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It's a question from a past calculus exam in my University which I can't formulate a proof to. Will be happy for a help!

Let $h:[a,b] \to \mathbb{R} $ a monotonic function, so there exists $\xi\in[a,b]$ such that:

$$\int_{a}^{b}h(t)dt = h(a)(\xi-a)+h(b)(b-\xi)$$

I tried defining a function $g(x) = \int_{a}^{x}h(t)dt+\int_{x}^{b}h(t)dt$ and trying to apply mean value theorem to it but that doesn't seems to take me nowhere. Also, besides the fact that if $h$ is monotonic on $[a,b]$ it's integrable there I didn't use that fact.

I know I should probably use integral mean value thm but I think I'm missing intuition as per how to use it.

Thanks for your help.

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Hint: Consider $f:\xi \mapsto h(a)(\xi - a) + h(b)(b-\xi)$. It is continuous, $f(a) = h(b)(b-a)$ and $f(b) = h(a)(b-a)$ ... –  martini May 18 '12 at 9:44

1 Answer 1

up vote 2 down vote accepted

well, as $\xi$ ranges from $a$ to $b$, $h(a)(\xi-a)+h(b)(b-\xi)$ ranges from $h(b)(b-a)$ to $h(a)(b-a)$, and these two values are, since $h$ is monotonic, upper and lower bounds for the true integral.

Since $h(a)(\xi-a)+h(b)(b-\xi)$ is continuous w.r.t. $\xi$, you can use the mean value theorem to conclude that it has to hit the right value somewhere inbetween.

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nice, thank you! –  Amihai Zivan May 18 '12 at 10:51

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