58.130 --> 00:02:18.270
The area π΄ of the region bounded by the curves π¦ equals π of π₯ and π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π, where π and π are continuous and π of π₯ is greater than or equal to π of π₯ in the closed interval π to π, is the definite integral of π of π₯ minus π of π₯ evaluated between the limits of π and π.
00:02:18.760 --> 00:02:26.040
Notice here that π of π₯ is greater than or equal to π of π₯ for all π₯ in between and including π and π.
00:02:26.420 --> 00:02:31.620
Weβll need to watch carefully for situations where this is not the case and apply some extra logic.
00:02:31.870 --> 00:02:34.450
For now though, weβll look at the application of this formula.
00:02:35.040 --> 00:02:42.640
Find the area of the region bounded by the curves π¦ equals three π₯ squared minus five π₯ and π¦ equals negative five π₯ squared.
00:02:42.950 --> 00:03:03.290
Weβll recall that the area of the region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π for continuous functions π and π, such that π of π₯ is greater than or equal to π of π₯ for all π₯ in the closed interval π to π, is the definite integral between the limits of π and π of π of π₯ minus π of π₯.
00:03:03.380 --> 00:03:14.840
Weβre, therefore, going to need to define the functions π of π₯ and π of π₯ really carefully and, of course, the values for π and π, ensuring that π of π₯ is greater than π of π₯ in the closed interval π to π.
00:03:15.420 --> 00:03:20.100
The lines π₯ equals π and π₯ equals π will mark the beginning and end of the region weβre interested in.
00:03:20.320 --> 00:03:22.320
So, what are the equations of these lines?
00:03:22.670 --> 00:03:26.550
Theyβre the π₯-coordinates at the points where the two graphs intercept.
00:03:26.720 --> 00:03:33.890
So, we can set the equations three π₯ squared minus five π₯ and negative five π₯ squared equal to each other and solve for π₯.
00:03:34.150 --> 00:03:36.800
We begin by adding five π₯ squared to both sides.
00:03:37.010 --> 00:03:41.690
And then, we factor the expression on the left-hand side by taking out that factor of π₯.
00:03:42.250 --> 00:03:46.250
And we obtain π₯ times eight π₯ minus five to be equal to zero.
00:03:46.920 --> 00:03:53.810
We know that for this statement to be true, either π₯ itself must be equal to zero or eight π₯ minus five must be equal to zero.
00:03:54.380 --> 00:03:58.170
To solve this equation on the right, we add five and then divide through by eight.
00:03:58.360 --> 00:04:00.650
And we obtain π₯ to be equal to five-eighths.
00:04:00.750 --> 00:04:06.530
So, we can see that the π₯-coordinates of the points of intersection of our two curves are zero and five-eighths.
00:04:06.660 --> 00:04:10.270
So, π is equal to zero and π is equal to five-eighths.
00:04:10.530 --> 00:04:15.270
Now, weβre going to need to decide which function is π of π₯ and which function is π of π₯.
00:04:15.700 --> 00:04:23.810
What we do next is sketch out the graphs of π¦ equals three π₯ squared minus five π₯ and π¦ equals negative five π₯ squared.
00:04:24.160 --> 00:04:28.300
Weβre looking to establish which of the curves is essentially on top.
00:04:28.560 --> 00:04:33.340
We know that the graph of π¦ equals three π₯ squared minus five π₯ is a U-shaped parabola.
00:04:33.550 --> 00:04:39.610
We can even factor the expression three π₯ squared minus five π₯, set it equal to zero, and solve for π₯.
00:04:39.850 --> 00:04:44.270
And we see that it passes through the π₯-axis at zero and five-thirds.
00:04:44.550 --> 00:04:46.870
So, it will look a little something like this.
00:04:47.100 --> 00:04:53.730
The graph of π¦ equals negative five π₯ squared is an inverted parabola which passes through the origin like this.
00:04:54.020 --> 00:04:56.180
And so, we obtain the region shaded.
00:04:56.520 --> 00:05:05.960
We can now see that in the closed interval of zero to five-eighths, the function thatβs on top, if you will, is the function defined by π¦ equals negative five π₯ squared.
00:05:06.230 --> 00:05:09.320
So, we can say that π of π₯ is equal to negative five π₯ squared.
00:05:09.550 --> 00:05:12.810
Meaning, π of π₯ is three π₯ squared minus five π₯.
