# Thread: Integral inequality

1. ## Integral inequality

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

2. Originally Posted by Coda202
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
There are not restrictions on a and b?

and is this your inequality?

$\displaystyle \frac{1}{3\sqrt{2}}\leqslant\int_a^{b}\frac{x^2}{\ sqrt{1+x^2}}\leqslant\frac{1}{3}$

3. ## Integral Inequality

Ahh I do apologize; [a,b] = [0,1], and yes that is the inequality which I meant.

4. 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 $\displaystyle \int_0^1 \frac{x^2}{2}~ dx \leq \int_0^1 \frac{x^2}{\sqrt{1+x^2}}~ dx \leq \int_0^1 x^2~ dx$

$\displaystyle \frac 16 \leq \int_0^1 \frac{x^2}{\sqrt{1+x^2}}~ dx \leq \frac 13$