integration of a continuous function $f(x) $ and $xf(x)$ is zero [duplicate]

Question

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Prove that $\exists a<b$ s.t. $f(a)=f(b)=0$ when $\int_0^1f(x)dx=\int_0^1xf(x)dx=0$

Suppose that $f:[0,1]\to \mathbb{R}$ is continuous, and that $$\int_{0}^{1} f(x)=\int_{0}^{1} xf(x)=0.$$

How does one prove that $f$ has at least two distinct zeroes in $[0,1]$?

Well, if not say $\forall a,b\in [0,1] \ni a<b$, but $f(a)\neq f(b), f(a)>0$ then there will be a neighborhood of $a$ say $(a-\epsilon,a+\epsilon)$ where $f(x)>0$ and hence the integral will not be equal to $0$, but I don't know where I am using the other integral condition. am I wrong anywhere in my proof? please help.

Answer 1

Define $F(x):= \int_0^xf(t)\,dt$ for $x \in [0,1]$. Then the second integral tells us $$ \begin{aligned} 0 = \int_0^1xf(x)\,dx \, & = \, xF(x)\Bigr|_0^1-\int_0^1F(x)\,dx \\ & = 1\,F(1) - 0\,F(0) - \int_0^1F(x)\,dx \\ & = \int_0^1f(x)dx - \int_0^1F(x)\,dx \\ & = -\int_0^1F(x)\,dx. \end{aligned} $$

By the mean value theorem and continuity of $F$, which follows from continuity of $f$, this tells us that there is $c \in (0,1)$ such that $F(c)=0$. That is, $$ \int_0^cf(x)\,dx = 0. $$ It follows that $$ \int_c^1f(x)\,dx = \int_0^1f(x)\,dx-\int_0^cf(x)\,dx = 0 - 0 = 0 $$ as well.

We then apply the mean value theorem two more times, using continuity of $f$, for the intervals $[0,c]$ and $[c,1]$ to find $a \in (0,c)$ and $b \in (c,1)$ respectively such that $f(a)=f(b)=0$.