00:05:13.150 --> 00:05:24.830
The area that weβre interested in must, therefore, be given by the definite integral evaluated between zero and five-eighths of negative five π₯ squared minus three π₯ squared minus five π₯ with respect to π₯.
00:05:25.060 --> 00:05:30.290
Distributing the parentheses, and our integrand becomes negative eight π₯ squared plus five π₯.
00:05:30.660 --> 00:05:36.020
But wait a minute, we know that when we evaluate areas below the π₯-axis, we end up with a funny result.
00:05:36.020 --> 00:05:37.080
We get a negative value.
00:05:37.490 --> 00:05:41.440
You might wish to pause the video for a moment and consider what that means in this example.
00:05:42.460 --> 00:05:43.810
Did you work it out?
00:05:44.390 --> 00:05:51.640
We can see that our entire region sits below the π₯-axis and weβre just working out the difference between the areas.
00:05:51.880 --> 00:05:58.330
So, the negative results that we would obtain from integrating each function individually will simply cancel each other out.
00:05:58.710 --> 00:06:01.300
So, all thatβs left is to evaluate this integral.
00:06:01.650 --> 00:06:06.660
The integral of negative eight π₯ squared is negative eight π₯ cubed over three.
00:06:06.920 --> 00:06:10.180
And the integral of five π₯ is five π₯ squared over two.
00:06:10.520 --> 00:06:20.700
We need to evaluate this between zero and five-eighths, which is negative eight-thirds of five-eighths cubed plus five over two times five-eighths squared minus zero.
00:06:21.200 --> 00:06:25.700
Thatβs 125 over 384 square units.
00:06:26.870 --> 00:06:38.550
This question was fairly straightforward as the curve of π¦ equals negative five π₯ squared was greater than or equal to the curve of π¦ equals three π₯ squared minus five π₯ in the interval weβre interested in.
00:06:39.070 --> 00:06:42.530
Letβs now look at what we might do if this wasnβt the case.
00:06:43.090 --> 00:06:47.570
The curves shown are π¦ equals one over π₯ and π¦ equals one over π₯ squared.
00:06:47.930 --> 00:06:50.040
What is the area of the shaded region?
00:06:50.220 --> 00:06:51.360
Give an exact answer.
00:06:51.850 --> 00:07:12.720
Remember, the area of a region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π for continuous functions π of π, when π of π₯ is greater than or equal to π of π₯ for π₯ in the closed interval π to π, is given by the definite integral evaluated between π and π of π of π₯ minus π of π₯.
00:07:12.940 --> 00:07:15.090
Now, we do have a little bit of a problem here.
00:07:15.420 --> 00:07:21.720
We can see quite clearly that the region is bounded by the vertical lines π₯ equals 0.5 and π₯ equals two.
00:07:22.060 --> 00:07:29.890
But in the closed interval π₯ from 0.5 to two, we can see that one of our functions is not always greater than or equal to the other.
00:07:30.120 --> 00:07:31.920
So, we canβt actually apply this definition.
00:07:32.260 --> 00:07:38.050
We can see, however, that if we split our region up a little more, we do achieve that requirement.
00:07:38.460 --> 00:07:41.660
Iβve added a third line at the point where the two curves intersect.
00:07:41.840 --> 00:07:43.590
This has the equation π₯ equals one.
00:07:44.000 --> 00:07:50.610
In the closed interval 0.5 to one, the values on the red line are always greater than or equal to those on the green line.
00:07:50.960 --> 00:07:55.190
And in the closed interval π₯ between one and two, the reverse is true.
00:07:55.480 --> 00:07:59.340
So, all we do is split our region up and then add the values at the end.
00:07:59.650 --> 00:08:03.090
Letβs find the area of our first region, π
one.
00:08:03.650 --> 00:08:06.930
To do so, weβre going to need to double check which line is which.
00:08:07.020 --> 00:08:10.840
We can probably deduce that the red line is more likely to be one over π₯ squared.
00:08:11.130 --> 00:08:14.670
But letβs choose a coordinate pair and substitute these values in just to be safe.
00:08:15.180 --> 00:08:19.800
We can see that the curve passes through the point with coordinates 0.5, 4.
00:08:20.200 --> 00:08:24.870
So, letβs substitute π₯ equals 0.5 into the equation π¦ equals one over π₯ squared.
00:08:25.190 --> 00:08:32.740
When we do, we get π¦ equals one over 0.5 squared, which is one over 0.25, which is four as required.
00:08:32.990 --> 00:08:39.410
So, the red line has the equation π¦ equals one over π₯ squared and the green line has equation π¦ equals one over π₯.
