# Question on Mean Value Theorem & Intermediate Value Theorem

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1. Aug 18, 2015

### Titan97

1. The problem statement, all variables and given/known data
for $0<\alpha,\beta<2$, prove that $\int_0^4f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]$

2. Relevant equations
Mean value theorem: $f'(c)=\frac{f(b)-f(a)}{b-a}$

3. The attempt at a solution
I got the answer for the question but I have made an assumption but I don't know if it's correct.

Let $g(x)=\int_0^{x^2}f(t)dt$ and let $h(x)=g'(x)=2xf(x^2)$

now, applying intermediate value theorem for $h(x)$ in $x\in (\alpha,\beta)$, there exists a point $x=k$ such that,
$$h(k)=\frac{h(\alpha)+h(\beta)}{2}=\alpha f(\alpha)+\beta f(\beta)$$
$$g'(k)=\alpha f(\alpha)+\beta f(\beta)$$
assumption: Now, for g(x), if $x=k$ also satisfies Mean value theorem and since $0<k<2$,

$$g'(k)=\frac{f(2)-f(0)}{2}=\frac{1}{2}\int_0^{4}f(t)dt=\alpha f(\alpha)+\beta f(\beta)$$

Hence, $$\int_0^{4}f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]$$

Last edited by a moderator: Aug 19, 2015
2. Aug 19, 2015

### HallsofIvy

Staff Emeritus
What, exactly, are you to prove? You say "for $0<α,β< 2$ $\int_0^4 f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]$"
That reads "for all $\alpha$ and $\beta$ strictly between 0 and 2, and for any integrable function, f" this equality is true.

But that is clearly NOT true. In your "proof" you appear to determine that there exist numbers $\alpha$ and $\beta$ that work.

3. Aug 19, 2015

### pasmith

What exactly is the question? To prove that there exist $0 < \alpha < 2$ and $0 < \beta < 2$ such that $$\int_0^4 f(t)\,dt = 2(\alpha f(\alpha) + \beta f(\beta))?$$ Are there any conditions on $f$?

Unfortunately
$$\frac{h(\alpha) + h(\beta)}{2} = \alpha f(\alpha^2) + \beta f(\beta^2)$$ which doesn't help you.

EDIT: It occurs to me that $\alpha f(\alpha) + \beta f(\beta)$ doesn't involve the value of $f$ on $(2, 4]$ at all, which seems suspicious.

Last edited: Aug 19, 2015
4. Aug 21, 2015

### Titan97

I am very sorry. My teacher gave me a wrong question.