Answer
$\frac{1}{3} \leq \int_{3}^{5}\frac{1}{x+1}dx \leq \frac{1}{2}$
Work Step by Step
$3 \leq x \leq 5$
$3+1 \leq x+1 \leq 5+1$
$4 \leq x+1 \leq 6$
$\frac{1}{6} \leq \frac{1}{x+1} \leq \frac{1}{4}$
Using the property $8$ it follows:
$\frac{1}{6}(5-3) \leq \int_{3}^{5}\frac{1}{x+1}dx \leq \frac{1}{4}(5-3)$
$\frac{2}{6} \leq \int_{3}^{5}\frac{1}{x+1}dx \leq \frac{2}{4}$
$\frac{1}{3} \leq \int_{3}^{5}\frac{1}{x+1}dx \leq \frac{1}{2}$
