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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 $\Delta x_{i}$?

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    This may not help, since it just glosses over the point of arbitrary partitions, but you might want to take a look at http://math.stacke$x$change.com/questions/15294/why-is-the-area-under-a-curve-the-integral/15301#153012011-02-26

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The "convergence" of partitions, associated with defining a Riemann integral through a limiting process, is a bit trickier than convergence of a sequence.

We want two notions: that one partition "refines" another partition, and that a partition of an interval [a,b] produces subintervals of width at most h > 0. You seem to have a good grasp on the second of these, namely that even if the partition is not constructed to have n equal length subintervals, the subintervals may still have some "small" bound on their maximum width. We can specify that "mesh size" is small in this way, setting the stage for proving something about the limit as mesh size tends to zero.

It is the first concept, that of one partition refining another, which typically causes more difficulty. The notes you linked to do not seem to treat that topic. Here is a PDF on the Riemann integral which does.

For a given function f on interval [a,b] we define the lower Riemann integral as the sup over all partitions of their lower Riemann sums, and the upper Riemann integral as the inf over all partitions of their upper Riemann sums. If these agree, then we say the function f is Riemann integrable on [a,b], and the common value of the upper and lower Riemann integrals is the value of the Riemann integral.

A partition $Q$ of [a,b] is said to refine partition $P$ when the endpoints of $P$ are also endpoints of $Q$. The theory of Riemann integration uses a construction that given two partitions $P_1$ and $P_2$ produces a common refinement of both $Q$ by taking the union of the two sets of endpoints. Just having a smaller mesh size is not enough to guarantee that one partition refines another.

See the notes of the link above for more details of how the theory of Riemann integration is developed using the concept of partition refinement.

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    This [vid$e$o](http://fr$e$evideolectures.com/Course/2267/Mathematics-I/17) was also useful.2011-03-05