Question on Mean Value Theorem & Intermediate Value Theorem

[Titan97]

Homework Statement

for $0<\alpha,\beta<2$, prove that $\int_0^4f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]$

Homework Equations

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

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

 

[HallsofIvy]

What, exactly, are you to prove? You say "for 0&lt;α,β&lt; 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.

 

[pasmith]

Homework Statement

for $0<\alpha,\beta<2$, prove that $\int_0^4f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]$

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

Homework Equations

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

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,
$$h(k)=\frac{h(\alpha)+h(\beta)}{2}=\alpha f(\alpha)+\beta f(\beta)$$

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.

 

[Titan97]

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