I am trying to understand how definite integral works when the partitions of the function are of unequal length. I found this link and am stuck here: $$I = \int_a^b f(x)dx = \lim_{mesh(P) \to 0} R(f,P,T)$$
I understand that $\int_a^b f(x)dx = \text{Area under the curve}$ but not how that is equal to $\lim\limits_{mesh(P) \to 0} R(f,P,T)$.
The way I see this, in the case of equal partitions, the mesh is $\Delta x_{i} = \frac{1}{n}(b-a)$ so a 0 mesh means that there will be infinitely many thin partitions so that the sum of the area of each partition equals the area under the curve.
In the case of unequal partitions, since the mesh is $\Delta x_{i} = x_i - x_{i-1}$, if each mesh goes to 0, it appears that the number of partitions $n$ remains fixed so that the sum of the area of each partition definitely won't cover the area under the curve. Am I visualizing this correctly? If the number of partitions increase, how is $n$ related to $\Delta x_{i}$?