Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd love your help with proving that for $f:[0,1] \to \mathbb{R}$ monotonically decreasing function ,for every $\alpha \in (0,1)$ :$\int_{0}^{\alpha} f(x)dx \geq \alpha\int_{0}^{1}f(x)dx$. I tried couple of inequalities but I didn't conclude what I should.

Thanks a lot.

share|cite|improve this question
up vote 6 down vote accepted

$\int_0^{\alpha} f(x) dx = \alpha \int_0^1 f(\alpha t) d t \geq \alpha \int_0^1 f(t) d t$

First equality is integration by substitution with $x=\alpha t$, then inequality holds since $\alpha t \leq t$ and $f$ is decreasing.

share|cite|improve this answer

You can also note that: $\alpha \int_\alpha^1{f(t)dt}\leq\alpha(1-\alpha)f(\alpha)\leq (1-\alpha)\int_0^\alpha{f(t)dt}$

Rearranging the inequality follows. I think you are able to figure out why the inequalities above holds.

share|cite|improve this answer

Divide both sides by $\alpha$. The inequality becomes $${1\over \alpha} \int_0^\alpha f(x)\, dx \ge \int_0^1 f(x)\,dx. $$

$$A(\alpha) = {1\over \alpha} \int_0^\alpha f(x)\, dx\qquad \alpha\in[0,1].$$

Then is the average value of $f$ on $[0,\alpha]$. Since $f$ is decreasing, this average must decrease as a function of $\alpha$, so $A(\alpha) \ge A(1)$ for $\alpha\in[0,1]$. The inequality follows immediately.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.