Answer
$2 \leq \int_{-1}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$
Work Step by Step
Note the following:
$\int_{-1}^{1} \sqrt{1+x^2}~dx = \int_{-1}^{0} \sqrt{1+x^2}~dx+\int_{0}^{1} \sqrt{1+x^2}~dx = 2\int_{0}^{1} \sqrt{1+x^2}~dx$
On the interval $0 \leq x \leq 1$:
$0 \leq x^2 \leq 1$
$1 \leq 1+x^2 \leq 2$
$\sqrt{1} \leq \sqrt{1+x^2} \leq \sqrt{2}$
$1 \leq \sqrt{1+x^2} \leq \sqrt{2}$
Therefore, by the Comparison Property 8:
$1 \leq \int_{0}^{1} \sqrt{1+x^2}~dx \leq \sqrt{2}$
$2 \leq 2\int_{0}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$
$2 \leq \int_{-1}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$