00:08:39.530 --> 00:08:46.580
And when evaluating the area of π
one, π of π₯ is therefore one over π₯ squared and π of π₯ is equal to one over π₯.
00:08:47.390 --> 00:08:55.220
The area is, therefore, given by the definite integral between the limits of 0.5 and one of one over π₯ squared minus one over π₯.
00:08:55.440 --> 00:08:57.810
So, all thatβs left here is to evaluate this integral.
00:08:58.010 --> 00:09:04.600
This is much easier to do if we rewrite one over π₯ squared as π₯ to the power of negative two and then recall some standard results.
00:09:04.850 --> 00:09:09.280
To integrate π₯ to the power of negative two, we add one to the power and then divide by this new number.
00:09:09.500 --> 00:09:14.770
That gives us π₯ to the power of negative one over negative one, which is negative one over π₯.
00:09:14.960 --> 00:09:19.130
The integral of one over π₯, however, is the natural log of the absolute value of π₯.
00:09:19.290 --> 00:09:24.070
So, our integral is negative one over π₯ minus the natural log of the absolute value of π₯.
00:09:24.230 --> 00:09:27.870
Weβre going to now evaluate this between π₯ equals 0.5 and π₯ equals one.
00:09:28.270 --> 00:09:35.930
Thatβs negative one over one minus the natural log of one minus negative one over 0.5 minus the natural log of 0.5.
00:09:36.180 --> 00:09:41.200
And notice, Iβve lost the symbol for the absolute value because one and 0.5 are already positive.
00:09:41.560 --> 00:09:43.490
The natural log of zero is one.
00:09:43.720 --> 00:09:45.580
Negative one over one is negative one.
00:09:45.580 --> 00:09:48.490
And negative one over 0.5 is two.
00:09:48.920 --> 00:09:54.600
Iβve also rewritten the natural log of 0.5 as the natural log of a half and distributed the parentheses.
00:09:54.790 --> 00:09:58.310
And this simplifies to one plus the natural log of one-half.
00:09:58.650 --> 00:10:03.230
A really important skill, though, is to be able to spot when we can further simplify a logarithmic term.
00:10:03.530 --> 00:10:07.560
If we rewrite the natural log of a half as the natural log of two to the power of negative one.
00:10:07.670 --> 00:10:12.490
And then, use the fact that the natural log of π to the πth power is equal to π times the natural log of π.
00:10:12.660 --> 00:10:17.610
We see that the exact area of the first region π
one is one minus the natural log of two.
00:10:18.520 --> 00:10:21.680
Letβs clear some space and repeat this process for region two.
00:10:22.150 --> 00:10:29.820
This time, the green line is above the red line, so weβre going to let π of π₯ be equal to one over π₯ and π of π₯ be equal to one over π₯ squared.
00:10:30.190 --> 00:10:40.810
Our area is the definite integral between one and two of one over π₯ minus one over π₯ squared which, when we integrate, gives us the natural log of the absolute value of π₯ plus one over π₯.
00:10:41.080 --> 00:10:51.100
Evaluating between the limits of one and two, and we get the natural log of two plus a half minus the natural log of one plus one, which is equal to the natural log of two minus a half.
00:10:51.350 --> 00:10:55.300
We want to find the area of the whole region, so we add these two values.
00:10:55.300 --> 00:11:01.510
Itβs one minus the natural log of two plus the natural log of two minus one-half, which simplifies to one-half.
00:11:02.070 --> 00:11:05.810
The area of the shaded region is a half square units.
00:11:07.200 --> 00:11:24.110
In this example, we saw that the area formula can be applied to find the area between two curves where one curve is above the other for part of the integration interval and the opposite in the second part of the interval, as long as we remember to split the region up at the point where the curves intersect.
00:11:24.760 --> 00:11:29.630
Weβll now see how we can develop this formula further to help us find the region bounded by three curves.
00:11:30.510 --> 00:11:38.040
Find the area of the region bounded by the curves π¦ equals four minus π₯ squared, π¦ equals negative π₯, and π¦ equals the square root of π₯.
00:11:38.190 --> 00:11:40.280
Give your answer correct to one decimal place.
00:11:40.560 --> 00:11:59.260
Remember, for continuous functions π and π, the area of the region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π, as long as π of π₯ is greater than or equal to π of π₯ in the closed interval π to π, is given by the integral evaluated between π and π of π of π₯ minus π of π₯.
