How to find the left riemann sum using a for loop?

I need to find the left sum of a function handle f, on the interval [a,b] with n subintervals. So far what I have is:

 function r=myleftsum(f,a,b,n)
r=0;
dx=(b-a)/n;
for k=1:n
    c=a+k*dx;
    r=r+f(c);
end
r=dx*r;
end

I think this is taking the right sum but I need the left sum. I am not sure which line to change or what will make this code take the left sum.

Answer 1

[Geoff Hayes]

Renee - since you are calculating the Left Riemann Sum, then the code needs to use the left-end point of each sub-interval. The left-end points are a,a+dx,a+2dx,...,a+(n-1)dx. So your code becomes

function r=myleftsum(f,a,b,n)

    dx=(b-a)/n;

    % initialize r to f(a) (the left-end point of the first sub-interval
    % [a,a+dx])
    r=f(a);

    % need only consider the n-1 remaining sub-intervals
    for k=1:n-1
      c=a+k*dx;
      r=r+f(c);
    end 
    r=dx*r;
  end

Try the above and see what happens!

[Shea Brockman]

maybe start from k = 0?

[Geoff Hayes]

Yes, I guess we could start from k=0 but then we'd need to initialize r to be zero instead of a.