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In the next few lectures, let us look at various types of integration schemes. And this helps
us to get an idea why Gaussian quadrature is preferred for finite element calculations.
To start with it, let us look at what are called Newton-cotes rules. And under that
we have the first one, one-point rule or rectangle rule. And rectangle rule basically for numerical
integration purpose we are going to evaluate function at one point, so n is equal to 1.
Basically, in rectangle rule if we have a function f (x) to evaluate integral f (x) within the limits x1 to x 2, it is approximated
as the function value at one of the points that is lower bound that is x1 function value
at x1 multiplied by a weight. Here, for rectangle rule weight is going to be the distance from
x 2 minus x1 is going to be the weight, and function evaluated at a single point that
is x1. By doing this basically, what we are doing is whatever with the type of function
whether it is constant, linear, quadratic or any order of function where basically approximating
that function as a constant function Q naught as shown in the schematic there.
Basically rectangle rule is good for evaluating integrals where f (x) is a constant. Since
we are evaluating function at only one point as you can see in the figure, the difference
between the area enclosed by f(x) between points x1, x 2. And the area enclosed by the
curve q not between the points x1, x2. This difference in area is going to be the error
that is associated with this rectangle rule, that is the accuracy that we are losing when
we adopt rectangle rule if the function is not a constant function, but for a constant
function rectangle rule or one-point rule or Newton-cotes rules with n is equal to 1
results in accurate solution.
Now, let us to look at two-point rule or trapezium rule. The function is going to be evaluated
at two points n is equal to 2 as in the figure function f (x) is shown. Basically, in trapezium
rule to integrate f (x) between x1 and x2 what we are going to do is? We are going to
evaluate this function f (x) at two points. The two points are pre-decided and two points
are the limits between, which we want to find this integral this f( x) is going to be evaluated
at lower limit x1 and f x is going to be evaluated at upper limit x2.
Once we have this function values, the function value at x1 is multiplied with a weight which
is half the distance between x1 and x2 or x2 and x1. Similarly, the function value at
x2 is multiplied with a weight which is equal to half the distance between x2 and x1 is
h over 2. Basically in trapezium rule, we are replacing
whatever function that function we are replacing with a linear function q1 as shown in this
schematic. Whatever a error associated with this trapezium rule is the difference in area
between the area enclosed between the area enclosed by the curve between the points x1,
x2. And the area enclosed by q1 between points x1 and x2. The difference between these two
areas is going to be the error associated with trapezium rule, and trapezium rule is
going to be accurate, if function f (x) is a linear function or trapezium rule is going
to be accurate, if the function order of the function is 0 or 1. Even for constant function
trapezium rule is going to result in accurate solution, so this is about trapezium rule.
The formula will be exact for function of degree one or less.
To illustrate the procedure let us try to evaluate this integral sine x between the
limits phi over 4 and phi over 2, so lower limit is phi over 4, upper limit is phi over
2 using trapezium rule. Basically what we require is, we need to find what is the distance
between upper limit and lower limit phi over 2 minus phi over 4 that is going to be h.
Once we have h, we need to evaluate function that is sine x at phi over 4, and sine x at
phi over 2 multiplied with corresponding weights, which is going to be h divided by 2. In this
case it is going to be phi over 4 times half which is going to be phi over 8 all those
details are given here.
When h is equal to phi over 4 for this particular case, therefore the integral is approximated
in this manner and the approximate solution turns out to be 0.6704. And this function
can also be integrated exactly. So, the exact solution if we compare with exact solution,
the approximate solution that we just obtained using trapezium rule it turns out that exact
solution is 0.7071 that is the accuracy that we get using trapezium rule.
