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$$\int_a^b\big[\;f(x)-\int_a^b f(\xi) d\xi\;\big]dx=0$$

Is this fact true?

I think yes, it is sufficient ti change $x$ with $\xi$.

Is it correct? Thank you

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    $\begingroup$ Notice that the LHS is $(1-b+a) \int_{a}^{b} f(x) \, dx$. So the equation does not hold if $b-a \neq 1$ and $ \int_{a}^{b} f(x) \, dx \neq 0$. $\endgroup$ – Sangchul Lee Jan 4 '18 at 9:56
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    $\begingroup$ It is not true... $\endgroup$ – Shashi Jan 4 '18 at 9:56
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No, this is wrong, try with simple functions :

$f(x)=1, a=0, b=2$ gives you $-2$

But it is true that $$\int_a^b\left(f(x)-\frac{1}{b-a}\int_a^b{f(u)du}\right)=0$$

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Using the linearity of the integral, we can deduce $$ \int_a^b\left[\;f(x)-\int_a^b f(\xi) d\xi\;\right]dx=\int_a^bf(x)~dx-\int_a^b\int_a^bf(\xi)~d\xi~dx\\=\int_a^bf(x)~dx-(b-a)\int_a^bf(\xi)~d\xi $$ So you get $$ \int_a^b\left[\;f(x)-\int_a^b f(\xi) d\xi\;\right]dx=0\Leftrightarrow \int_a^bf(x)~dx=(b-a)\int_a^bf(\xi)~d\xi $$ You see, the statement is true if either $b-a=1$ or $\int_a^bf(x)~dx=0$.

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Well, a counterexample. When $\text{f}\left(x\right):=\sin\left(x\right)$:

$$\int_\text{a}^\text{b}\left\{\sin\left(x\right)-\int_\text{a}^\text{b}\sin\left(t\right)\space\text{d}t\right\}\space\text{d}x=\left(1+\text{a}-\text{b}\right)\cdot\left(\cos\left(\text{a}\right)-\cos\left(\text{b}\right)\right)\tag1$$

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