Proof of $\int_a^b f(nx) \;dx=\frac{1}{n} \int_{na}^{nb} f(x) \;dx $ > Let $f:\Bbb{R} \rightarrow \Bbb{R}$ be continuous. How to prove this ? $$\int_a^b f(nx) \;dx=\frac{1}{n} \int_{na}^{nb} f(x) \;dx $$ where $n \in N$ Really I don't know where to start. Any help?

I would substitute $$nx=t$$ then we get $$x=\frac{t}{n}$$ and $$dx=\frac{1}{n}dt$$ and $$t_1=na$$ and $$t_2=nb$$