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

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.

• nbd? pla?  
– Did
Jul 7 '12 at 20:38
• could you tell me where I am wrong making the negation of the fact that I need to show? Jul 7 '12 at 20:38
• The mean value theorem for integrals will get you at least one such point. Jul 7 '12 at 20:40
• $\int_{a}^{b}f(x)dx= f(c)(b-a)$,for some $c\in (a,b)$ This one? Jul 7 '12 at 20:44
• yah so $\int_{0}^{1}f(x)=f(a)=0$ and $\int_{0}^{1}xf(x)=bf(b)=0$ for some $a\,b in(0,1)$ but how to show $a<b$? Jul 7 '12 at 20:49

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 $f(a)=f(b)=0$.