Answer
$$\overline f \left( x \right) = 3x - 2$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 3x + 1 \cr
& {\text{The }}n{\text{ - unit moving average of a function }}f{\text{ is}} \cr
& \overline f \left( x \right) = \frac{1}{n}\int_{x - n}^x {f\left( t \right)} dt \cr
& {\text{The 2 - unit moving average of }}f\left( x \right) = 3x + 1{\text{ is}} \cr
& \overline f \left( x \right) = \frac{1}{2}\int_{x - 2}^x {\left( {3t + 1} \right)} dt \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\frac{3}{2}{t^2} + t} \right]_{x - 2}^x \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\frac{3}{2}{t^2} + t} \right]_{x - 2}^x \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\left( {\frac{3}{2}{x^2} + x} \right) - \left( {\frac{3}{2}{{\left( {x - 2} \right)}^2} + x - 2} \right)} \right] \cr
& {\text{Simplifying}} \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\frac{3}{2}{x^2} + x - \left( {\frac{3}{2}\left( {{x^2} - 4x + 4} \right) + x - 2} \right)} \right] \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\frac{3}{2}{x^2} + x - \left( {\frac{3}{2}{x^2} - 6x + 6 + x - 2} \right)} \right] \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\frac{3}{2}{x^2} + x - \left( {\frac{3}{2}{x^2} - 5x + 4} \right)} \right] \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {\frac{3}{2}{x^2} + x - \frac{3}{2}{x^2} + 5x - 4} \right] \cr
& \overline f \left( x \right) = \frac{1}{2}\left[ {6x - 4} \right] \cr
& \overline f \left( x \right) = 3x - 2 \cr} $$
