How to prove the following integral equation? $\int_{0}^{c}x^2f(x)=0$ Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function such that $\int_{0}^{1}f(x)(1-x)dx=0$. Prove that there exists $c\in(0, 1)$ such that $$\int_{0}^{c}x^2f(x)=0$$
Didn't try anything because I don't know how to approach it.
 A: First, we show that there exists a number $a\in (0,1)$ such that $\int_{0}^{a}f(x)\,dx=0$. Let $F(x) := \int_0^x f(t)\,dt$. Doing integration by part to $\int_{0}^{1}f(x)(1-x)\,dx$ we get
$$0=(1-x)F(x)\Bigr\rvert_{0}^1 + \int_{0}^{1}F(x)\,dx = \int_{0}^{1}F(x)\,dx$$
Hence, by mean value theorem, there exists $b\in (0,1)$ such that $F(a)=0$, i.e,  $\int_{0}^{a}f(x)\,dx=0$.
Now, using the fact that $\int_{0}^{a}f(x)\,dx=0$, we will show there exists $b\in (0,a)$ such that $\int_{0}^{b}xf(x)\,dx=0$.
Let $\displaystyle G(x)=\int_0^x\int_0^t f(u)\,du\,dt\quad$ and $\displaystyle g(x)=\int_0^x tf(t)\,dt$
Since $f(x)=G''(x)$, applying integration by part to $g(x)$ we get
$$\displaystyle g(x)=\int_0^x tG''(t)\,dt=\bigg[tG'(t)\bigg]_0^x-\int_0^x G'(t)\,dt=xG'(x)-G(x)$$
Now we know that $G'(0)=0$ and $G'(a)=\int_{0}^{a}f(x)\,dx=0$ so by Flett's theorem, there exists $b\in (0,a)$ such that
$$G'(b)=\dfrac{G(b)-G(0)}{b-0}=\dfrac{G(b)}{b}$$
Hence, $g(b)=bG'(b)-G(b)=0$.
Lastly, note that the above argument work for the continuous function $\tilde{f}(x):=xf(x)$ as well, namely, since $\int_{0}^{b}\tilde{f}(x)\,dx=\int_{0}^{b}xf(x)\,dx=0$, there exists $c\in (0,b)$ such that
$$0=\int_{0}^{c}x\tilde{f}(x)\,dx=\int_{0}^{c}x^2f(x)\,dx$$
hence the claim.
Remark: Clearly by the same argument, for any $n\in \mathbb{N}$, there exists $c_n\in (0,1)$ such that $$\int_{0}^{c_n}x^nf(x)\, dx=0$$
A: COMMENT.-A realization of such a function is $f(x)=\dfrac{\sin(2\pi x)}{1-x}$ that admits an extension by continuity at the point $x=1$ because $\lim_{x\to 1}\dfrac{\sin(2\pi x)}{1-x}=-2\pi$. It is easily verified that $\int_0^1f(x)(1-x)dx=0$.
On the other hand $$\int_0^{0.5}x^2f(x)dx\approx 0.0358981 \text { 
 and }\int_{0.5}^1x^2f(x)dx\approx -1.2948947$$ (the function $x^2f(x)$ being positive on $[0,0.5)$ and negative on $(0.5,1]$) so $$\left|\int_{0.5}^1x^2f(x)dx\right|\gt\int_0^{0.5}x^2f(x)dx$$ Consequently by mean value theorem there exists $c$ such that $\int_{0.5}^cx^2f(x)dx=-\int_0^{0.5}x^2f(x)dx$ thus $\int_0^cx^2f(x)dx=0$
