## Calculus (3rd Edition)

$$\frac{1}{3} \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{2}$$
Since $f(x)=\dfrac{1}{x}$ has a maximum value on $[4,6]$, then $M=f(4)=\frac{1}{4}$; also, $f(x)$ has a minimum value on $[4,6]$, so $m=f(6)=\frac{1}{6}$: $$m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a)$$ Hence, for $a=4,\ b=6$ \begin{aligned} \frac{1}{6}(6-4) & \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{4}(6-4) \\ \frac{1}{3} & \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{2} \end{aligned}