Calculate $\int_1^x \frac{1}{t} dt$ using the definition of integral I am being asked to calculate the integral $\int_{1}^{x}\frac{1}{t}~dt$ using the definition of integral (i.e. expressing it as limit of Riemann sums)
Here's what I did:
Let's divide the interval $[1,x]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.
Let $x_i=1+\frac{(x-1)i}{n}$ where $i=0,1,2...,n-1. $ be the start points of each subinterval such that $x_0=1$ and $x_n=x$
Let $\Delta x_i=x_i-x_{i-1}=\frac{x-1}{n}$
I defined the integral like this:
$$\begin{align}\int_{1}^{x}\frac{1}{t} \, dt&=\lim_{n\rightarrow \infty}\sum_{i=o}^{n-1}f(x_{i})\Delta x_{i}\\
&=\lim_{n\rightarrow \infty}\frac{x-1}{n}\sum_{i=0}^{n-1}\frac{1}{\left(1+\frac{(x-1)i}{n}\right)}\\
\end{align}$$
From here, is there a way to go forward without Taylor series?
 A: Here is a method that your markers probably won't like but  I find it entertaining.  The question is rather silly and deserves a perverse answer.
Let us "use the definition of the Riemann integral" to find $\int_1^x \frac1t\,dt$  for $x>1$ with $0<x<1$ similar.  The function  $f(t)=\frac1t$ is continuous on $[1,x]$ so the integral exists and (thanks to Cauchy) can be computed as a limit of a sequence of Riemann sums.
[By the way Riemann was unborn at the time that Cauchy used "Riemann sums" and  when Euler did much earlier.]
Having studied some calculus we all know the relation between this integral and logarithms. Some texts define $\int_1^x \frac1t\,dt =\ln x$ while others prove it from some other definition.
Start as you did with an integer $n$ and the points
$$ 1= 1 + a_0 < 1 + a_1 < \dots < 1+a_n =  x$$
where $a_i = \frac{i}n [x-1]$.
On the interval  $[1+ a_0,1+a_1]$  we can choose any associated point $\xi_1$ in that interval .   We use the mean-value theorem to do so:
$$\frac{\ln(1+a_1) - \ln (1+a_0)}{(1+a_1) -  (1+a_0)} = \frac{1}{\xi_1}$$
Do the same on each interval of the partition to get `appropriate' associated points.
This is because $\frac{d}{dx} \ln x = \frac 1x.$  On any interval $[a,b] \subset (0,\infty)$ we know that $\frac{\ln b -\ln a}{b-a} = \frac{1}{\xi}$  for some point $a<\xi< b$.
The Riemann sum now is easy to compute:
$$ \sum_{i=1}^n  \frac{1}{\xi_i}[(1+a_i) -  (1+a_{i-1})
=  \sum_{i=1}^n \left[   \ln(1+a_i) - \ln (1+a_{i-1}) \right]
=  \ln x - \ln 1 = \ln x.
$$
Take the limit as $n\to\infty$ [rather easy now] and this proves that
$$ \int_1^x \frac1t\,dt = \ln x  \ \ \ (x>1).$$

Even Cauchy knew that you did not have to use the endpoints in computing a 'Riemann sum.'  Riemann knew that any associated points would work for continuous functions and he made it a requirement for 'integrable' functions.


P.S.  The reason I call it perverse is that this method can defeat many calculus assignments.
E.g., Use the definition of integral to compute $\int_3^7 x^2\,dx$.
Consider $f(x)=x^2$, $F(x)=x^3/3$, $F(7)-F(3)=316/3$.  Take any partition
$3=x_0<x_1< \dots<x_{n-1}<  x_n=7$ and select, by the mean-value theorem, associated points
$\xi_i\in (x_{i-1},x_i)$ so that $F(x_i)-F(x_{i-1})=f(\xi_i)(x_i-x_{i-1})$.
Then, clearly,
$$\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1})= \sum_{i=1}^n[F(x_i)-F(x_{i-1})]=F(7)-F(3) =  316/3$$
for all such Riemann sums.  QED.
