Answer
$$F\left( x \right) = \frac{1}{6}{x^4} + x + 3$$
Work Step by Step
$$\eqalign{
& F'''\left( x \right) = 4x,{\text{ }}F''\left( 0 \right) = 0,{\text{ }}F'\left( 0 \right) = 1,{\text{ }}F\left( 0 \right){\text{ = 3 }} \cr
& F''\left( x \right) = \int {F'''\left( x \right)} dx \cr
& {\text{Substitute }}F'''\left( x \right) \cr
& F''\left( x \right) = \int {4xdx} \cr
& F''\left( x \right) = 2{x^2} + C \cr
& {\text{Use the initial condition }}F''\left( 0 \right) = 0 \cr
& 0 = 2{\left( 0 \right)^2} + C \cr
& C = 0,{\text{ then}} \cr
& F''\left( x \right) = 2{x^2} \cr
& \cr
& F'\left( x \right) = \int {F''\left( x \right)} dx \cr
& {\text{Substitute }}F''\left( x \right) \cr
& F'\left( x \right) = \int {2{x^2}dx} \cr
& F'\left( x \right) = \frac{2}{3}{x^3} + C \cr
& {\text{Use the initial condition }}F'\left( 0 \right) = 1 \cr
& 1 = \frac{2}{3}{\left( 0 \right)^3} + C \cr
& C = 1,{\text{ then}} \cr
& F'\left( x \right) = \frac{2}{3}{x^3} + 1 \cr
& \cr
& F\left( x \right) = \int {F'\left( x \right)} dx \cr
& {\text{Substitute }}F'\left( x \right) \cr
& F\left( x \right) = \int {\left( {\frac{2}{3}{x^3} + 1} \right)dx} \cr
& F\left( x \right) = \frac{1}{6}{x^4} + x + C \cr
& {\text{Use the initial condition }}F\left( 0 \right) = 3 \cr
& 3 = \frac{1}{6}{\left( 0 \right)^4} + \left( 0 \right) + C \cr
& C = 3,{\text{ then}} \cr
& F\left( x \right) = \frac{1}{6}{x^4} + x + 3 \cr} $$
