Answer
$$-\frac{35}{2}$$
Work Step by Step
Since
$$\int_{1}^{b} f(x) d x =1-b^{-1} $$
Then
\begin{align*}
\int_{1}^{6}(3 f(x)-4) d x&=3 \int_{1}^{6} f(x)-\int_{1}^{6} 4 d x\\
&=3\left(1-6^{-1}\right)-4(6-1)\\
&=3\left(1-\frac{1}{6}\right)-4(5)\\
&=3\left(\frac{5}{6}\right)-20\\
&=\frac{5}{2}-20\\
&=-\frac{35}{2}
\end{align*}
