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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!

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4 Answers 4

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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.

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  • $\begingroup$ Oh, I get what you mean. Thanks a lot! $\endgroup$
    – sugaarrrr
    Commented Aug 7, 2020 at 16:34
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$ \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 : $$\int_{0}^1\sqrt{1+x^6}dx\leq \int_{0}^1\sqrt2dx=\sqrt2$$

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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.

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

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    $\begingroup$ +1. The picture is a very good hint. (Can we change in Desmos $1.414$ to $\sqrt{2}$?) $\endgroup$
    – user9464
    Commented Aug 8, 2020 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
    Commented Aug 8, 2020 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
    Commented Aug 8, 2020 at 17:47
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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.

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