simpsons rule for integration with multivariable

I am having trouble setting up the equation.

supposedly, I am trying to fine the integral of 0.7051*r.*T from 0.308 to 0.478

both r and T are vectors.

so far I have @f = @(r,T) 0.7051*r.*T

a = 0.308

b = 0.478

n = length(r) to be number of subdivisions

h = (0.478-0.308)/10 to be step size

The following is my approach so far I= h/3*(f(r(1),T(1))+2*sum(f(r(3:2:end-2),T(3:2:end-2)))+4*sum(f(r(2:2:end),T(2:2:end)))+f(r(end),T(end)))

I am not sure if I need to use a for loop to include all a,b,and h input or this (I) is good....

Answer 1

[Roger Stafford]

There is some confusion here as to what your variable of integration is to be. In the way you define ‘h’ it would appear to be some variable that varies from 0.308 to 0.478 in ten discrete steps. But how does that relate to the two vectors ‘r’ and ’T’ with their ’n’ values? You would need for 'n' to equal eleven to be compatible with the ten in the definition of ‘h’.

Also the term “+4*sum(f(r(2:2:end),T(2:2:end)))” looks incorrect. It should be “+4*sum(f(r(2:2:end-1),T(2:2:end-1)))”.