I'm having difficulty with proving this inequality: $$\frac{1}{20\sqrt{2}}<\int_0^1 \frac{x^{19}}{\sqrt{1+x^2}}dx<\frac1{20}$$
I thought this task should be fairly easy to prove with the following theorem:
If $f(x)$ and $g(x)$ are Riemann-integrable and $f(x) \le g(x)$ for each $x$ in $[a,b]$ then $\int_a^b f(x)dx \le \int_a^bg(x)dx$
Especially with the specific version of the above lemma:
If $f(x)$ is Riemann-integrable then it is necessarily bounded on this interval. Thus there are real numbers m and M such that $m \le f(x) \le M$ for all $x$ in $[a,b]$. Since the lower and upper sums of $f$ over $[a,b]$ are therfore bounded by $m(b-a)$ and $M(b-a)$ we have $m(b-a)\le \int_a^b f(x)dx \le M(b-a)$
However, this bound seems to be too weak. ($\frac{x^{19}}{\sqrt{1+x^2}}<\frac1{20}$ doesn't hold for all $x$ in $[0,1]$) Any help is appreciated. Thanks.
HINT:
If $x\in[0,1],$ then $$\frac{x^{19}}{\sqrt{2}}\leq\frac{x^{19}}{\sqrt{1+x^2}}\leq x^{19}$$