You could use the fact that if $f$ is integrable on $[a,b]$ and $[b,c]$ then $f$ is integrable on $[a,c]$ and satisfies
$$ \int_{a}^{c} f = \int_{a}^{b} f + \int_{b}^{c} f $$
Now $\int_{1}^{2} x - 1 \, dx = \int_{0}^{1} u \, du$ and therefore it suffices to show that $\int_{0}^{1} x \, dx$ is integrable.
The lower sums are
$$ L = \sum_{i=1}^{n} f(t_{i-1})(t_{i} - t_{i-1}) $$
and the upper sums are
$$ U = \sum_{i=1}^{n} f(t_{i})(t_{i} - t_{i-1}) $$
If we use the uniform partition where $t_{i} = a + \frac{b-a}{n}i = \frac{i}{n}$ then these become
\begin{align*}
L
& = \sum_{i=1}^{n} \frac{i-1}{n} \frac{1}{n} \\
& = \frac{1}{n^{2}} \sum_{i=1}^{n} (i - 1) \\
& = \frac{(n-1)(n)}{2n^{2}} \\
& = \frac{n^{2} - n}{2n^{2}} \\
& = \frac{1 - \frac{1}{n}}{2} \\
& \rightarrow \frac{1}{2}, \, \text{ as } n \rightarrow \infty
\end{align*}
You should be able to show that $U = \frac{n^{2}+n}{2n^{2}} \rightarrow \frac{1}{2}$. Since the upper sum and lower sum both converge to the same value, the integral is defined.