Answer
$$\frac{1}{3} \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{2}$$
Work Step by Step
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}