Now, let us go and see what is called Simpson’s rule or three point-rule where we are going
to evaluate function at three points. So, n is equal to 3 integral f (x) between the
limits x1, x 3 intentionally the upper limit is named as x 3. If one wants to evaluate
this function integral of this function between limits x1 and x3 that is approximated using
Simpson’s rule in this manner weight function at x1 multiplied by function value at x1 plus,
weight function at x2 multiplied by function value at x2, weight function at x3 multiplied
by function value at x3. So, x1 let us see where this x2 is this x2 should be between
x1 and x3. Once we know upper limit and lower limit we can easily figure out what is x1
value, what is x2 value, and what is x3 value. We can easily evaluate function at these points
and only thing unknowns are W1, W2, and W3. Let us see how we can evaluate this W1, W2,
W3 .And the procedure that we are going to follow for evaluating these weights is going
to be a general procedure which also helps us to understand in gauss quadrature how locations
of integration points and weights are obtained. The Simpson’s rule is going to be exact
for a function of degree 012. And this gives three equations in three unknown waiting functions
W1, W2, and W3. Basically, if you see three-point rule, it is going to be exact for a function
of order two. If you take trapezium rule, it is going to be exact for a function of
order one. And if you take rectangle rule, it is going to be exact for function of order
zero. So, what we can conclude from here is Newton-cotes rules is that if you adopt n
points, we can integrate a function of order up to n minus 1 accurately using Newton-cotes
rules. Newton-cotes rules can either be rectangle
rule, trapezium rule or Simpson’s rule or we can go for n is equal to 4 or n is equal
to 5, whatever it may be Newton-cotes rules are going to be exact for a function with
n points. Newton-cotes rules with n points are going to be exact for a function of order
n minus 1. Now, let us see the procedure how to find
these weights W1, W2, W3 to get to solve for 3 unknowns, we know that we require 3 equations
to get unique solution for 3 unknowns. To evaluate this W1, W2, and W3 we are going
to come up with 3 equations in terms of W1, W2, W3. And solve these 3 unknowns using these
3 equations that is the basically the procedure to get these weights W1, W2, and W3.
Now, let us look at the procedure. Basically in Simpson’s rule a function f (x) is replaced
or it is approximated as a quadratic function, q2 as shown in the figure. And this function
is evaluated at 3 points x1, x2, and x3. And x2 is midway between x1 and x3 and the distance
between x1 and x2 is h, x2 x3 is h. So, our job is to evaluate this function integral
f(x) between the limits x1 and x3. Since x1 corresponds to minus h and x2 corresponds
x3 corresponds to h. If you take x2 as origin, x1 is at a distance minus h and x2 is x3 is at a distance h. We
can replace these limits x1 x3 with minus h and h. Similarly, since the function is
initially is with respect to the origin that is shown with respect to the x axis. Since,
we are defining a new axis this function also needs to be expressed in terms of new origin
that is x2. So, this f(x) is replaced with F(X) it can be easily verified that this F(X)
and f(x) values at any point along the curve are going to be the same. So, the distance
between x3 and x1 is 2h. The procedure goes like this we need to determine
3 unknowns W1, W2, and W3. We will start with a F (X) is equal to 1. When F(X) is equal
to 1 integral minus h to h .And 1dx becomes 2h and the function value taken at any point
that is x1, x2, x3 is going to be the same which is going to be equal to 1. The previous
approximation W1 times function evaluated at x1 plus W2 times function evaluated at
x2 plus W3 times function evaluated at x3 it becomes 2h is equal to W1 plus W2 plus
W3 we got one equation now.
Let us say F(x) is equal to x, again apply that equation which we have already seen.
So, replace f (x) with x integrate between the limits minus h and h, x d x and since
x is a odd function. And we are integrating odd function over a symmetrical domain minus
h to h. The integral turns out to be 0. And this should be equal to or this is approximated
as W1 times function evaluated at x1here in this case x1 is nothing but minus h. So, W1
times function itself is x f(x) is equal to x function evaluated at x1 is equal to minus
h, function evaluated at x2 is equal to 0. Function evaluated at x3 is equal to h. Using
this information, we get this equation W1 times minus h plus W2 to times 0 plus W3 times
h. If we simplify it, we get equation in terms of W1 and W3 and since the W2 or h function
evaluated at x2 is equal to 0. Now, let us get another equation by letting
f is equal to x square .And that results in the second equation that is shown that is
minus h to h x square d x, x square even function. And we are integrating over a symmetrical
domain. So, we are going to get some non-zero value which turns out to be 2 over 3 h q is
equal to W1 times function evaluated at x1. And since x1 is same as minus h that is going
to be minus h square.W2 times function evaluated at x2, x2 is same as 0. It is going to be
0 square. W3 times function evaluated at x3 function evaluated at x3, function evaluated
x3 square is h .It becomes W3 times h square that is how we obtained this equation. We
got 3 equations since we need to determine 3 unknowns. We use this condition that f (x)
is equal to 1, f(x) is equal to x, f(x) is equal to x square. And using these conditions
we got 3 equations. Now, we can solve these three equations for
these three unknowns W1, W2, and W3. We imposed while deriving these weights W1, W2, W3 itself
we impose these conditions that the function has to be exact up to quadratic, that is x
square Simpson’s rule is going to be exact for a function of order 2 that is quadratic
function.
