Answer
$\frac{3}{7} \leq \int_{0}^{3}\frac{1}{x+4}dx \leq \frac{3}{4}$
Work Step by Step
$0 \leq x \leq 3$
$0+4 \leq x+4 \leq 3+4$
$4 \leq x+4 \leq 7$
$\frac{1}{7} \leq \frac{1}{x+4} \leq \frac{1}{4}$
Using the property $8$ it follows:
$$\frac{1}{7}(3-0) \leq \int_{0}^{3}\frac{1}{x+4}dx \leq \frac{1}{4}(3-0)$$
$$\frac{3}{7} \leq \int_{0}^{3}\frac{1}{x+4}dx \leq \frac{3}{4}$$
