Answer
$$C\left( x \right) = \frac{{0.2}}{3}{x^3} + \frac{5}{2}{x^2} + 10$$
Work Step by Step
$$\eqalign{
& C'\left( x \right) = 0.2{x^2} + 5x;{\text{ fixed cost is }}\$ 10 \cr
& {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr
& {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr
& C\left( x \right) = \int {C'\left( x \right)} dx \cr
& {\text{replacing }}0.2{x^2} + 5x{\text{ for }}C'\left( x \right) \cr
& C\left( x \right) = \int {\left( {0.2{x^2} + 5x} \right)} dx \cr
& {\text{integrating }} \cr
& C\left( x \right) = 0.2\left( {\frac{{{x^{2 + 1}}}}{{2 + 1}}} \right) + 5\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right) + K{\text{ }} \cr
& C\left( x \right) = 0.2\left( {\frac{{{x^3}}}{3}} \right) + 5\left( {\frac{{{x^2}}}{2}} \right) + K{\text{ }} \cr
& C\left( x \right) = \frac{{0.2}}{3}{x^3} + \frac{5}{2}{x^2} + K{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find }}K,{\text{we know that the fixed cost is }}\$ 10{\text{ then }}C\left( 0 \right) = 10 \cr
& 10 = \frac{{0.2}}{3}{\left( 0 \right)^3} + \frac{5}{2}{\left( 0 \right)^2} + K \cr
& 10 = K \cr
& {\text{then substituting }}K = 10{\text{ into the equation }}\left( {\bf{1}} \right){\text{ we obtain}} \cr
& C\left( x \right) = \frac{{0.2}}{3}{x^3} + \frac{5}{2}{x^2} + 10 \cr} $$
