<s>[INST] Prove that $\frac{1}{7\sqrt{2}} \leq \int_{0}^1 \frac{x^6}{\sqrt{1+x^2}}dx \leq \frac{1}{7}$ 
Prove that $\displaystyle \dfrac{1}{7\sqrt{2}} \leq \int_{0}^1 \dfrac{x^6}{\sqrt{1+x^2}}dx \leq \dfrac{1}{7}$.

My book says let $f(x) = \dfrac{1}{\sqrt{1+x^2}}$ and $g(x) = x^6$. Then $\displaystyle \int_0^1 \dfrac{x^6}{\sqrt{1+x^2}}dx = \dfrac{1}{\sqrt{1+\xi^2}} \int_{0}^1 x^6 dx$ where $0 \leq \xi \leq 1$. Thus $$\dfrac{1}{7\sqrt{2}} \leq \dfrac{1}{\sqrt{1+\xi^2}} \leq \dfrac{1}{7}.$$
I don't understand how they are getting all these results. The problem did say to use other results in the book, but I didn't see any that related. For reference it is in chapter 13 of Michael Spivak's calculus book.
 [/INST] We have $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1$$ for all $x \in [0,1]$ since $0 \le x^2 \le 1$. Indeed, then $1 \le 1+x^2 \le 2$ and so $$\frac{1}{2} \le \frac{1}{1+x^2} \le 1.$$ Thus $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1.$$
Then just multiply by $x^6$ and integrate.
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