With solving these three equations, we get W1 is equal to W3 which turns out to be h
over 3 W2 that is turns out to be four h over 3. And the function or the integral f(x) between
the limits minus h and h is approximated as one third h times function evaluated at minus
h, four third times function evaluated at 0 plus one third times function evaluated
at h. The first one should be at minus h or after returning to the original function and
its limits we get this equation. And where we need to keep a note at x2 is midway between
x1 and x3. Basically, this is the procedure that we can generalize for any number of points
any number of points. Newton-cotes rules of any number of points
to obtain the weights this is the procedure that we can generalize. If one wants to derive
Newton-cotes rules, weights for a Quartic function that is a fourth order function then
one needs to start getting four sorry five equations. We need to start with f(x) is equal
to 1, f (x) is equal to x, f(x) is equal to x square, f (x) is equal to x cube, f(x) is
equal to x power 4. If it is a cubic function, we need to impose the condition f(x) is equal
to 1, f(x) is equal to x, f (x) is equal to x square, f (x) is equal to x cube. If it
is Quartic function we already discussed .And if it is fifth order function, we need to
impose conditions f(x) is equal to 1 2. And f (x) is equal to x power 5. We get six equations
there we need to use six weights or we need to use Newton-cotes rules n is equal to 6.
Now, let us take this example which we already solved using trapezium rule integral sine
x between the limits phi over 4, and phi over 2 using Simpson’s rule. x1 is going to be
phi over 4, x 3 is going to be phi over 2, x2 is going to be between that is midway between
phi over 4 and phi over 2.
Solution h is equal to phi over 8 and function value evaluated at these 3 points, and multiplied
by the corresponding weights results which can be simplified to this. And we and we compare
this solution with exact solution which is 0.7071, we can observe that it is very closely
matching with exact solution, because Simpson’s rule we actually evaluated function at more
number of points. Now, let us look at what is the associated
error with each of these 3 rules, 3 Newton-cotes rules that is rectangle rule, trapezium rule
and Simpson’s rule to do that. To do that lets consider a Taylor series expansion f
of x about the lower limit of integration x1. If we do Taylor series expansion of f(x)
about the about x1 we get this equation. And integrating this equation on both sides between
the limits x1 and x2 results in this which can be simplified to truncated by rectangle
rule.
Basically, if you recall for rectangle rule to integrate function between the limits x1,
x2 what we are doing is we are evaluating function at x1 lower limit .And multiplying
with a weight which is equal to x2 minus x1. And we are not considering the higher order
terms that are shown in the series there. So, all the higher order terms are truncated
by the rectangle rule. The terms that are truncated by the rectangle rule are shown
in the equation. Rectangle rule has a dominant error term of order or of the form h o, x
square over two first derivate of f evaluated at x1. So, this is the error associated with
rectangle rule. Similarly, we can derive or we can find what is the error associated with
trapezium rule.
Consider Taylor series expansion about x1 to obtain x2, we get this and multiplying
by h over 2 and rearranging, we get this equation. Now, what we are going to do is? We are going
to substitute this; that is half h square times first derivative of f evaluated x1 into
the previous equation that we already looked for when we are deriving error term for rectangle
rule, which is reproduced here.
We are going to substitute the previous equation into this equation. And that results in this
equation. And if you see the second equation, basically it is same as the formula that we
looked for trapezium rule. Except we have some higher order terms which are truncated
by trapezium rule. Trapezium rule has a dominant error term of whatever term highest term that
we did not consider in the trapezium rule; that is minus one over twelve h cube second
derivative of function evaluated at x1. This is the dominant error term associated with
trapezium rule.
Similarly, we can derive error term dominant error term associated with Simpson’s rule.
Basically in Simpson’s rule we approximated function in this manner. And where higher
order terms are grouped into the form or the dominant error term is shown as alpha times
fourth derivative of function. Now, let us take where α is a constant coefficient.
