Answer
$$F\left( x \right) = \frac{1}{2}{x^2} + 3x + 4$$
Work Step by Step
$$\eqalign{
& F''\left( x \right) = 1,{\text{ }}F'\left( 0 \right) = 3,{\text{ }}F\left( 0 \right) = 4,{\text{ }} \cr
& F'\left( x \right) = \int {F''\left( x \right)} dx \cr
& {\text{Substitute }}F''\left( x \right) \cr
& F'\left( x \right) = \int {dx} \cr
& F'\left( x \right) = x + C \cr
& {\text{Use the initial condition }}F'\left( 0 \right) = 3 \cr
& 3 = 0 + C \cr
& C = 3,{\text{ then}} \cr
& F'\left( x \right) = x + 3 \cr
& \cr
& F\left( x \right) = \int {F'\left( x \right)} dx \cr
& {\text{Substitute }}F'\left( x \right) \cr
& F\left( x \right) = \int {\left( {x + 3} \right)dx} \cr
& F\left( x \right) = \frac{1}{2}{x^2} + 3x + C \cr
& {\text{Use the initial condition }}F\left( 0 \right) = 4 \cr
& 4 = \frac{1}{2}{\left( 0 \right)^2} + 3\left( 0 \right) + C \cr
& C = 4,{\text{ then}} \cr
& F\left( x \right) = \frac{1}{2}{x^2} + 3x + 4 \cr} $$