Proving a Definite-Integral Lies Between Two Values

Question

I'm trying to solve the following inequation but not sure what I'm supposed to do, I know it's an integral of an even function over a symmetric interval, so I can double the area over the positive x-axis. but I don't know if that helps me here.

It seems like I'm not supposed to actually calculate the integral but just prove it's in between those values.

not asking for an answer here, just a direction please.

$$ \frac{2}{3} \leq \int_{-1}^{1} \frac{1}{{2x^4+1}} \ dx \leq 2$$

Answer 1

$-1≤x≤1$

$0≤x^4≤1$

$0≤2x^4≤2$

$1≤2x^4+1≤3$

$\frac{1}{3}≤\frac{1}{2x^4+1}≤1$

$\int_{-1}^1 \frac{1}{3}≤ \int_{-1}^1 \frac{1}{2x^4+1}≤ \int_{-1}^1 1$

$ \frac{2}{3}≤ \int_{-1}^1 \frac{1}{2x^4+1}≤ 2$

Answer 2

Since the function $f(x) =2x^4+1$ is continuous, it has a maximum of $3$ and a minimum of $1$ in $[-1,1]$, so $$\frac{1}{3}\leq\frac{1}{2x^4+1}\leq 1,$$ in $[-1, 1]$, what can you conclude from here?

Answer 3

The previous answers totally address the issue. I'll just add that if you consider upper and lower bounds for the integrand function that are not constant, you may obtain tighter bounds for the integral. For instance, $$ \int_{-1}^1 \frac{dx}{2x^4+1} \geq 2 \int_0^1\dfrac{dx}{1+2x^2}=\frac{2}{\sqrt{2}}\left[\arctan(\sqrt{2} x) \right]_0^1 = \sqrt{2} \arctan \sqrt{2} > 1.351 $$