When learning Riemann Integral, I was introduced to the concept of partitions and refining them. It stated that refining a partition (e.g. controlling its norm) increases its efficiency. What I mean by efficiency is the ability to measure the area beneath the curve. My teacher said it's intuitive. But, that's hardly satisfying. Also, can anyone give me a rigorous proof (without any graphical methods) of the fact that Riemann sum actually gives the area under a curve?
Any help will be appreciated.
At your "level of sophistication" it is assumed without proof that
(i) all "reasonable" domains in ${\mathbb R}^2$, in particular domains bounded by the $x$-axis, a continuous curve $y=f(x)>0$, and two ordinates $x=a$, $x=b$ have "area";
(ii) rectangles have area $\ell\cdot h$;
(iii) "area" is additive if two domains have at most parts of their boundaries in common; and
(iv) $A\subset B$ implies ${\rm area}(A)\leq{\rm area}(B)$.
These things are dealt with in detail in (geometric) measure theory. Integration is about computing the area for domains $A$ of the kind described in (i). Note that there are surprises in store, insofar as domains as simple as a circular disk of radius $1$ can have very strange area values.
Given any partition $P$ of the interval $[a,b]$ the lower and upper sums are areas of unions of rectangles. From the principles (ii)–(iv) it then follows that $$L(f,P)\leq {\rm area}(A)\leq U(f,P)\ .$$ Exploiting the continuity of $f$ one proves that the lower and the upper sums have a common limit under refinement, and one denotes this limit by $$\int_{[a,b]}f(x)\>{\rm d}x\ .$$ A squeeze argument (the lower sums are increasing, the upper sums decreasing under refinement of $P$) then enforces $$ {\rm area}(A)=\int_{[a,b]}f(x)\>{\rm d}x\ .$$ On the other hand, purely analytic reasoning shows that $$\int_{[a,b]}f(x)\>{\rm d}x=F(b)-F(a)\ ,\tag{1}$$ whereby $F$ is any function satisfying $F'=f$. Formula $(1)$ is a great miracle, and is called the fundamental Theorem of Calculus.