I have to prove that $$ \int_a^b (cf)(x)dx=c\int_a^b f(x)dx)$$
This is my approach:
If $c=0$ it's trivial.
Let $\varepsilon>0$, $c\neq0$, there exists $\delta>0$ such that: ($S(f,P)$ denotes Riemman's sum),
$$||P||<\delta\implies| S(f,P)-\int_a^bf(x)dx|<\frac{\varepsilon}{|c|}$$
Observation: $$S(c.f,P,\{x_i\})=\sum_{i=1}^n(cf)(x_i)(t_i-t_{i-1})=c\sum_{i=1}^nf(x_i)(t_i-t_{i-1})=c\cdot S(f,P,\{x_i\})$$
Then, $$| S(cf,P)-c\int_a^bf(x)dx|=|c\cdot S(f,P)-c\int_a^bf(x)dx|=|c||\displaystyle S(f,P)-\int_a^bf(x)dx|<\frac{\varepsilon}{|c|}|c|=\varepsilon$$
QED.
Is this correct?