# Prove $\int_a^b f(x)dx=(b-a)f\{a+\theta(b-a)\}$ for some $\theta\in(0,1)$.

Assume $f$ is Riemann integrable on $[a,b],$ and that $F'=f$ on $[a,b].$ Using Fundamental theorem of integral calculus, show that $\int_a^b f(x)dx=(b-a)f\{a+\theta(b-a)\}$ holds for some $\theta\in(0,1)$.

I am using the "Fundamental theorem of integral calculus" as

Let $f: [a,b] \to \mathbb{R}$ be a Riemann integrable function. If $F: [a,b] \to \mathbb{R}$ is an antiderivative of $f$, then $$\int_a^b \! f(x) \, \mathrm{d}x = F(b)-F(a).$$

How to use this theorem to show the above?

• It would seem to me that you need to add, that there exists $\theta$ satisfying what you wrote above. It currently seems to imply that for all $\theta \in (0,1)$. – Keen-ameteur Mar 11 '18 at 15:29
• @Keen-ameteur Edited, thanks, – user1942348 Mar 11 '18 at 15:33
• I assume you meant to ask the question as in my answer. The way you have it is confusing. Please restate. – zhw. Mar 11 '18 at 18:52
• @zhw. Restated, as you have suggested. Please check – user1942348 Mar 11 '18 at 19:04
• Much better.$\,\,$ I'm going to edit your title to make it simpler. – zhw. Mar 11 '18 at 19:21

I assume the problem was supposed to be this: Assume $f$ is Riemann integrable on $[a,b],$ and that $F'=f$ on $[a,b].$ Then the conclusion follows as stated.

Proof: Let $I = \int_a^b f.$ Define

$$G(x) = F(x)-\frac{I}{b-a}x,\,\,\,x\in [a,b].$$

Note that $G'(x) = f(x) - \dfrac{I}{b-a}$ on $[a,b].$ Also note $G(b) - G(a) = 0,$ here using the version of the FTC you stated. Thus by the mean value theorem, $G'(c) = 0$ for some $c\in (a,b).$ This implies

$$0=G'(c) = f(c)-\frac{I}{b-a} \, \implies\, I = (b-a)f(c).$$

Since $c,$ like any point in $(a,b),$ can be writen as $a+\theta(b-a)$ for some $\theta \in (0,1),$ we're done.

• G(b)−G(a)=0, here using the FTC -- please explore – user1942348 Mar 11 '18 at 18:56
• Just carry through the calculation, using $F(b)-F(a) = \int_a^bf.$ – zhw. Mar 11 '18 at 19:08
• @user1942348 I added a small bit to make it clearer. – zhw. Mar 11 '18 at 19:16

Consider the function $F$ as you mentioned. What you want to show translates by the Fundamental theorem to:

$F(b)-F(a)=(b-a) \cdot f\big(a+ \theta(b-a) \big)$

Which again rewrites as:

$\dfrac{F(b)-F(a)}{b-a}=f\big(a+ \theta(b-a) \big)$

If you know that $f$ is continuous, then $F$ is differentiable on $(a,b)$ and it's derivative is $f$ . By Lagrange's theorem there exists $c\in (a,b)$ such that:

$\dfrac{F(b)-F(a)}{b-a}= F'(c)$

But $c$ can be written as $a+\theta(b-a)$, for some $\theta \in (0,1)$.

• But we're not given $f$ is continuous. – zhw. Mar 11 '18 at 18:26
• I assumed that this assumption was ommited, since changing $f$ in a finite amount of points does not change the integral. – Keen-ameteur Mar 11 '18 at 18:33
• But it could destroy the assumption that $f$ has an antiderivative. – zhw. Mar 11 '18 at 18:50
• @Keen-ameteur f is continuous -- is it required? – user1942348 Mar 11 '18 at 19:01
• I'm not sure, in the formulation I'm familiar with it is. – Keen-ameteur Mar 11 '18 at 19:05

It sounds as though you're trying to prove this for Riemann integrable $f$. It's false in that context: Define $f:[0,2]\to\Bbb R$ by $$f(t)=\begin{cases}1&(0\le t<1),\\-1&(1\le t\le 2).\end{cases}$$Then $\int_0^2 f(t)=0$ although there is no $t$ with $f(t)=0$.

There must be another hypothesis in the problem that you haven't told us about...

• No more hyp were there other that I have mentioned. – user1942348 Mar 11 '18 at 15:37
• @user1942348 That's hard to believe. If so then the problem is simply wrong. – David C. Ullrich Mar 11 '18 at 15:38
• @user1942348 In fact there must be something you haven't stated. I was just guessing that you're given that $f$ is Riemann integrable. You did't say so. You must be given something about $f$ or the problem makes no sense at all. Exactly what are you given? – David C. Ullrich Mar 11 '18 at 15:43
• @david-c-ullrich Ok. I understand, thanks, Yes f should be R-integrable. – user1942348 Mar 11 '18 at 15:44