Let $f:[0,3]\to\mathbb R$ given, such that $\int_0^3 f(x)\,dx $ exists.
Give a function g and a constant C such that $\int_0^3 f(x)dx=C\int_a^b g(x)dx$ while $g$ is only depending on $f$ and $C$ is only depending on $a$ and $b$ .
How does one handle this problem? This task was an old exam question.
On
One way to do this is to switch the $a\to b$ integral to one on $0\to 3$. So let's try
$$\int_a^bg(x)~dx=\int_0^3 g(u)~du$$ Now we need to find the substitution for $u$. When $x=a$ in the first integral we need $u=0$, and when $x=b$ we need $u=3$. So try $u=\frac{3}{b-a}(x-a)$. So we can reverse this to obtain $g$ in terms of $f$. Then the constant $C$ will be to take care of the change of variables, since $du=\frac{3}{b-a}~dx$. I leave it to you to put everything together.
Take $g(x) = \int_0^3f(t)dt$ (yeap, a constant function) and $C = \frac{1}{b-a}$.