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Prove: $\frac{1}{11\sqrt{2}} \leq \int_0^1 \frac{x^{10}}{\sqrt{1+x}}dx \leq \frac{1}{11}$

Hint: Use the (weighted) Mean Value Theorem for Integrals.

The MVT for Integrals:

Suppose that $u$ is continuous and $v$ is integrable and nonnegative on $[a,b]$

Then $\int_a^b u(x)v(x)dx=u(c)\int_b^a v(x)dx$

for some $c$ in $[a,b]$.

I plan on using $u(x)$ as $x^{10}$ as it is continuous and $v(x)$ as $\frac{1}{\sqrt{1+x}}$ as it is integrable on [0,1]. I'm not sure how to go from there and find $c$.

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  • $\begingroup$ Second mean value? $\endgroup$ – BCLC Nov 6 '16 at 11:39
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    $\begingroup$ First mean value @BCLC $\endgroup$ – angelo086 Nov 6 '16 at 11:41
  • $\begingroup$ If you just swap your choice of $u$ and $v$, you will find that everything falls out nicely. $\endgroup$ – TonyK Nov 6 '16 at 13:23
  • $\begingroup$ @TonyK What's the direct relevance of MVT anyway? I think what's relevant is the inequality in the proof of MVT $\endgroup$ – BCLC Nov 6 '16 at 14:04
  • $\begingroup$ @angelo086 Oh I didn't know/forgot about that MVT. Interesting $\endgroup$ – BCLC Nov 6 '16 at 14:05
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Hint. One has, by a direct approach, $$ \frac{x^{10}}{\sqrt{1+1}}\le \frac{x^{10}}{\sqrt{x+1}}\leq \frac{x^{10}}{\sqrt{0+1}}, \qquad 0\le x\le1. $$

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    $\begingroup$ Short and elegant! (+1) $\endgroup$ – Brian Cheung Nov 6 '16 at 11:41
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Choosing $v(x) = \frac{1}{\sqrt{x+1}}$ doesn't work, I think:

I didn't use MVT directly. I used something from the proof of MVT:


enter image description here


$$m \int_0^1 \frac{1}{\sqrt{x+1}} dx \le \int_0^1 \frac{x^{10}}{\sqrt{x+1}} \le M \int_0^1 \frac{1}{\sqrt{x+1}} dx$$

The best $m$ and $M$ I got are $0$ and $1$ resp as $0 \le x^{10} \le 1$ for $0 \le x \le 1$


So let's try

$u(x) = \frac{1}{\sqrt{x+1}}$

Now $\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{x+1}} \le 1$ for $0 \le x \le 1$

So we have

$$\frac{1}{\sqrt{2}} \int_0^1 x^{10} dx \le \int_0^1 \frac{x^{10}}{\sqrt{x+1}} \le 1 \int_0^1 x^{10} dx$$


Remark: I think the 11's in the denominator may be a hint to use $u(x) = \frac{1}{\sqrt{x+1}}$

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By the Cauchy-Schwarz inequality $$ I=\int_{0}^{1}\frac{x^{10}}{\sqrt{1+x}}\,dx\leq \sqrt{\int_{0}^{1}x^{10}\,dx\int_{0}^{1}\frac{x^{10}\,dx}{1+x}}=\sqrt{\frac{1}{11}\left(\log2-\frac{1627}{2520}\right)}$$ and by integration by parts $$ I = \frac{1}{11 \sqrt{2}}+\frac{1}{22}\int_{0}^{1}\frac{x^{11}\,dx}{\sqrt{(1+x)^3}}\geq\frac{1}{11\sqrt{2}}+\frac{1}{528\sqrt{2}}$$ so we have the much better inequality $$ \color{red}{0.065}621652\leq I \leq \color{red}{0.065}721354.$$

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