Let $f: [0,3] \to \mathbb{R}: x \mapsto \begin{cases}5 \quad x \in [0,1] \\ 0 \quad x \in (1,3]\end{cases}$
Using only the definition of integrability, prove that $f$ is integrable.
Source problem: The real numbers and real analysis, Ethan Block, p241 chapter 5 exercice 5.2.4.
My attempt: Let $\epsilon > 0$. Choose $\delta := \min\{\epsilon/5,1/2\}$. Let $P= (x_0, x_1, \dots, x_n)$ be a partition of $[a,b]$ with $\Vert P \Vert< \delta$ and let $T= (t_1, \dots, t_n)$ be a representative tupel of $P$ (that is $t_i \in [x_{i-1}, x_i]$).
We have
$$|S(f,P,T)- 5| = |\sum_{i=1}^n f(t_i)(x_i-x_{i-1}) - 5|$$
$$= | 5\sum_{i:t_i \in [0,1]} (x_i-x_{i-1})-5| = 5 |x_{\max\{i\mid t_i \in [0,1]\}}-1| < 5 \epsilon/5 = \epsilon$$
Note that we took $\delta \leq 1/2$ so that there is an index $i$ to sum over, because then $x_1 - x_0 = x_1 \leq 1/2$, so $t_1 \leq 1/2$.
Is this correct?