The integral of some function of $[x]$

Question

I am trying to take the following integral which sounds easy but I really got confused and need help.

$f: [0,2]\to \mathbb{R}$ is integrable. Show that $$\int_0^2(x-1)f[(x-1)^2]dx=0.$$

For me the answer is $$\int_0^1f[u]du= f(0)!!$$

which is not necessarily zero. I appreciate any suggestion.

P.S. By $[x]$ the greatest integer less than $x$.

Answer 1

The function is symmetric through $x=1.$
Set $t=x-1,$ then $$\int_0^2(x-1)f([x-1]^2)dx=\int_{-1}^1tf([t]^2)dt=f(0)\cdot \int_{-1}^1t\,dt=0$$

Answer 2

I am getting: $$\int_0^{1^-}(x-1)f(0)dx+\int_{1}^{2^-}(x-1)f(0)dx$$ Which , I think would evaluate to zero

Answer 3

$$\int_0^2(x-1)f[(x-1)^2]dx = \int_0^1(x-1)f(0)\ dx + \int_1^2(x-1)f([x-1]^2)\ dx$$

Using $u = x-1$ in the second integral, $$ = \int_0^1(x-1)f(0)\ dx + \int_0^1uf(0)\ du$$ $$ = 2\int_0^1xf(0)\ dx - \int_0^1f(0)dx$$ $$ = f(0)\left.\left(x^2 - x\right)\right|_0^1$$ $$ = 0$$

Answer 4

Since $\lfloor (x-1)^2\rfloor=0$ on the open interval $(0,2)$ we have

\begin{eqnarray} \int_0^2(x-1)f\left(\lfloor(x-1)^2\rfloor\right) dx&=&\lim_{\epsilon\to0^+}\int_\epsilon^{2-\epsilon}(x-1)f(0)\,dx\\\ &=&f(0)\cdot\lim_{\epsilon\to0^+}\frac{(x-1)^2}{2}\bigg|_\epsilon^{2-\epsilon}\\\ &=&f(0)\cdot0\\\ &=&0\end{eqnarray}