How to transform $\int_{x_0}^x\frac{y(x)}{x}dx$ from an integral to a sum? I need to transform this integral
$\int_{x_0}^x\frac{y(x)}{x}dx$
to a sum for computational purposes but it is years since I had calculus, so I have forgotten (if I even knew it in the first place) how to do it.
I am primarily interested in a numerical method to solve this problem (thank you for your comment, @Doug M).
 A: Hint: $$\int_{a}^{b}f(x)dx=\lim_{n\to \infty} \sum_{k=1}^{n}f(x+k\Delta x)\Delta x$$ where $\Delta x=\frac{b-a}{n}.$ This is the formula for representing a definite integral as the limit of a Riemann sum.
A: I completely agree with Pendronator's answer and advise the OP to use it as a starting point.  Then the OP will basically have two choices:
(1) using math 
(2) using a computer
..........
(1)
Based on the OP's query, this does not involve attempting to find the antiderivative.  Instead, it involves setting up the summation as a function of $n$, and then using math to compute the limit of the summation, as $n$ goes to infinity.
The classic problem where this is an apt approach is 
$\int_0^1 x^2 dx.$ 
Here, you would (arguably) need that 
$\sum_{i=1}^n i^2 = \frac{n (n+1) (2n+1)}{6}.$
(2)
The classic problem where this is an apt approach is 
$\int_0^1 \frac{1}{x^2 + 1}dx.$ 
In fact, the pertinent antiderivative is arctan$(x)$, so the
exact answer is $\pi/4.$
However, I am assuming that the OP would want to use the computer
to compute an approximate answer.  If you set $n$ equal to the
fixed value of (for example) $n = 100,$ set up your summation accordingly, and then use a computer to calculate the sum, you will get an answer that approaches $\pi/4.$
You might then repeat this trial with $n = 1000,$ and compare the accuracy of this approximation with that from $n = 100.$  This will give you a reasonable idea of how large to make $n$ (i.e. setting $n$ to some fixed large integer) in order to achieve the desired accuracy when approximating the area under the curve for your specific function $y(x)$, whatever that function happens to be.
