5
$\begingroup$

I am trying to prove that $$2≤\int_{-1}^1 \sqrt{1+x^6} \,dx ≤ 2\sqrt{2} $$ I learned that the equation $${d\over dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))h'(x) - f(g(x))g'(x) $$ is true due to Fundamental Theorem of Calculus and Chain Rule, and I was thinking about taking the derivative to all side of the inequality, but I am not sure that it is the correct way to prove this. Can I ask for a help to prove the inequality correctly? Any help would be appreciated! Thanks!

$\endgroup$
10
$\begingroup$

So there are two inequalities to be proved. You can use that $\sqrt{1+x^6} \leq \sqrt{2}$ for all $x \in [-1,1]$ for the upper bound, as it follows $\int_{[-1,1]} \sqrt{1+x^6} dx\leq \int_{[-1,1]} \sqrt{2} dx\leq 2 \sqrt{2}$. The lower bound follows very similarly.

$\endgroup$
1
  • $\begingroup$ Oh, I get what you mean. Thanks a lot! $\endgroup$
    – sugaarrrr
    Aug 7 '20 at 16:34
8
$\begingroup$

$ \int_{-1}^1 \sqrt{1+x^6}dx=2\int_{0}^1 \sqrt{1+x^6} dx$ because $\sqrt{1+x^6}$ is an even function. so we must show: $$2≤2\int_{0}^1 \sqrt{1+x^6} dx ≤ 2\sqrt{2}$$ or we must show: $$1≤\int_{0}^1 \sqrt{1+x^6} dx ≤ \sqrt{2}$$

$1≤\sqrt{1+x^6}$ then $\int_{0}^11dx\leq\int_{0}^1 \sqrt{1+x^6} dx$ we have $1≤\int_{0}^1 \sqrt{1+x^6} dx$ $$$$

we have $1+x^6\leq2$ if $0\leq x \leq1$ and then we have $\sqrt{1+x^6}\leq \sqrt2$ if $0\leq x \leq1$ therefore : $$\int_{0}^1\sqrt{1+x^6}dx\leq \int_{0}^1\sqrt2dx=\sqrt2$$

$\endgroup$
7
$\begingroup$

enter image description here

[Edited after the meta question and this conversation.]

Let $f(x)=\sqrt{1+x^6}$. It is evident that the argument of the square root $1+x^6\ge 1\ \forall\ x\in\mathbb R$ and there is only one point of global minima $(0, 1) $. Hence, $f(x)$ is monotically decreasing and increasing for $x\le 0$ and $x\ge 0$ respectively.

$$\text{Area}(\square CEFD)\le \text{Area under curve} \le \text{Area}(\square ABDC) \\ \implies 2\le \int_{-1}^{1} \sqrt{1+x^6}\ dx\le 2\sqrt 2$$

$\endgroup$
3
  • 2
    $\begingroup$ +1. The picture is a very good hint. (Can we change in Desmos $1.414$ to $\sqrt{2}$?) $\endgroup$
    – user9464
    Aug 8 '20 at 15:42
  • $\begingroup$ @T.S, unfortunately we can't. I'm using Geogebra mobile, where the captions are not supported. If you want, you may add. $\endgroup$
    – SarGe
    Aug 8 '20 at 15:55
  • $\begingroup$ I'm OK with that. Anyone who understand the hint should be able to identify that $1.414$ is supposed to be $\sqrt{2}$. $\endgroup$
    – user9464
    Aug 8 '20 at 17:47
1
$\begingroup$

Surprised not to see this technique yet.

For $|z|< 1$ (the most conservative convergence case), the generalized binomial theorem states $$ (1+z)^a = \sum_{k=0}^{\infty} \binom{a}{k}z^{k} $$In particular, for $z=x^6$ and $a=1/2$, we get convergence for $|x|\le 1$, $\binom{1/2}{k}$ is well-known, and we have $$ (1+x^6)^{1/2} = \sum_{k=0}^{\infty} \binom{1/2}{k}x^{6k} = 1 + \frac{1}{2}x^6 - \frac{1}{8}x^{12}+\frac{1}{16}x^{18}-\frac{5}{128}x^{24}+\cdots $$Now we can power-rule our way out of it: $$ \frac{773}{364}=\int _{-1}^{1}1 + \frac{1}{2}x^6 - \frac{1}{8}x^{12}\,dx \le \int _{-1}^{1} \sqrt{1+x^6}\,dx \le \int _{-1}^{1} 1 + \frac{1}{2}x^6\,dx = \frac{15}{7} $$The inequalities follow because the series is alternating, so if we end on a positive term we overestimate it and if we end on a negative term we underestimate it. The inequalities could be improved by adding more terms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.