.
So what you are finding out here is that we are
getting similar type of formulas for each of those . . . for each of those
integrals, and what we now need to do is to be able to bundle them
together into some kind of a compact form.|What you can also see is that this is
nothing but what?|This is nothing but 2 times h, this is
also 2 times h, because the difference between the upper and the lower limit is
always 2 times h here,
because each segment is width h, so since this is
the . . . this is the last, let's suppose this is the last
number here, and this is the second-last . . . second-last node here,
so that's why the difference between them is 2 h.|So if we do that, what we're going
to do is we're going to get 2 h, divided by 6, because 2 h, divided by 6 will be a
common factor amongst all of these terms which we have written there, and then what we have to recognize
is that which of the terms can be bundled together.|Now, one of the things which you're going to see is that you have
f of x0 here, and that's just by itself, okay?
Now, what you are also seeing is that you have certain terms which are multiplied by 4, look at this, 4,
4, and 4 here, so I'm going to bundle those together.|So I'll get 4 times
f of x1, and the next term which I see is f of x3,
and then so on and so forth, all the way up to f of
xn-minus-3, plus f of xn-minus-1.
So that's how I am bundling the terms which are multiplied by
4 here.|And then I'm going to see that what I am seeing here is that
I have this f of x2 here once, and then I have f of x2 here once.
So what's going to happen is that any of these subscripts which are even, those
are going to be one here and one here, so there'll be two times.|Same thing, look at this f of x4,
there'll be an f of x4 in the term which I have not shown here, that'll be two times.|Same thing, look at this,
this is one time here, and this is one time here, it becomes two times, so what that means
is that I'll have 2 times f of x2, plus 2 times f of
x3, x4, plus all the way up to f of
xn-minus-2, so those will be 2
times, so let me just use the proper
parentheses here . . . or brackets here, so I get this here,
and then the only which I am left with now is this
last term, which is f of xn.|So I think what you should do is you should go ahead
and try it, not just by just watching the video, but do it on paper,
and convince yourself that the way I have bundled the terms, that those are correct, and that's
the way it should be done.|One of the things which you're already finding out is that these terms which are
getting multiplied by 4, the subscripts are odd, 1, 3,
n minus 3, n minus 1, they're all odd, because if you look at n minus 1, because since n is even,
n minus 1 will be odd.|Again, same thing here, the subscripts here are even, you have
2, 4, n minus 2, n, and since n is even, n minus 2 will be even, too.
So that gives us some . . . let's forget about this f of xn, that's not
to be included in this discussion.|We have x-sub-n-minus-2, which is even, because n is even, so
n minus 2 will be even, too.|So what that gives us, it gives us some form of that, hey, what we
are adding together are simply the subscripts of odd number of terms . . .
with odd number subscript, and then we're going to add here the even numbered subscripts.
And also what we're going to do is we're going to take this 2, and we're going to do h divided by 3.
So you can very well see that you have this h divided by 3, and that's why this particular
method is called Simpson's 1/3 rule, because you're getting h divided by 3.|If you take h
inside, you get 1 divided by 3 outside, and that's why it's called Simpson's 1/3 rule.
So let's go ahead and bundle these terms together, and be able to see that whether we can write this is a compact form.
So we get h divided by
3, times the value of the function at x0, which is a term by
itself, then all the terms which are multiplied by 4 can be summed from
1 to n minus 1, f of xi,
where i is simply odd, because it's 1,
3, 5, like that, then plus 2 times the summation, i is equal to 1
to n minus x, but we're only including i being even,
so it'll be those terms, then plus
f of xn.|So that is the compact form of Simpson's
1/3 rule.|Some people will substitute h by
b minus a, divided by n.|So in that case, you will get
b minus a, divided by 3 n, if you substitute h to be b minus a,
divided by n, times the value of the function at x0, plus 4
times the value of the function at these odd subscripts
of x,
plus 2 times summation, i is equal to 1 to n minus 2,
f of xi, and i being even,
plus f of xn.|So that is the
compact form of multiple segment Simpson's 1/3 rule.|In the next segment, we will
take an example, and see how we can apply this formula now algorithmically,
as opposed to just dividing our intervals into as many segments as we want.
So this is something which can be programmed in a computer program
in any of the languages, so as to be able to develop approximate values of integrals.
And that's the end of this segment.
.