# determining the number of divisions in riemann sums

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WILLBES BANDA on 26 Apr 2020
Commented: Aniket Devgun on 26 Jul 2020
Hi, since riemann sum is all about adding smaller divided rectangles below the graph. I developed a code which calculates the difference between present sum and previous sum, when the difference is greater than 1e-5, the code must store the number of rectangles(terms) added to reach the difference and store the number of terms in n. c is the final area, the problem is that my code is not giving me the number of divisions nor the difference between sums, please help me find the (1).difference between sums and (2).the number of rectangles/divisions when the difference is greater than 1e-5
format longG
syms x
f = 2*pi*(0.16*(0.25-(x-1)^2)+44.52)*(1+(-0.32*x - 0.32)^2)^1/2;
a = int(f,5,17.688);
b = sym(a);
c = double(b)
if diff(c) > 0.00001
n = length(c)
end

David Hill on 26 Apr 2020
You did not say if you wanted left, right, middle riemann sum. This is left.
f = @(x)2*pi*(0.16*(0.25-(x-1).^2)+44.52).*(1+(-0.32*x - 0.32).^2).^1/2;
a=1;%whatever bounds you want
b=3;%whatever bounds you want
Rs(1) = f(a)*(b-a);
Rs(2) = sum(f(linspace(a,b,2)).*((b-a)/2));
n=2;
while abs(Rs(n)-Rs(n-1))>1e-5
n=n+1;
Rs(n) = sum(f(linspace(a,b,n)).*((b-a)/n));
end
Aniket Devgun on 26 Jul 2020
What would it look like if it was a middle riemann sum?

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