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I am working out a homework problem about Riemann Integrals and the question is as follows:

Suppose that $f$ is integrable on $[a, b]$, then $\exists \ x \in [a, b] s.t. \int_a^{x}f=\int_x^{b}f$. Is it always possible to choose $x$ to be in $(a, b)$?

I have managed to prove the first part and now I am attempting the second part of the question.

This is my reasoning:

Let $f$ be a function such that choosing $x=a$ means $\int_a^{a}f=\int_a^{b}f$.

Now $\int_a^{a}f=0 \implies \int_a^{b}f=0$ and for this to be true, $f$ must be a function defined at only one point, ie $a=b$, which brings me to my question: is a function defined at a only one point Riemann Integrable and is the rest of my reasoning correct?

Dylan Zammit
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  • Well, $\int_a^b f=0$ does not mean that the function is defined only at one point. For example $$\int_{0}^{2 \pi} \sin x \ \mathrm d x = 0$$ and $\sin x$ is defined on the interval $[0, 2 \pi]$. – Crostul Mar 30 '18 at 09:03
  • But I have chosen $x$ in such a way that $\int_a^xf=\int_x^bf \implies \int_a^af=\int_a^bf$ since $x=a$ – Dylan Zammit Mar 30 '18 at 09:04

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Yes, provided that $\int_a^b f\ne 0$.

For example, if $f(x)=x$, and $[a,b]=[-1,1]$, then $\int_{-1}^x t\,dt=\frac{1}{2}(x^2-1)$, while $\int_x^1 t\,dt=\frac{1}{2}(1-x^2)$ and $$ \int_{-1}^x t\,dt=\int_x^1 t\,dt\quad\Longrightarrow\quad x=\pm 1, $$ and hence $x\not\in (-1,1)$.

Yiorgos S. Smyrlis
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