Show using the formal, limited based definition of integral I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $\int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $\mathbb{R}$). However, I already seem to get stuck on the formal definition of the integral, which is (I believe) $\lim_{n \rightarrow \infty}\sum_{i=0}^{n-1}f(t_i)\Delta t$. Can anyone explain to me how to proceed? Or show me the steps I need to follow to get to the right answer?

Divide the interval from x= 3 to x= 7 into n equal length subintervals.  Each has length (7- 3)/n= 4/n.  The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$.  At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)\Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= \frac{4}{n}+ \frac{32i}{n^2}$.  Add those: $\sum_{i= 0}^{n-1} \frac{4}{n}+ \frac{32i}{n^2}= \frac{1}{n}\sum_{i=0}^{n-1} 4+ \frac{32}{n^2}\sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum?  To start with lets call it S(n).  That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1.  Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0.  Add those two "vertically"!  That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1].  Every term  is "n-1" and there are n terms.  2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n:  with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10.  with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $\frac{32}{n^2}$: $\frac{32}{n^2}\frac{n(n-1)}{2}= \frac{32}{n^2}\frac{n^2- n}{2}= 16(1- \frac{1}{n})$.
Putting those together  the "Riemann sum" with n intervals is $4+ 16(1- \frac{1}{n})$.  As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight  line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height  7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before. 
