Answer
$\frac{3}{7} \leq \int_{0}^{3} \frac{1}{x+4}~dx \leq \frac{3}{4}$
Work Step by Step
On the interval $0 \leq x \leq 3$:
$\frac{1}{7} \leq \frac{1}{x+4} \leq \frac{1}{4}$
Therefore, by Property 8:
$\frac{1}{7}(3-0) \leq \int_{0}^{3} \frac{1}{x+4}~dx \leq \frac{1}{4}(3-0)$
$\frac{3}{7} \leq \int_{0}^{3} \frac{1}{x+4}~dx \leq \frac{3}{4}$
