Answer
$$C\left( x \right) = 3{e^{0.01x}} + 5$$
Work Step by Step
$$\eqalign{
& C'\left( x \right) = 0.03{e^{0.01x}};{\text{ fixed cost is }}\$ 8 \cr
& {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr
& {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr
& C\left( x \right) = \int {C'\left( x \right)} dx \cr
& {\text{replacing }}0.03{e^{0.01x}}{\text{ for }}C'\left( x \right) \cr
& C\left( x \right) = \int {\left( {0.03{e^{0.01x}}} \right)} dx \cr
& {\text{integrating by the formula }}\int {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C,\,\,\,\,\,k \ne 0 \cr
& C\left( x \right) = 0.03\left( {\frac{{{e^{0.01x}}}}{{0.01}}} \right) + K{\text{ }} \cr
& C\left( x \right) = 3{e^{0.01x}} + K{\text{ }} \cr
& C\left( x \right) = 3{e^{0.01x}} + K{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find }}K,{\text{we know that the fixed cost is }}\$ 8{\text{ then }}C\left( 0 \right) = 8 \cr
& 8 = 3{e^{0.01\left( 0 \right)}} + K \cr
& 8 = 3 + K \cr
& K = 5 \cr
& {\text{then substituting }}K = 5{\text{ into the equation }}\left( {\bf{1}} \right){\text{ we obtain}} \cr
& C\left( x \right) = 3{e^{0.01x}} + 5 \cr} $$
