Answer
$F\left( x \right) = \frac{{{x^2}}}{4} + x{\text{ }}$
Work Step by Step
$$\eqalign{
& F\left( x \right) = \int_0^x {\left( {\frac{1}{2}t + 1} \right)} dt \cr
& {\text{Integrate}} \cr
& F\left( x \right) = \left[ {\frac{1}{2}\left( {\frac{{{t^2}}}{2}} \right) + t} \right]_0^x = \left[ {\frac{{{t^2}}}{4} + t} \right]_0^x \cr
& F\left( x \right) = \left[ {\frac{{{x^2}}}{4} + x} \right] - \left[ {\frac{{{0^2}}}{4} + 0} \right] \cr
& F\left( x \right) = \frac{{{x^2}}}{4} + x{\text{ }}\left( {{\text{accumulation function}}} \right) \cr
& \cr
& \left( {\text{a}} \right)F\left( 0 \right) \cr
& F\left( 0 \right) = \frac{{{{\left( 0 \right)}^2}}}{4} + \left( 0 \right) \cr
& F\left( 0 \right) = 0,{\text{ }}\left( {{\text{Graph shown below}}} \right) \cr
& \cr
& \left( {\text{b}} \right)F\left( 2 \right) \cr
& F\left( 2 \right) = \frac{{{{\left( 2 \right)}^2}}}{4} + \left( 2 \right) \cr
& F\left( 2 \right) = 3,{\text{ }}\left( {{\text{Graph shown below}}} \right) \cr
& \cr
& \left( {\text{c}} \right)F\left( 6 \right) \cr
& F\left( 6 \right) = \frac{{{{\left( 6 \right)}^2}}}{4} + \left( 6 \right) \cr
& F\left( 6 \right) = 15,{\text{ }}\left( {{\text{Graph shown below}}} \right) \cr
& \cr
& {\text{Graphs}} \cr} $$
