Prove $\int_{na}^{nb}f(x)dx=n\int_a^bf(x)dx$ > Prove that $$ \int_{na}^{nb}f(x)dx=n\int_a^bf(x)dx $$ This is given as a general property of definite integrals in my reference. Is it true for all cases ? If it is true I have no clue of how to prove it ? $$ t=nx\implies dt=n.dx\\\ \int_{na}^{nb}f(x)dx=\int_{na}^{nb}f\big(\frac{t}{n}\big).\frac{dt}{n} $$

This is not true as Brevan points out in comments. But there are two changes I can think of you can make to get a true statement. The first is a change of variables, but you typed it wrong. It should be $t=x/n$ to remove the $n$ multiplying $a,b$: then $dt = dx/n$ and $x=nt$, so $$ \int_{na}^{nb} f(x) dx = n \int_a^b f(nt) dt$$ or equivalently from MathematicsStudent1122's comment $$ \int_{na}^{nb} f(x/n) dx = \int_a^b f(t) dt$$ The second is if $n$ is a special number for $f$ such that $f(t) = f(nt)$. Nontrivial (i.e. nonconstant) such functions exist; see Alex's counterexample from here: $$f(x) = \sin\left(\frac{\log x}{\frac{\log2}{2\pi}}\right)$$ I've made a moving graph, showing $f(x)$ and $f(kx)$ as $k$ varies between $1$ and $2$...here's what this $f$ looks like: ![enter image description here](