Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function on $[a,b]$. Prove that there exists $c \in (a,b)$ such that $\frac{1}{b-a}\int_a^b d(s)\,ds = f(c)$.
I understand there is an easy way of doing this, I just want to check if my working to a more complicated method is correct.
$f$ continuous on $[a,b] \implies $ Riemann Integrable on $[a,b]$ $\implies $ makes sense to write $\int_{a}^{b}f(s) ds$
Define $A(x)=\int_{a}^{x}f(s) ds$ then by the Fundamental Theorem of Calculus $A$ is differentiable on $(a,b)$ and $A'(x)=f(x)$ ($f$ is continuous on $[a,b]$)
$A$ differentiable on $(a,b) \implies A$ continuous on $(a,b)$
$\lim\limits_{x \to a+} A(x)=0= A(a) \implies A$ is right continuous at $a$
$\lim\limits_{x \to b-} A(x)=\int_{a}^{b}f(s) ds$ (which exists as $f$ is Riemann Integrable on $[a,b])$$
= A(b) \implies A$ is left continuous at $b$
So $A$ continuous on $[a,b]$ and differentiable on $(a,b)$
By Mean Value Theorem $\exists c \in(a,b)$ st.
$\frac{A(b)-A(a)}{b-a}=A'(c)=f(c) $ (F.T.C) $\implies$
$\frac{\int_{a}^{b}f(s) ds}{b-a}=f(c)$
Best Answer
$f$ is a continuous function on the compact $[a,b]$ so it's bounded and then there's $m,M$ such that
$$m\le f(x)\le M,\quad\forall x\in[a,b]$$ so
$$m=\frac1{b-a}\int_a^b mds\le \frac1{b-a}\int_a^b f(s)ds\le \frac1{b-a}\int_a^b Mds=M$$ and the result follows by applying the intermediate value theorem.