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

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?

• Because you've said that this number lies between $m$ and $M$, the minimum and maximum values of $f$ on $[a,b]$. – Ted Shifrin Feb 27 '19 at 21:31

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 $$\lambda = \displaystyle \frac{1}{b-a} \int_a^b f(x) \, dx.$$
Assuming $$a Let $$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}.$$ By the Intermediate Value Theorem for derivatives there exists $$c \in (a,b)$$ such that $$\frac {F(b)-F(a)}{b-a}=F'(c).$$
And $$F'=f$$ so $$F'(c)=f(c).$$
Your approach using $$m$$ and $$M$$ is also perfectly valid.