Proving Mean Value Theorem for Integrals and I am struggling to see how the intermediate value theorem shows $f(c)=\frac{1}{b-a}\int_{a}^{b}f$??

Question

I have so far written the following proof in terms I understand for the Mean Value Theorem for Integrals:

'If $f$ is continuous on an interval $I = [a,b]$, then for some $c \in I$ we have: $$\int_{a}^{b}f(x) \cdot dx = f(c)(b-a)$$ Let $m$ and $M$ denote the minimum and maximum values of $f$ on $I$. Then: $m \leq f(x) \leq M$ for all $x \in [a,b]$. Integrating the inequalities we should find that: $$\int_{a}^{b}m \leq \int_{a}^{b}f(x) \leq \int_{a}^{b}M$$ $$m(b-a) \leq \int_{a}^{b}f(x) \leq M(b-a)$$ $$m \leq \frac{1}{b-a}\int_{a}^{b}f(x) \leq M$$

However, the book I am using now says: 'label $$\frac{1}{b-a}\int_{a}^{b}f(x)$$ as $A(f)$. But now the intermediate value theorem says $A(f)=f(c)$ for some $c \in I$. I am confused how the intermediate value theorem shows this and was wondering if anyone could help me understand or clarify it?

Answer 1

There exists a point $x \in [a,b]$ satisfying $f(x) = m$. There exists a point $y \in [a,b]$ satisfying $f(y) = M$. For every value $\lambda \in (m,M)$ there exists a point $c$ in between $x$ and $y$ satisfying $f(c) = \lambda$.

Take $\lambda = \displaystyle \frac{1}{b-a} \int_a^b f(x) \, dx.$

Answer 2

Assuming $a<b.$ Let $F(y)=\int_a^yf(x)dx.$ Since $f$ is continuous, the Fundamental Theorem of Calculus implies that $F$ is differentiable and that $F'=f.$ And (obviously) $F(a)=0.$ So $$\frac {1}{b-a}\int_a^bf(x)dx=\frac {F(b)}{b-a}=\frac {F(b)-F(a)}{b-a}.$$ By the Mean Value Theorem for derivatives there exists $c \in (a,b)$ such that $$\frac {F(b)-F(a)}{b-a}=F'(c).$$

And $F'=f$ so $F'(c)=f(c).$

I think the book should have been clearer about this.

Your approach using $m$ and $M$ is also perfectly valid.