Finding a function g such that $\int_0^3 f(x) dx = C\int_a^b g(x) dx$

Let $f:[0,3]\rightarrow \mathbb{R}$ be a function and let $\int_0^3 f(x) dx$ exist. Find a function g such that $\int_0^3 f(x) dx = C\int_a^b g(x) dx$.

I'm not sure what I am supposed to do to solve the problem. Do I need to use one of the four rules : rectangle rule, trapezoidal rule, Gaussian quadrature or the Simpson's rule ? And what about the error ?

• Hint: Substitute $g=f\circ \phi$ and determine $\phi$ to match the boundaries. – Nephente Feb 18 '17 at 17:58
• Change variable $u=a+\dfrac{b-a}{3}x$ in left integral. – Nosrati Feb 18 '17 at 18:25
• $$\int_0^3 f(x) dx = F(3) -F(0) =C\int_a^b g(x) dx = C\int_a^b f(\phi(x)) dx = C\int_a^b f(\phi(x))\phi^{-1}(x)\phi(x) dx=C\int_{\phi(a)}^{\phi(b)} f(z)\phi^{-1}(z)dz$$ How do I solve the integral from here on? – Septime44 Feb 19 '17 at 15:23
• And sorry I don't get the 2nd tip ... – Septime44 Feb 19 '17 at 15:24
• I mean $C\int_{\phi(a)}^{\phi(b)} f(z)(\phi^{-1})'(z)dz$ – Septime44 Feb 19 '17 at 17:04