With naive intuition one can obviously see that the definite integral as infinite subdivisions of an area under a curve, within the finite interval "a to b", from which the function of integration must be continuous between the points "a to b" on the x axis.
With all that being said, I am having a hard time getting a hold of this concept being explained in a rigorous fashion. Specifically the screenshot below where the author introduces the expression with the curly braces,
h{a+{a+h}+{a+2h}+...
I am completely lost at that point on how to understand what he has conveyed here. Can you guys clear things up for me? You may have to click the image into an external tab to be enlarged honestly.
First partition the interval $[a,b]$ into $n$ equal parts. Denote each equal sized interval by $h$ and note that $$h = \frac{b - a}{n}$$ Now the author wishes to show you the lower sum of function over these distinct parts. If $f : [a,b] \to \mathbb{R}$ is an increasing function then the lower sum may be written as $$ U(f, [a,b]) = \sum_{k = 0}^{n - 1} f(a + kh)h$$ Note that what we are doing here is taking the value of the function at the left edge of an interval, and multiplying it by the width of that slice. This is what the equation in the curly braces is trying to tell you, he just skipped the part where he states that he uses the function $f(x) = x$ for this purpose! Now note that if $f$ is defined by $f(x) = x$ then we have $$ U(f, [a,b]) = h \sum_{k=0}^{n-1}(a + kh) = h\left((n-1)a + h \sum_{k=0}^{n-1}k\right)$$ The sum on the right is the sum of natural numbers from $1$ to $n - 1$ so we have $$ \sum_{k=0}^{n-1} k= \sum_{k=1}^{n-1}k = \frac{n(n - 1)}{2} $$ To verify this first derive the fact that $$\sum_{k=1}^N = \frac{N(N+1)}{2}$$ and then set $N = n - 1$. The lower sum then becomes $$U(f, [a,b]) = h \left( (n-1)a + h \frac{n(n-1)}{2}\right) = \frac{h(n-1)}{2} \left( 2a + hn\right)$$ Now by substituting the value of $h$ we have $$ U(f, [a,b]) = \frac{(b-a)(n - 1)}{2n} \left( 2a + \frac{b - a}{n}n\right) = \frac{(b-a)(n - 1)}{2n}(b + a) = \left( \frac{1}{2} - \frac{1}{2n}\right)(b - a)(b + a) = \left( \frac{1}{2} - \frac{1}{2n}\right) (b^2 - a^2)$$ Then the result follows from $n \to \infty$.