$I=[a,b]$ with $a<b$ is compact range in $\mathbb R $. It is also given that $f,g \in C(I)$ and $r,s\in I$. $C(I)$ is quantity of all in $I$ continuous functions. Show that:
$Re\int_{r}^{s} \! f(t) \, dt= \int_{r}^{s} \! Re f(t) \, dt$
and
$Im\int_{r}^{s} \! f(t) \, dt= \int_{r}^{s} \! Im f(t) \, dt$
With what should I start?
If you have that the integral is additive, write $f=\text{Re}(f)+i\text{Im}(f)$. Then since both of Re and Im are continuous both functions are still integrable and $$\int f(x)dx=\int \text{Re}(f(x))dx+i\int \text{Im}(f(x))dx$$ take the real and imaginary parts of both sides to get the result.