And then let us take a function let f(x) be x power 4 when we substitute this fourth derivative
of x power 4 is four times, three times two that is going to be fourth derivative of f(x)
is going to be 24. And alpha is there, so 24 plus 24 alpha that is the error dominant
error term associated with Simpson’s rule. Now, we can also evaluate this function exactly
that is minus h to h x power 4 d x x power 4 is an even function. And we are actually
integrating it over a symmetrical domain, so the integral value turns out to be a non-zero
value which is going to be 2 over 5, h power 5, so we can easily back calculate, what is
alpha?
The dominant error term associated with this Simpson’s rule is minus 1 over 90, h power
4, fourth derivative of function. So, this is how we can actually find what is the dominant
term associated with any of these Newton-cotes rules. To summarize a function between the
limits or integral between the limits a to b f x dx can be approximated using this Newton-cotes
rules using this formula. The second one second formula sorry second term of this formula
is associated with the dominant error term for that particular rule. And where the constant
C’s c naught W’s and c 1 all those details are given in this table.
If one wants to select n is equal to 5 they can easily go through this table. And find
what is c not and corresponding W’s, and also C1 gives you idea about the a dominant
error term this is about Newton-cotes rules that is rectangle rule, trapezium rule, Simpson’s
rule. Sometimes, we may not evaluate any integral at a single stretch using any of these rules
instead of that we can use them repeatedly.
Now, let us look at what is called repeated trapezium rule. Instead of integrating a function
between the limits A to B at a single stretch, we can actually divide this into number of
strips here. For example, function f (x) needs to be evaluated between the limits A to B.
And the region A to B is divided into three strips. And width of each strip is B minus
A divided by k, where k is the number of strips. Here in this case, it is 3 in the previous
figure whatever we have seen k is equal to 3, but it can be something else for a different
problem. For this each strip, we need to apply trapezium rule. And sum it up then we are
going to get the repeated trapezium rule for this problem or for this integral. And that
is going to be as shown in the equation. And this equation can be rearranged in this manner,
so this is about repeated trapezium rule.
The previous examples example, which we already considered can be solved using repeated trapezium
rule with k is equal to 3 applying this formula only thing is we need to know what is h here,
h is going to be upper limit minus lower limit divided by k. Doing calculation it turns out
that k is equal to phi over 12 for this problem. And applying this formula we get approximate
value of the integral as given in this equation. And simplification of this results in this
value 0.7031. And a comparison of this with exact solution which is 0.7071 shows that
the accuracy is improved. It may be remembered that a single application of trapezium rule
gave 0.6704.
Now, let us look at repeated Simpson’s rule. We need to apply same l logic instead of integrating
at a single stretch a function or integral between the limits A to B, we divide this
width A to B into number of strips of width h. So, h value depends on the number of strips,
and the distance between A to B when we apply this repeated Simpson’s rule integral f(x)
between the limits a to b turns out to be this one or which can be further simplified.
Here I think, because of fund problem approximation symbol is not coming properly.
Now let us apply Simpson’s rule to estimate this integral taking k is equal to 4 width
of each strip is going to be upper limit minus lower limit divided by 4. And it turns out
that h is equal to phi over 16. And applying repeated Simpson’s rule results in this
approximation of the given integral which can be further simplified as 0.7071. And exact
solution matches very accurately with exact solution the four significant digits that
are shown there so, this is about Newton-cotes rules.
In summary, Newton-cotes rule of n number of point results in exact solution of integral
of order n minus 1 or if we want to integrate a function of order n exactly, we need to
use a Newton-cotes rules having n plus 1 number of points. If we want to integrate a function
of order 0, that is a constant function we need to use 1 point. And if we want to integrate
a linear function, we need to use 2 point. If we want to integrate a quadratic function,
we need to use 3 points.
Now, let us look at Gaussian quadrature also called as gauss Legendre rules. We will understand
why Legendre word is there. Now in gauss Legendre rule basically for one-point rule or its also
called midpoint rule that is n is equal to 1.We are going to evaluate function at a single
point which is midpoint or this is the equation that is shown here is a general equation for
gauss Legendre rules or gauss quadrature that is integral f (x) between the limits A to
B is approximated. As function evaluated at certain points multiplied by the corresponding
weights for one-point rule n is going to be equal to 1.Function sorry integral minus 1
to 1 f x dx is approximated as some weight times function evaluated at midpoint of minus
1 to 1. What is midpoint of minus 1 to 1? It is 0. So, function is evaluated at 0.
