# Proving the integral inequality $2≤\int_{-1}^1 \sqrt{1+x^6} \,dx ≤ 2\sqrt{2}$

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!

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.

• Oh, I get what you mean. Thanks a lot! Aug 7, 2020 at 16:34

$$\int_{-1}^1 \sqrt{1+x^6}dx=2\int_{0}^1 \sqrt{1+x^6} dx$$ 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 : $$\int_{0}^1\sqrt{1+x^6}dx\leq \int_{0}^1\sqrt2dx=\sqrt2$$ [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.

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

• +1. The picture is a very good hint. (Can we change in Desmos $1.414$ to $\sqrt{2}$?)
– user9464
Aug 8, 2020 at 15:42
• @T.S, unfortunately we can't. I'm using Geogebra mobile, where the captions are not supported. If you want, you may add. Aug 8, 2020 at 15:55
• I'm OK with that. Anyone who understand the hint should be able to identify that $1.414$ is supposed to be $\sqrt{2}$.
– user9464
Aug 8, 2020 at 17:47

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.