Welcome to an example of Numerical Integration.
We want to use Simpson's Rule with n equals six
to approximate the given definite integral.
The formula for Simpson's Rule is given here below.
And, there are several things to recognize here.
First, notice a is the lower limit of integration
and b is the upper limit of integration.
So, for this example a equals one
and b equals six.
We're also told that n equals six, so we'll have
n sub-intervals on the interval from one to six.
So, now let's determine delta x
which will give us the width of each sub-interval.
Delta x is equal to b minus a divided by n,
or six minus one divided by six which is 5/6.
So, delta x equals 5/6.
So, x sub zero would be equal to one,
that lower limit of integration.
X sub one would be one plus delta x
or one plus 5/6, that's 11/6.
Next would be x sub two which would be 11/6 plus 5/6
that's 16/6 or 8/3.
X sub three would be 8/3 plus 5/6
which would be 21/6 or 7/2.
Next, x sub four would be 7/2 plus 5/6
which would be 13/3.
X sub five is equal to 13/3 plus 5/6
which is 31/6.
If we do this correctly x sub six
should be the upper limit of integration.
So, 31/6 plus 5/6 is 36/6 or six.
And, before we set this up though
remember when using Simpson's Rule
we're actually using quadratic functions
to make our estimate.
And, we form quadratic functions
over two sub-intervals, which is the reason why
when using Simpson's Rule n must be even.
So, we'll take a look at the graph of our function
over these sub-intervals before we use our formula.
So, here's the interval from one to six
divided up into our sub intervals,
x sub zero through x sub six.
These green points would be the points
that we would use to form the quadratic functions
to make our estimate.
So, the first quadratic function we would use
the first three green points,
or the first two sub-intervals.
So, the first quadratic might look something like this.
The second quadratic would be over the interval,
from x of two x sub four, may be something like this.
And then, the third quadratic would be
over the sub-interval from x sub four to x sub six,
they might look something like this.
So, because the way this function curves
you can see how quadratic functions
would be a good estimate for this definite integral.
Now, let's set up our formula.
The definite integral is going to be approximately
equal to b minus a divided by three n,
that's going to be six minus one
divided by three times six
times, notice here we have f of x sub zero
then we have the coefficients four, two, four, two, four
and then we'll have f of x sub n,
or in our case f of x sub six.
If we begin with f of x sub zero ref of one
plus four times f of x sub one,
that's four time f of 11/6
plus two times f of x sub two,
or f of 8/3
plus four times f of x sub three
or f of 7/2
plus two times f of x sub four,
or f of 13/3
plus four times f of x sub five,
which is f of 31/6.
And then, we end with just f of x sub six,
so f of six.
Let's begin simplifying this.
This will be 5/18
times, we have cosine one divided by one
plus four times cosine
11/6 divided by 11/6
plus two times cosine 8/3
divided by 8/3
plus four times cosine 7/2
divided by 7/2 plus, where are we here?
Two times cosine 13/3
divided by 13/3
and we're almost there.
Plus four times, we'd have cosine
31/6 divided by 31/6
and then finally, plus just cosine six
divided by six.
And then I'll show how they can use the calculator
to approximate these function values.
First thing we should do is make sure we're in radian mode
so we'll push the mode key,
notice that radian is highlighted.
If it wasn't we would highlight radian and then press enter.
Then, from the home screen what we're going to do
is type in our function cosine x divided by x
into y sub one.
So, we'll press y equals, crit out any old functions.
Then, type in cosine x
divided by x.
Go back to the home screen press second mode.
And then, for cosine
one divided by one
that's just going to be y sub one of one,
so we'll press vars right arrow enter, enter.
Then, instead of f of one we'll enter y
sub one of one.
So in parenthesis we enter one,
it'll give us the function value.
So, with four decimal places this would be
approximately 0.5403.
For this next product this is four times f of 11/6
we'll type in 4 times y sub one of 11/6.
So, four times vars right arrow enter, enter
then in parenthesis 11/6.
This method would avoid any entering errors
if we try to enter this expression in this form here.
So, it's approximately negative 0.5663.
Plus, here this would be two times f of 8/3
So, two vars right arrow enter, enter.
And then in parenthesis just 8/3.
So, this would be to four decimal places
negative 0.6670.
And, to save some time I've already determined
the rest of these values.
We have plus negative 1.0702
plus negative 0.1708
plus 0.3397
and finally, plus 0.1600.
And then, 5/8 times this sum would give us approximately
negative 0.3984.
This would be our approximation of the definite integral
using Simpson's Rule when n equals six.
I hope you found this explanation helpful.