How can I convert the Riemann Sum into an Integral?

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To convert the Riemann sum into an integral, the expression provided can be analyzed as the limit of the sum as n approaches infinity, which aligns with the definition of a definite integral. The function c(x, Δx) represents the area between x and x + Δx, and the sum captures the total area under the curve from the lower limit to the upper limit b. The formula for c(x, Δx) involves trigonometric functions and the area of a triangle, indicating a geometric interpretation of the integral. The goal is to express the limit of the sum in terms of the integral of the function f(z) over the specified interval. Ultimately, the conversion hinges on recognizing that as Δx approaches zero, the Riemann sum converges to the integral of f(z) with respect to z.
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If I have a function c(x,Δx) that gives the area between x and x + Δx of a function.
The area under the function can be given by:
Sum from j = 0 to n-1 of c(b/n j,c/b)
As n tends to infinity and b is the upper limit of integration.

How can I convert this from a sum into a integral? I'm not sure if this is already in the form of a Riemann integral or not.

Thankyou in advance
 
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c(b/n j,c/b)

Please clarify this expression.
 
Well c(x, Δx) is
1/2 Csc(x) Csc(x + Δx) s(x) s(x + Δx) Sin(Δx)
(Formula for the area of a triangle where Csc(x) s(x) are the length sides.

Where s(x) is the solution for z of f(z) == z Cot(x).

Where f(z) is the function I want to integrate. (I don't want to just integrate it f(z) dz)
 
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