Using simpson 1/3 rule for double integration I want to give first only y limit then after getting this integration I want to give x limit f(x,y)=sin​(x)(A1sinh​(y)+B1sinh​(b-y))+ C1sin(x)*s​inh(b-y)+D​1sin(x)*si​nh(y)+E1si​nh(Lx)* sin(y)+sinh(x)*sin(y)

is there a method in simpson 1/3 rule for double integration that first i give only y limit after getting this integration i give x limit and function has both x and y variable in multiplication so i cant use simpson 1/3 rule for single integration i am using a simpson 1/3 rule for double integration the code but in this code i have to give x and y limit both at a same time the code is

 syms x y;
f = inline(f(x,y));
c = inline('1');
d = inline('2');
a =0; 
b = x;
m = 4;
n = 4;
hx = (b-a)/n;
An = 0;
Ae = 0;
Ao = 0;
for i = 0:n
x = a + i*hx;
ya = c(x);
yb = d(x);
hy = (d(x) - c(x))/m;
Bn = f(x,ya) + f(x,yb);
Be = 0;
Bo = 0;
for j = 1:m-1
y = ya + j*hy;
z = f(x,y);
if rem(j,2)==0
Be = Be + z;
else
Bo = Bo + z;
end
end
A1 = (Bn + 4*Bo + 2*Be)*hy/3;
if i == 0 | i == n
An = An + A1;
else
if rem(i,2)==0
Ae = Ae + A1;
else
Ao = Ao + A1;
end
end
end
Dint = (An + 4*Ao + 2*Ae)*hx/3;
end

[Torsten]

is there a method in simpson 1/3 rule for double integration that first i give only y limit after getting this integration i give x limit

Why do you want to do this ?

Best wishes

Torsten.

[NILESH PANDEY]

because in my integration problem first i have to integrate w.r.t. y than after getting this i have to integrate dx/f(x) so if i give both x and y limit at same time it will integrate w.r.t x also to clarify this i am attaching a file i hope this will help

Answer 1

[Torsten]

Write down the Simpson-rule approximation for

integral_{y=c}^{y=d) exp(f(x,y)) dy .

This will be a finite sum including the variable x. Call it sum(x).

Thus under the first integral you have an expression of the form 1/sum(x) to integrate over [a:b].

Now apply Simpson's rule again in x-direction.

Best wishes

Torsten.

[NILESH PANDEY]

can you give the code of Simpson -rule that first integrate w.r.t y as you write but also have variable x.it will help me lot

[Torsten]

I assume you have a routine "simpson" for 1d-integration of the form

function integral = simpson(f,left,right,delta)

where f is a function handle and you want to integrate over [left:right] with a step size of delta.

Then your problem can be solved as follows:

function main
result = simpson(@(x)I(x),a,b,deltax);

function inner = I(x)
inner = simpson(@(y)exp(f(x,y)),c,d,deltay);
inner = 1/inner;

Best wishes

Torsten.

[NILESH PANDEY]

sir i cant understand it what is I(x) and result = simpson(@(x)I(x),a,b,deltax) what is the meaning of result and simpson. and is there a difference in answer if i use this q = integral(fun,xmin,xmax) not simpson 1d code

[Torsten]

In function I(x), you evaluate

1/integral_{y=c}^{y=d) exp(f(x,y)) dy

for a given value of x.

Thus by calling

result = simpson(@(x)I(x),a,b,deltax);

you integrate

1/integral_{y=c}^{y=d) exp(f(x,y)) dy

from x=a to x=b which should give you the integral you try to evaluate.

If you have problems, you may want to post the complete code you are using such that we can check it for errors.

Best wishes

Torsten.