Answer
$2 \leq \int_{-1}^{1}\sqrt {1+x^{4}}dx \leq 2\sqrt 2$
Work Step by Step
$$-1 \leq x \leq 1$$
$$0 \leq x^{2} \leq 1$$
$$0^{2} \leq (x^{2})^{2} \leq 1^{2}$$
$$0 \leq x^{4} \leq 1$$
$$1 \leq 1+x^{4} \leq 2$$
$$\sqrt 1 \leq \sqrt {1+x^{4}} \leq \sqrt 2$$
$$1 \leq \sqrt {1+x^{4}} \leq \sqrt 2$$
Using the property $8$ it follows:
$$1(1-(-1)) \leq \int_{-1}^{1}\sqrt {1+x^{4}}dx \leq \sqrt 2(1-(-1))$$
$$2 \leq \int_{-1}^{1}\sqrt {1+x^{4}}dx \leq 2\sqrt 2$$
