Answer
$4 \leq \int_{1}^{3}{\sqrt {x^{2}+3}}~dx \leq4 \sqrt{3}$
Work Step by Step
$1 \leq x \leq 3$
$1 \leq x^{2} \leq 9$
$4 \leq x^{2}+3 \leq12$
$2 \leq \sqrt {x^{2}+3} \leq2 \sqrt{3}$
So using the property $8$ it follows:
$2(3-1) \leq \int_{1}^{3}\sqrt {x^{2}+3}dx \leq2 \sqrt{3}(3-1)$
$4 \leq \int_{1}^{3}{\sqrt {x^{2}+3}}~dx \leq4 \sqrt{3}$
