College Calculus: Level II Simpson's Rule

We are going to work out a couple more examples on Simpson's Rule.0000

We are going to start with the example of the integral from 1 to 2 of x log(x) using Simpson's Rule with n = 4.0004

We are integrating from 1 to 2, and we are uisng n=4.0014

Remember, for Simpson's Rule, n always has to be even.0022

So we break the interval from 1 to 2 into 4 equal pieces.0026

So, that means that our breakpoints are 1, 5/4, 3/2, 7/4, and 2.0030

Then, that means that δx which is the width of the intervals, is 1/4.0040

We are going to invoke Simpson's Rule formula.0047

It starts out with x/3, so that is 1/12.0050

Then you plug these points into the function you are integrating, in this case x (log(x)).0059

Then you attach these coefficients, remember that funny pattern, 1, 4, 2, 4, 2, 4, as many times as you need until the last is 4 and then 1.0066

The first point is 1.0083

So, 1 × ln(1) + 4 × 5/4 ln(5/4) + 2 × 3/2 ln(3/2) + 4 × 7/4 ln(7/4) + 1 × 2 × ln(2).0086

This can be a little simplified, because ln(1) is 0.0124

The other values are not common values we know easily.0139

This would be something you would plug into a calculator.0144

I have already worked it out on my own calculator.0148

Here is what you get, 0.67233487080150

To reiterate there, we are estimating an integral as the area under a curve using the Simpson's Rule formula.0160

So, that is the answer.0176

In our last example today we are going to compare the accuracy of using the midpoint rule and Simpson's Rule,0000

Both with n = 4, on the integral from 1 to 2 of x ln(x) dx.0006

We just worked out what the answer was using Simpson's Rule.0015

The Simpson's Rule gave us the answer 0.6723348708.0023

Th midpoint rule is something we did earlier in another lecture.0037

I am going to remember what the answer was there.0046

That was done as an example in the previous lecture on the trapezoidal rule, the midpoint rule, and the left and right endpoint rules.0051

The answer of this integral using n=4 was 0.6344928081.0056

Those are the two answers we get using the different estimation rules.0071

You can also do this integral directly, the integral from 1 to 2 of x ln(x) dx.0079

I am not going to show the details of this but I will tell you how you can figure it out.0091

You can use integration by parts,0095

And the way you would break it down using integration by parts,0100

If you remember the LIATE rule, that was the order in which you look at the functions to use integration by parts.0106

LIATE stands for ln, inverse functions, algebraic functions, trigonometric functions, exponential functions.0110

Very first on that list is natural log.0120

Remember LIATE means you try to make whatever function comes first be the u.0124

For integration by parts on this, we see a ln(x), so a good choice is to make u be ln(x), and dv be x dx.0130

Then you can work out this integral using integration by parts.0144

It takes a couple steps so I will not show the details.0147

What you get as the final answer is -x²/4 + x²/2 × ln(x).0151

All of that evaluated from x = 1 to x = 2.0164

Then if you plug in those values, you get -3/4 + ln(4).0170

If you plug that into a calculator, you get 0.6362943611.0182

That is as close as we can come to the true value of the integral.0196

Let us see how accurate our two approximation techniques were.0202

Let us start with our midpoint approximation.0205

If we look at the midpoint - the true value of the integral, remember this was the true value over here.0210

If we do the midpoint approximation - the true value, we get - 0.001801553.0221

That looks very accurate, because our error here, if we subtract the true value from the midpoint rule approximation.0235

Our error is just 0.001.0245

Let us look at Simpson's Rule though.0252

If we do the Simpson's Rule approximation minus the true value, we get 0.000015469.0257

That error is really tiny, very small error using Simpson's Rule.0272

In fact, when we calculated the integral using Simpson's Rule, we did not have to do anymore work than the midpoint rule.0283

We still were just plugging in those values at 1, 5/4, 3/2, 7/4 and 2.0292

So, it was about the same amount of work, but Simpson's Rule is much more accurate.0300

Using about the same amount of work.0310

So, the moral of the story there is that we have various approximation techniques.0325

The trapezoid rule, the midpoint rule, the left and right endpoint rules, and the Simpson's Rule.0337

However, by far the most powerful of all of these is the Simpson's Rule.0340

It gives us extremely accurate results.0346

So that is the end of this lecture.0350

Today we are going to talk about Simpson's Rule.0000

Simpson's Rule is a way of estimating the value of an integral when you cannot solve it by traditional integration techniques.0005

We already learned a few of those, the trapezoidal rule, the midpoint rule, and the left and right endpoint rules.0011

Simpson's Rule is based on the same kinds of ideas but it is a little more sophisticated.0017

We will see later on that the answers it gives are a lot more accurate.0023

The idea here is that we are trying to estimate the area under a curve, y = f(x).0029

Just as we did with the previous rules, what we are going to do is divide the input into partitions.0039

