Antiderivative Supremum inequality $F(b)-F(a) \le (b-a)\sup\{f(x):x \in [a,b]\}$

Question

Hello everyone I am suppose to show the following:

Let F be the antiderivative of f, show that $$ F(b)-F(a) \le \sup\{f(x):x \in [a,b]\} $$

I figured I could use the Mean-Value-Theorem since $F'(x)=f(x)$ so $\frac{F(b)-F(a)}{b-a}=f(x)$ for some $x \in [a,b]$ but then the $(b-a)$ part gets in the way. Maybe some trick with the more general form $\frac{F(b)-F(a)}{g(b)-g(a)}=\frac{f(x)}{g'(x)}$?

I think one could solve it using some facts about the Riemann Integral and upper Darboux sums like:

$$F(b)-F(a) = \int_a^b f(x) dx \le \sum^{n-1}_{i=0} \sup_{t\in[x_i,x_{i+1}]}f(t)(x_{i+1}-x_i)$$

since it has all the parts but we are not allowed to use this yet.

Thank you in advance!

Edit

There was an error in the assignment. It was to show that:

$$ F(b)-F(a) \le (b-a)\sup\{f(x):x \in [a,b]\} $$

Which is just the Mean-Value-Theorem:

$$\frac{F(b)-F(a)}{b-a}=f(x)\le\sup\{f(x):x \in [a,b]\}$$

and done.

Answer 1

That's not true, a couterexample is $f(x)=1$. So that $F(x)=x+c$, then if $b>a+1$ $F(b)-F(a)=b-a>1=\sup_{[a,b]}f(x)$.