And weight here is 2, so two times function evaluated at 0. Basically, whatever may be
the order of function whether it is constant, linear, quadratic, cubic, quadratic or fifth
order function. Basically it is replaced with a constant function as shown in the figure.
f (x) is replaced with q not and in one-point rule this integral minus 1 to 1 f(x) dx is
evaluated by using the function evaluated at the midpoint of the range over, which integration needs
to be performed multiplied by the weight which is turns out to be 2 here for this one-point
rule.
Now let us look at how this two is obtained, how zero is obtained for location, how is
obtained for the weight? We will understand it better when we look at three point rule.
Now, let us look at two-point rule that is n is equal to 2 again function f (x) or the
integral f (x) minus 1 to 1 dx is approximated as W1 times f (x1) function evaluated at x1
plus W2 two times function evaluated at x2. Now, let us understand how to get these weights, and weights W1,
W2 and x1, and x2. Just before I made a statement that the procedure will be cleared when we
take n is equal to 3, but let us look here itself for two point rule how to get this
weights and locations x1 and x2. There are four unknowns to be determine x1,
x2, W1, W2. So, one can easily guess we require four equations to uniquely determine these
W1 W2 and x1 x2, So, what we will do is we will play the same or we will use the same
logic as what we did for getting the weights for Simpson’s rule. Let us start with f(x)
is equal to 1. We need to get 4 equations, so let us start with f (x) is equal to 1,
f (x) is equal to x, f(x) is equal to x square, f(x) is equal to x cube and we get 4 equations.
When we substitute f(x) is equal to 1.We get the equation integral 1 dx between the limits
minus 1to 1 is going to be 2. And substituting f (x) function evaluated x1 is equal to 1,
function evaluated at x2 is equal to 1, we get this equation. Similarly, at f (x) is
equal to x again x is a odd function we are integrating over a symmetrical integral domain,
integral turns out to be 0 .And f(x) is equal to x means function evaluated at x1 is going
to be x1. And function evaluated x2 is going to be x2. We will get this equation 0 is equal
to W1, x1 plus W2 x2.
Now, let f(x) is equal to x square, again x square is an even function integrating this
over a symmetrical domain results in a non-zero value, which is going to be 2 over 3. And
function evaluated at x1 is going to be x1 square. And function evaluated at x2 is going
to be x2 square which gives us this equation 2 over 3 is equal to W1 x square plus W2 x
square, W1 x1 square plus W2 x2 square sorry, f (x) is equal to x cube, again x cube is
odd function where integrating over symmetrical domain integral turns out to be 0, f(x1) function
evaluated x1 is going to be x1 cube, function evaluated at x2 is going to be x2 cube that
results in this equation 0 is equal to W1 x1 cube plus W2 x2 cube. So, we got the four
equations that we actually looking for. We can solve these four equations for the four
unknowns x1, x2, W1, and W2. And the solution results in these equations x1 or x2 is plus
or minus x1, W1 is equal to W2 is equal to 1. And x1 square is equal to x2 square is
equal to one third. And simplification of previous equation results in this and W1 is
equal to 1 and W2 is equal to 1.
According to two-point gauss-Legendre rule, we are going to evaluate function at minus
1 over root 3 and multiply with corresponding weights. And add them together that is corresponding
weights are 1. It is basically function evaluated at minus 1 over root 3 plus function evaluated
at 1 over root 3. And this is what is schematically shown in the figure. The original function
f x is replaced with a linear function or function q1 original function is replaced
with a function q1. And the function is evaluated at two points x1 and x2. Estimate the value
of this integral and the highest order of integrand here is 3.
We need to evaluate function by substituting in the previous equation x is equal to minus
1 over root 3 and x is equal to 1 over 3, we get this. And it turns out that the approximate
solution is same as the exact solution. What we can learn from this is? If we use Gauss-Legendre
rule or if we adopt Gauss-Legendre rule, we can integrate a function of order two n Gauss-Legendre
rule of n points can integrate a function of order 2 n minus 1 exactly. We can integrate
a function up to order 2 n minus 1 exactly. If we adopt n number of points, so two Point
Gauss-Legendre rules integrates cubic exactly a function cubic function exactly. This procedure
can also be extended for more number of points.
