Prove the equation $\int_0^4 f(x(x-3)^2) \,dx=2\int_1^3 f(x(x-3)^2) \,dx$

Question

Let $f:\mathbb{R}→\mathbb{R}$ be a continuous function. Prove the equation $$\int_0^4 f(x(x-3)^2) \,dx=2\int_1^3 f(x(x-3)^2) \,dx.$$ I have tried substituting $$x(x-3)^2 = u.$$ But after that I couldn't figure out.

Answer 1

Consider the function $g(x)-t=x(x-3)^2-t$ and consider the respective inverse functions as $z=z(t),y=y(t),x=x(t)$ . Then using Vieta's formula on the equation $g(x)-t=0$ we have $x(t)+y(t)+z(t)=6$ . Therefore \begin{align*}\int_0^4f(x(x-3)^2)\,dx&=\left(\int_0^1+\int_1^3+\int_3^4\right)f(x(x-3)^2)\,dx\\&=\int_0^4f(t)\,dy(t)+\int_1^3f(x(x-3)^2)\,dx+\int_0^4f(t)\,dz(t)\\&=\int_0^4f(t)(y'(t)+z'(t))\,dt+\int_1^3f(x(x-3)^2)\,dx\\&=-\int_0^4f(t)\,dx(t)+\int_1^3f(x(x-3)^2)\,dx\\&=-\int_3^1f(x(x-3)^2)\,dx+\int_1^3f(x(x-3)^2)\,dx\\&=2\int_1^3f(x(x-3)^2)\,dx\end{align*} Done!

Answer 2

By using Kings Property (Equivalently substitute $4-t$ to $x$) We see that both LHS and RHS are equal except the factor of $2$ on the RHS. Hence, there's a typo in your question. There will not be $2$ and the equality holds.

Hence,

$$\int_0^4 f(x(x-3)^2) \,dx≠2\int_1^3 f(x(x-3)^2) \,dx$$

But,

$$\int_0^4 f(x(x-3)^2) \,dx=\int_1^3 f(x(x-3)^2) \,dx$$