Answer
$F\left( y \right) = 8\left( {{e^{y/2}} - {e^{ - 1/2}}} \right)$
Work Step by Step
$$\eqalign{
& F\left( y \right) = \int_{ - 1}^y {4{e^{x/2}}} dx \cr
& {\text{Integrate}} \cr
& F\left( y \right) = 8\left[ {{e^{x/2}}} \right]_{ - 1}^y \cr
& F\left( y \right) = 8\left( {{e^{y/2}} - {e^{ - 1/2}}} \right){\text{, }}\left( {{\text{accumulation function}}} \right) \cr
& \cr
& \left( {\text{a}} \right)F\left( { - 1} \right) \cr
& F\left( { - 1} \right) = 8\left( {{e^{ - 1/2}} - {e^{ - 1/2}}} \right) \cr
& F\left( { - 1} \right) = 0,{\text{ }}\left( {{\text{Graph shown below}}} \right) \cr
& \cr
& \left( {\text{b}} \right)F\left( 0 \right) \cr
& F\left( 0 \right) = 8\left( {{e^0} - {e^{ - 1/2}}} \right) \cr
& F\left( 0 \right) \approx 3.1477,{\text{ }}\left( {{\text{Graph shown below}}} \right) \cr
& \cr
& \left( {\text{c}} \right)F\left( 4 \right) \cr
& F\left( 4 \right) = 8\left( {{e^4} - {e^{ - 1/2}}} \right) \cr
& F\left( 4 \right) \approx 54.2602,{\text{ }}\left( {{\text{Graph shown below}}} \right) \cr
& \cr
& {\text{Graphs}} \cr} $$