An important difference between Simpson's Rule and the previous ones is that we have to use an even number of partitions.0050

Just as before, we have an x₀, x₁, x₂, up to x_n, but remember n has to be even now.0058

The principle of Simpson's Rule is that you look at those partitions two at a time.0072

Let me draw an expanded version of that.0073

So you look at two partitions here, and you have got the function above there.0078

What you do is look at the three values on the boundaries of those partitions and you draw a parabola through it, through those three values.0084

Then you find the area under that parabola and you add up those areas.0109

So, you find this area and you add all those up and that gives you your approximation of the value of the integral.0116

Finding the formulas to calculate that area would be a little bit tedious if you had to do it every time, so I will just give you the answer.0124

The answer turns out to be, for the area under one parabola, δx/3, remember δx is the width of one of these three partitions,0133

Times the value of f(x) at these three points multiplied by certain coefficients.0145

The coefficients are 1, 4, 1, which is kind of strange but that is just the way the math works out.0153

And then you go through, and you do this over each set of partitions, over the entire area.0160

What you end up doing is you end up adding 1, 4, 1, and then 1, 4, 1, for the next one but that overlaps here0168

Because remember the last endpoint of one set of partitions is the first endpoint of the next set.0179

And then 1, 4, 1 and so on for however many double partitions you have.0186

Then you add up those coefficients and you got 1, 4, 2, 4, 2, 4, 1.0193

You get this funny pattern 1, 4, 2, 4, and that keeps alternating until the final coefficient is 1.0202

So that is where we get the final formula for Simpson's Rule.0210

This δx/3 and then the coefficients 1, 4, 2, 4, 2, 4, 1, for however many partitions you have.0215

Let us try that out with some examples and you will see how it works.0225

The first example we are going to do is the integral of sin(x) as x goes from 1 to 2.0229

That is an integral that you could do without using Simpson's Rule.0236

Certainly we know what the integral of sin(x) is, but it is an easy function to practice using Simpson's Rule on.0239

What we are going to do is look at the interval from 1 to 2, and it says to use n = 4, so we are going to divide that into 4 sub-intervals.0248

That means that δx is (b - a)/4, so δx is 1/40265

Our points that we are going to check are x₀ = 1, x₁ = 5/4 because we are going over by 1/4 each time, x₂ = 3/2, x₃ = 7/4, and x₄ = 2.0276

The Simpson's Rule formula, so δx/3, so that is 1/12, times f(left endpoint),0302

So that is sin(1) + 4sin(5/4) + 2sin(3/2) + 4sin(7/4) and remember that the last coefficient is 1, so we get + sin(2).0310

That is our approximation for the area.0340

Now it is just a matter of plugging those values into a calculator.0347

Take your calculator and plug these values in and I have already actually worked it out on my calculator, and it turned out to be 0.9564700541.0352

That is our guess for this area using Simpson's Rule0383

So let us try that out on a slightly more complicated example.0391

It is the same integral, but now we are going to use n = 8, so we have a much finer partition there.0393

Now δx is 1/8 and our partition points are x₀ = 1, x₁ = 9/8, x₂ = 5/4, all the way up to x₈ = 2.0406

So, our Simpson's rule formula says we do δx/3,0434

So, that is 1/8/3, so 1/24 × (sin(1) + 4sin(9/8) + 2sin(5/4) + 4sin(11/8) + 2sin(3/2) + 4sin(13/8) + 2sin(7/4) + 4sin(15/8) + 1sin(2)).0436

Again, that is a complicated expression but it is easy enough if you just plug it into your calculator.0511

I already have worked it out and what I got was 0.9564504421.0519

That is what we get using Simpson's Rule, using more partitions, n=8 instead of n=4.0536

Remember n has to be even every time in order for this formula to work.0543

Our next example is to look back at the two previous examples and compare the accuracy of using n = 4 and n = 8.0548

Now, in order to do this, to say how accurate they are, we really need to know the true value of this integral.0558

This is an easy integral to solve using calculus 1 techniques.0564

The true value is the integral from 1 to 2 of sin(x) dx, which is, well the integral of sin is -cos.0569

Evaluated from x = 1 to x = 2, so -cos(2) + cos(1)0587

Again if you plug that into a calculator, you will get 0.9564491424.0600

That is the true value of the integral, but we are interested in figuring out how accurate our approximations using Simpson's Rules were.0617

Let us look at the n = 4 values, remember that was 0.9564700541, and if we subtract the true value from that, what you get is 0.000020912.0624

That is extremely accurate.0661

We get 4 decimal places of accuracy using n = 4.0665

But let us try n = 8.0671

The value we had was 0.9564504421, and if we subtract the true value from that, what you end up with is 0.000001300,0674

Which is much smaller than what we got with n = 4.0706

The point of that is that using n = 8 gave us a much more accurate result than using n = 4 did.0715

The conclusion there is that n = 8 is much more accurate.0723

OK, that finishes this section on Simpson's Rule.0739