Prove: $\frac{1}{11\sqrt{2}} \leq \int_0^1 \frac{x^{10}}{\sqrt{1+x}}dx \leq \frac{1}{11}$ 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 $c$.
 A: Hint. One has, by a direct approach,
$$
\frac{x^{10}}{\sqrt{1+1}}\le \frac{x^{10}}{\sqrt{x+1}}\leq \frac{x^{10}}{\sqrt{0+1}}, \qquad 0\le x\le1.
$$
A: 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:



$$m \int_0^1 \frac{1}{\sqrt{x+1}} dx \le \int_0^1 \frac{x^{10}}{\sqrt{x+1}} \le M \int_0^1 \frac{1}{\sqrt{x+1}} dx$$
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 $u(x) = \frac{1}{\sqrt{x+1}}$
A: 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
$$ \color{red}{0.065}621652\leq   I \leq \color{red}{0.065}721354.$$
