Prove that:
1/(3 square root(2)) less than or equal to integral from a to b of x^2/(square root (1 + x^2)) dx less than or equal to 1/3
Yop,
$\displaystyle \forall x \in [0,1],~ 1+x^2 \geq 1 \implies \frac{1}{\sqrt{1+x^2}} \leq 1$
And $\displaystyle \forall x \in [0,1],~ 1+x^2 \leq 1+1^2=2 \implies \frac 12 \leq \frac{1}{\sqrt{1+x^2}}$
Hence $\displaystyle \frac{x^2}{2} \leq \frac{x^2}{\sqrt{1+x^2}} \leq x^2$
Therefore $\displaystyle \int_0^1 \frac{x^2}{2}~ dx \leq \int_0^1 \frac{x^2}{\sqrt{1+x^2}}~ dx \leq \int_0^1 x^2~ dx$
$\displaystyle \frac 16 \leq \int_0^1 \frac{x^2}{\sqrt{1+x^2}}~ dx \leq \frac 13$