#### Answer

$$L = 5\sqrt {10} $$

#### Work Step by Step

$$\eqalign{
& y = 4 - 3x{\text{ on }}\left[ { - 3,2} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = 4 - 3x{\text{ and }}\left[ { - 3,2} \right] \to a = - 3{\text{ and }}b = 2.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {4 - 3x} \right] \cr
& f'\left( x \right) = - 3 \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_{ - 3}^2 {\sqrt {1 + {{\left( { - 3} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_{ - 3}^2 {\sqrt {10} } dx \cr
& {\text{integrate}} \cr
& L = \sqrt {10} \left( x \right)_{ - 3}^2 \cr
& L = \sqrt {10} \left( {2 - \left( { - 3} \right)} \right) \cr
& L = \sqrt {10} \left( 5 \right) \cr
& L = 5\sqrt {10} \cr} $$