Indefinite integration by parts

Question

I was given an excersice to solve wich asks you to prove:

$$\int_0^1 f(r) r dr = 0 $$

knowing that:

$$\int_0^1 f(t) dt = 0 $$

After doing integration by parts I ended up with:

$$\int f(r) r dr = (\int f(r)dr)r- \int\int f(r)drdr $$

My question is, how should I proceed in order to evaluate the integral with my integration limits between 0 and 1. I tried with Barrow, but I don't know what to do with the double integral and its limis

Answer 1

You can't prove it, since it is false. If $f(x)=\sin(2\pi x)$, then $\displaystyle\int_0^1f(x)\,\mathrm dx=0$, but $\displaystyle\int_0^1x f(x)\,\mathrm dx=-\frac1{2\pi}\neq0$

Answer 2

It does not work for a linear function: $$\int_0^1 \left(x-\frac12\right)dx=\left(\frac{x^2}{2}-\frac x2\right)|_0^1=0, \ \ \text{but}\\ \int_0^1 \left(x-\frac12\right)xdx=\left(\frac {x^3}{3}-\frac{x^2}{4}\right)|_0^1=\frac1{12}\ne 0.$$