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be f a integrable function in [a,b], proof that exists c in (a,b) such that $\int_a^cf(t)dt=\int_c ^bf(t)dt$.

I think that use the Fundamental theorem of calculus can help to proof that. Whath I did is this: $\int_a^bf(t)dt=F(b)-F(a)$ for FCT

$\int_a^bf(t)dt=\int_a^cf(t)dt+\int_c^bf(t)dt$ for linearity property.

But I don't know how to conclude the proof using that.

JimmyK4542
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    As written, the statement clearly is not true. Take $f(x)=1$, $a=0$, $b=1$. – Clayton Jun 12 '15 at 02:15
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    If $f$ is positive and continuous on [a,c] and your result is true, then $f(x)=0$ for all $x$ between $a$ and $c$. This means that the result is not true in general! Check its statement. – Idris Addou Jun 12 '15 at 02:36
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    You probably mean $c$ instead of $b$ in the first integral of the equality. – Aloizio Macedo Jun 12 '15 at 02:52

1 Answers1

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(Assuming that you want a $c$ such that $\int_a^cf=\int_c^b f$)

Define $F(x)=\int_a^x f-\int_x^bf$. Since $f$ is integrable, $F$ is continuous and well-defined.

Note that $F(a)=-\int_a^bf$, $F(b)=\int_a^bf$. It follows from the intermediate value theorem that there must be a $c$ such that $F(c)=0$, which is what you ask for.

Aloizio Macedo
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