00:11:59.430 --> 00:12:02.930
We are going to need to be a little bit careful here, as we have three curves.
00:12:03.170 --> 00:12:06.440
So, letβs begin by sketching this out and see what weβre dealing with.
00:12:06.610 --> 00:12:10.590
The area enclosed between the three curves looks a little something like this.
00:12:11.060 --> 00:12:15.360
Now, if weβre really clever, we can actually use the definition we looked up before.
00:12:15.630 --> 00:12:21.070
We can split this region into the region above the π₯-axis and the region below the π₯-axis.
00:12:21.370 --> 00:12:23.700
We then can split this up a little bit further.
00:12:23.840 --> 00:12:32.990
We see that we have π
one, thatβs the region between the π₯-axis and the curve π¦ equals root π₯ between π₯ equals zero and π₯ equals π.
00:12:33.170 --> 00:12:39.660
Where π is the π₯-coordinate at the point of intersection of the curve π¦ equals root π₯ and π¦ equals four minus π₯ squared.
00:12:39.990 --> 00:12:41.330
We then have π
two.
00:12:41.540 --> 00:12:47.200
Thatβs the region between π¦ equals four minus π₯ squared, π₯ equals π, and π₯ equals two.
00:12:47.510 --> 00:12:54.700
And the reason weβve chosen π₯ equals two as our upper limit is thatβs the π₯-value at the point at which the curve crosses the π₯-axis.
00:12:54.970 --> 00:12:59.840
We can even split our three up into two further regions to make life easier.
00:13:00.010 --> 00:13:03.050
But letβs deal first with the area of π
one and π
two.
00:13:03.370 --> 00:13:05.040
We need to work out the value of π.
00:13:05.180 --> 00:13:10.680
We said itβs the π₯-coordinate at the point of intersection of the two curves four minus π₯ squared and root π₯.
00:13:10.680 --> 00:13:13.380
So, we set these equal to each other and solve for π₯.
00:13:13.560 --> 00:13:18.380
That gives us an π₯-value of 1.648, correct to three decimal places.
00:13:18.520 --> 00:13:21.750
So, π is equal to 1.648.
00:13:22.050 --> 00:13:27.100
We can either do this by hand or use our graphical calculators to evaluate each of these integrals.
00:13:27.400 --> 00:13:32.230
The area of π
one becomes 1.4104 and so on.
00:13:32.750 --> 00:13:37.100
And the area of π
two is 0.23326 and so on.
00:13:37.380 --> 00:13:39.390
Letβs now consider the area of π
three.
00:13:40.080 --> 00:13:45.730
Itβs the integral between zero and two of negative π₯ evaluated with respect to π₯.
00:13:45.910 --> 00:13:52.160
We need to be a little bit careful here, since this is below the π₯-axis and therefore will yield a negative result on integration.
00:13:52.580 --> 00:13:54.580
In fact, it gives us negative two.
00:13:54.870 --> 00:13:57.470
So, we can say that the area is the absolute value of this.
00:13:57.470 --> 00:13:57.970
Itβs two.
00:13:58.200 --> 00:14:02.330
And notice, we could have actually used the formula for area of a triangle to work this area out.
00:14:02.600 --> 00:14:10.470
Now, the area of π
four is the definite integral between two and π of four minus π₯ squared minus negative π₯.
00:14:10.580 --> 00:14:17.380
And here, π is the π₯-coordinate of the point of intersection of the lines π¦ equals negative π₯ and π¦ equals four minus π₯ squared.
00:14:17.610 --> 00:14:22.140
We can once again set four minus π₯ squared equal to negative π₯ and solve for π₯.
00:14:22.250 --> 00:14:27.710
And we find, correct to three decimal places, that they intersect at the point where π₯ equals 2.562.
00:14:28.220 --> 00:14:35.050
We type this into our calculator and we find that the area of this region is 0.59106.
00:14:35.260 --> 00:14:44.680
We find the total of these four values, which gives us 4.2347, which is 4.2 square units, correct to one decimal place.
00:14:45.920 --> 00:15:01.690
In this video, weβve seen that we can use the formula area is equal to the definite integral between π and π of π of π₯ minus π of π₯ with respect to π₯ for continuous functions, π and π as long as π of π₯ is greater than or equal to π for all π₯ in the closed interval π to π.
00:15:02.010 --> 00:15:15.860
We also saw that for more complicated regions such as those bounded by three or more curves, those which involve regions above and below the π₯-axis, or those where π of π₯ and π of π₯ switch, we might need to split this region up a little further.