Vrftjkry

I am being asked to calculate the integral $int_{a}^{b}ln x~dx$ using the definition of integral (i.e. expressing it as limit of Riemann sums)

Here's what I did:
Let's divide the interval $[a,b]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.

Let $x_i=a+frac{(b-a)i}{n}$ where $i=0,1,2...,n-1. $ be the start points of each subinterval such that $x_0=a$ and $x_n=b$

Let $Delta x_i=x_i-x_{i-1}=frac{b-a}{n}$

I defined the integral like this:

$$begin{align}int_{a}^{b}ln x , dx&=lim_{nrightarrow infty}sum_{i=o}^{n-1}f(x_{i})Delta x_{i}\
&=lim_{nrightarrow infty}frac{b-a}{n}sum_{i=0}^{n-1}lnleft(a+frac{(b-a)i}{n}right)\
&=lim_{nrightarrow infty}frac{b-a}{n}lnleft(prod_{i=0}^{n-1}left(a+frac{(b-a)i}{n}right)right)end{align}$$

And I'm stuck here, I don't know how to find this product, any hint would be appreciated , thank you !

Some hints:

Let ${bover a}=:rho>1$, and choose an $Ngg1$. Then use the partition $a=x_0<x_1<ldots<x_N=b$ given by
$$x_k:=a,rho^{k/N}qquad(0leq kleq N) .$$
Then
$$R_N:=sum_{k=0}^{N-1}log(x_k)(x_{k+1}-x_k)$$
is an admissible Riemann sum for the given integral, i.e., we can be sure that $$lim_{Ntoinfty} R_N=int_a^blog x>dx .$$
Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $sum_{k=0}^{N-1} k, q^k$.