I tried to approach integrating by filling the space with infinitely many infinitely narrow rectangles.
$a$ is the left bound
$b$ is the right bound
$n$ means the number of rectangles, approaches $\infty$
$k$ means the k-th rectangle
My integral is described by the following formula:
$$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{b-a}{n} \cdot f(a + \frac{k(b-a)}{n}) $$
The rough idea in the picture below:
Let's solve for $f(x) = \frac{1}{x}$, a = 1, b = 3
$$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2}{n} \cdot \frac{1}{1 + \frac{2k}{n}} = \lim_{n \to \infty} \left(\frac{2}{n+2}+\frac{2}{n+4}+\frac{2}{n+6}+ \dots\right) $$ The sum clearly diverges, but I don't know why.. any ideas? Is this way of integrating possible? If not, why? Are my formulas correct?
Your sum should only go up to $n$. You have otherwise properly described the right hand Riemann sum. The sum has a finite number of terms, so it does not diverge. You take the limit after you do the sum and should get $\log 3$.