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  1. For every $x \geq 0$ prove that $\sqrt{x+1} - \sqrt{x} = \frac{1}{2\sqrt{x+\theta(x)}}$ for $\theta(x) \in (0,1)$. Also prove that $\theta(x) \in (\frac{1}{4}, \frac{1}{2})$ with $lim_{x \rightarrow 0} \theta(x) = \frac{1}{4}$ and $lim_{x \rightarrow \infty} \theta(x) = \frac{1}{2}$

  2. Given a continuous real valued function $f$ on $[0,1]$, show that for some $a \in [0,1]$, $\int^1_0x.f(x)dx = \frac{1}{2}f(a)$.

In Q.1, the first part is the direct application of the Mean value theorem, what I don't get is how to limit $\theta(x)$ to $(\frac{1}{4}, \frac{1}{2})$. I have no clue how to approach Q.2.

Any clues or references are welcomed.

Thanks in advance.

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2 Answers 2

up vote 2 down vote accepted

(1) $$ \sqrt{x+1}-\sqrt{x}=\frac{1}{2\sqrt{x+\theta(x)}}\implies \sqrt{x+\theta(x)}=\frac{1}{2(\sqrt{x+1}-\sqrt{x})}\implies\cdots $$ (2) $$ m\le f(x)\le M\implies\int^1_0 mx\,dx\le\int^1_0 x\,f(x)\,dx\le\int^1_0 Mx\,dx $$ and use the theorem of Bolzano.

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Thanks a lot for your help. –  Morty Apr 14 at 9:00

If we apply the mean value theorem for the function $f(x)=\sqrt{x}$ on the interval $[x,x+1]$ then we have $$\sqrt{x+1} - \sqrt{x} = \frac{1}{2\sqrt{a}}=\frac{1}{2\sqrt{x+(a-x)}}$$ where $a\in(x,x+1) $ and so $(a-x)=\theta(x)\in(0,1)$.

If you take the limit as x tends to zero and infinity then you get that $lim_{x \rightarrow \infty} \theta(x) = \frac{1}{2}$ and $lim_{x \rightarrow \infty} \theta(x) = \frac{1}{2}.$

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