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 ?

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

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