Making a Riemann Sums function with multiple n values

1 次查看(过去 30 天)
clc;clear;
fun = input('what is the function: ');
a = input('what is a: ');
b = input('what is b: ');
n = input('what is n: ');
N=length(n);
h = (b-a)/N;
sum = 0;
for j=1:N
h(j) = (b-a)/n(j);
sum(j) =fun(a+h(j)*(j-1));
end
sum(n) = sum(n).*h(n);
disp(sum)
So, I am trying to create a Riemann sums function and want the final value to be an array of the chosen n values, e.g. n = 1:10. So, the values as you increase should become more accurate to the actual integral. But, When I look at the results, they just get smaller over time not even close to the actual value. When I was trying to find the integral with just 1 n, I got a pretty close value to the actual integral. I am not sure what I did wrong in converting it to an array format.

回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Mathematics 的更多信息

标签

产品


版本

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by