Now let us see three-point Gauss-Legendre rule. That is n is equal to 3 integral minus
1 to 1, f(x) d x is approximated as function evaluated at three points x1, x2, and x3 multiplied
with corresponding weights W1, W2, and W3 and sum them up, this is how we can approximate
this integral. We need to determine these weights and locations W1, W2, W3, and x1,
x2, x3. What about a the accuracy? this formula will be exact for a function up to order 5
because a n point Gauss-Legendre rule is going to be exact in evaluating an integral having
integrand of order 2 n minus 1. How to determine these weights and this locations, we need
to adopt similar procedure that we already looked at. We need to get six equations to
determine the six unknowns uniquely that is f (x) is equal to 1, f (x) is equal to x,
f (x) is equal to x square, f (x) is equal to x cube, f (x) is equal to x power 4, f(x)
is equal to x power 5. We impose these conditions.
And we determine the six unknowns. And it turns out that the solution of six equations
results in this that is W1 is equal to W3 x1 is equal to minus x3 and x2 is equal to.
In the figure, the locations of these integration points is shown that is x1 is equal to minus root 3 over 5,
x2 is equal to 0. And x3 is equal to root 3 over 5 integral f (x) between the limit
minus 1 to one is basically approximated as 5 over 9 times function evaluated at function
evaluated at minus root 3 over 5 plus 8 over 9 times function evaluated at 0 plus 5 over
9 times function evaluated at root 3 over 5. So, this is a procedure that we can follow
even if one desires to find what are the locations, and weights for more points in Gauss-Legendre
rules. Basically, we have the weights and locations
for this Gauss-Legendre rules for any number of points we can derive there is no problem.
And also all these are well documented in any of the text books on finite elements or
any numerical methods. The formulas or the weights, and locations are available for the
limits minus 1 to 1. If limits are something else we need to discuss what are what will
be the procedure changing limits of integration.
If one wants to evaluate this integral A to B, f (x) d x we need to somehow bring into
the form minus 1 to 1 some integrand. The procedure is like this, we can use this formula
to change the limits of integration that is limits A to B or changed to minus 1 to 1.
Using this formula, x is equal to B minus A times t plus B plus A divided by 2, and
when we take derivative of this equation on both sides. We are going to get d x is equal
to B minus A over two times d t. So, this is how integral with limits other than minus
1 to 1 can be changed. Let us estimate but the rest of the procedure once we get into
the form minus 1 to 1, integral minus 1 to 1 the rest of the procedure similar to what
we already discussed. Now, let us estimate the value of integral 1 to 4 x cos x d x.
First job is to change the limits of integration. And here also it is mentioned that we need
to use three-point Gauss-Legendre rule. So, first job is to change the limits of integration.
We need to make a substituent x is equal to 3 t plus 5 over 2 and d x needs to be replaced
with 3 over 2 dt. So, making these substitutions we can proceed as we already discussed by
taking three points. We already know what are the locations? And what are the corresponding
weights. This integral when we make this substitution of x in terms of t and d x in terms of d t
this integral that is integral 1 to 4 x cos x d x becomes or the x cos x is plotted here
as a function of x. And the locations of the integration points
are shown in this figure. So, we need to evaluate the function at these points x1, x2 that is
t the integration point which is located at minus root 3 over 5, it corresponds to x1
is equal to 1.34. Integration point located at x2 is equal to 0 corresponds to the point
x2 is equal to 2.55. Integration point located at root 3 over 5 corresponds to the integration
point 3.66. Using the previous change of limits of integrations, we get these integration
points. So, we need to evaluate function at these integration points.
Substituting the change of limits into the previous equation of integral, we get this
equation. And substitution of weights and sample points we get this approximation. Actually
there is some problem with font again as brackets are shown as some blocks there. And when the
integral is simplified in turns out that these integral values minus 5.0611 and we compare
this with the exact solution, exact solution turns out to be minus 5.0626, which
is very close to the exact solution.
Now we can summarize Gauss-Legendre integration points and corresponding weights in a table
like this. Here only up to n is equal to 4 is shown, but as I mentioned in any of the
standard text book on finite element method or any numerical method. We can find integration
point n is equal to a up to 10 or more. And also these values of location, and weights
are can also be obtained in any of the commercial software like math lab. Why this word Legendre
is? Actually these locations or functions of these are it can be shown that these locations
that is exercise or roots of this Legendre polynomial that is why? It is called Gauss-Legendre
integration rule or Gauss-Legendre rules or for that matter most of the finite element
text books prefer this as gauss quadrature in a very simple form.