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Let $f : [0, a] → \mathbb{R}$ be a continuous function, where $a ∈ \mathbb{R^+}$ and $\int_0^a f(x)dx=0$. Prove that $∃ c ∈ (0, a) $ such that $\int_0^c xf(x) dx = 0.$

I feel I've to use the Fundamental Theorem of Calculus and then Lagrange's mean value theorem but I can't work out the details. From the condition we get, $F(a)=F(0)$ where $F'(x)=f(x)$. Then I tried considering the function $g(x)=xF(x)$ but I can't proceed.

Can a solution without using double integrals or flett's theorem be found? Using high school math.

Tapi
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1 Answers1

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Let's have some names for the anti-derivatives of $f(x)$ and $xf(x)$.

  • $\displaystyle F(x)=\int_0^x\int_0^t f(u)\,du\,dt\quad$ thus $F''=f$ and by hypothesis $F'(a)=0$
  • $\displaystyle g(x)=\int_0^x tf(t)\,dt$

We also have trivially that $F(0)=F'(0)=g(0)=0$ since the lower bound of all the integrals involved is zero.

Applying integration by parts to $g$ we get:

$\displaystyle g(x)=\int_0^x tF''(t)\,dt=\bigg[tF'(t)\bigg]_0^x-\int_0^x F'(t)\,dt=xF'(x)-F(x)$


Now remark that $F'(0)=F'(a)=0$ so by application of Flett's theorem there exists $c\in(0,a)$ such that

$$F'(c)=\dfrac{F(c)-F(0)}{c-0}=\dfrac{F(c)}{c}$$

Therefore we have our result: $$g(c)=cF'(c)-F(c)=0$$

zwim
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