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Let $f:[a,b] \to \Bbb R $ be a continuous function. Prove that there's a c, $c \in (a,b)$ with the following property: $$\int_a^btf(t)dt=a\int_a^cf(t)dt+b\int_c^bf(t)dt$$ Thanks in advance.

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I will use a lemma that I hope you are familiar with. If not, you can try to prove it, it's pretty easy.

Lemma: if $f:[a,b]\to\mathbb{R}$ is continuous then there is a point $c\in (a,b)$ such that $\int_a^b f(x)dx=f(c)(b-a)$.

Alright, now I'll show how to prove your problem. Let $F(x)=\int_a^x f(t)dt$. Since $f$ is continuous we know by the fundamental theorem of calculus that $F'(x)=f(x)$. So now from integration by parts we get:

$\int_a^b tf(t)dt=tF(t)|_a^b-\int_a^b F(t)dt=bF(b)-aF(a)-\int_a^b F(t)dt=bF(b)-F(c)(b-a)=$

$aF(c)+b(F(b)-F(c))=a\int_a^c f(t)dt+b\int_c^b f(t)dt$

I used the lemma I stated and also the trivial facts that $F(a)=0$ and $F(b)-F(c)=\int_c^b f(t)dt$.

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We know that $$a\int_a^bf(t)dt=\int_a^baf(t)dt < \int_a^btf(t)dt < \int_a^bbf(t)dt = b\int_a^bf(t)dt.$$

Let $F(x)= a\int_a^xf(t)dt+b\int_x^bf(t)dt$ for $x\in [a,b]$. Note that $F(a)=b\int_a^bf(t)dt$, $F(b)=a\int_a^bf(t)dt$, and $F(x)$ is continuous. Therefore, by IVT, there exists some $c$ such that $a<c<b$ and $F(c)=\int_a^btf(t)dt.$

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