Use the MVT for Integrals to bound $\int_0^1\frac{x^6}{\sqrt{1+x^2}}dx$

Question

I have an exercise to use the mean value theorem for integrals to show that

$$\frac{1}{7\sqrt 2}\le\int_0^1\frac{x^6}{\sqrt{1+x^2}}\ dx\le\frac{1}7$$

I've determined that the integrand is increasing on the given interval and therefore minimized at $x=0$ and maximized at $x=1$.

By the integral MVT there exists some point $c\in [0,1]$ such that

$$\int_0^1\frac{x^6}{\sqrt{1+x^2}}\ dx = f(c)(1-0) = f(c)$$

Since $0\le f(c)\le 1/\sqrt 2$ I get a bound but not as tight as the one requested.

But at this point I'm not sure where a factor of 7 comes from. I could guess that I should start actually integrating some stuff but then that seems like I'm not using the integral MVT as instructed.

If I tried doing just a little, I could set $u=1+x^2$ so that the integral becomes

$$\int_1^{2}\frac{x^6}{u^{1/2}} \left(\frac{du}{2x^2}\right) = \frac 1 2 \int_1^2 \frac{x^4\ du}{u^{1/2}} = \frac 1 2 \int_1^2\frac{(u-1)^2\ du}{u^{1/2}}$$

But like ... now I'm just computing the integral.

Answer 1

Let $f(x)=\frac1{\sqrt{1+x^2}}$ and $g(x)=x^6$. Then, $$\int_0^1f(x)g(x)dx=c\int_0^1g(x)dx$$ were $c\in[\min f([0,1]),\max f([0,1])]=[1/\sqrt2,1]$.

Answer 2

By MVT, we have some $c\in (0,1)$ such that

$$ \int_0^1 \frac{x^n}{\sqrt{1+x^2}} d x =\frac{1}{\sqrt{1+c^2}} \int_0^1 x^n d x =\frac{1}{(n+1) \sqrt{1+c^2}} $$ Meanwhile, $$ \begin{aligned} & \frac{1}{\sqrt{2}} \leqslant \frac{1}{\sqrt{1+c^2}} \leqslant \frac{1}{1}=1 \\ \Rightarrow & \frac{1}{(n+1)\sqrt{2}} \leqslant \frac{1}{(n+1) \sqrt{1+c^2}} \leqslant \frac{1}{n+1} \end{aligned} $$ Hence, in general,$$ \boxed{\frac{1}{(n+1) \sqrt{2}} \leqslant \int_0^1 \frac{x^n}{\sqrt{1+x^2}} d x \leqslant \frac{1}{n+1}} $$