# Show $f'(\beta) \int\limits_\beta^\alpha 1 dt \leq \int\limits_\beta^\alpha f'(t) dt \leq f'(\alpha) \int\limits_\beta^\alpha 1 dt$

Assuming $f(x)$ is a function of single variable, and $f'(x)$ is monotonically increasing

Then claim: $f'(\beta) \int\limits_\beta^\alpha 1 dt \leq \int\limits_\beta^\alpha f'(t) dt \leq f'(\alpha) \int\limits_\beta^\alpha 1 dt$

How do you show this is true?

Attempt:

$\int\limits_\beta^\alpha f'(t) dt \leq f'(\alpha) \int\limits_\beta^\alpha 1 dt$ is equivalent to $f(\alpha) - f(\beta) \leq f'(\alpha)(\alpha - \beta)$

Which theorem guarantees that this is true?

• That is not true unless you mean that $f'$ is monotonically increasing. A simple counter-example is $f(x) = \sqrt{x}$ on $[1,2]$. btw the mean value theorem is the theorem you are looking for (and to use it to show what you want you need $f'(x) \leq f'(\alpha)$ for all $\beta \leq x \leq \alpha$). Nov 7, 2015 at 18:00

I suggest not to use the mean value theorem. Also, the derivatives are just confusing (and so is the interval $[\beta,\alpha]$), so let me reformulate the statement as:
Given $g$ monotonically increasing on $[a,b]$, show that $$g(a)\int_a^b1\,dt\leq \int_a^b g(t)\,dt\leq g(b)\int_a^b1\,dt.$$
Let us look at the left inequality. Since $g(t)\geq g(a)$ for all $t\in[a,b]$, $g(t)-g(a)\geq 0$, and the integral of a non-negative function is non-negative (this follows by definition, since no undersums/Riemann sums can give negative contribution), i.e. $$\int_a^b g(t)-g(a)\,dt\geq 0.$$ Next, by linearity of the integral, the above is equivalent to $$\int_a^b g(t)\,dt-g(a)\int_a^b 1\,dt\geq 0.$$ Moving the things around, we get $$g(a)\int_a^b 1\,dt\leq \int_a^b g(t)\,dt.$$ I leave it to you to show the second inequality in the same manners.