Answer
$4 \leq \int_{1}^{3} \sqrt{x^2+3}~dx \leq 4\sqrt{3}$
Work Step by Step
On the interval $1 \leq x \leq 3$:
$2 \leq \sqrt{x^2+3} \leq 2\sqrt{3}$
Therefore, by the Property 8:
$2(3-1) \leq \int_{1}^{3} \sqrt{x^2+3}~dx \leq 2\sqrt{3}(3-1)$
$4 \leq \int_{1}^{3} \sqrt{x^2+3}~dx \leq 4\sqrt{3}$
