Answer
$$\eqalign{
& \left( {\text{a}} \right)\$ 9687.5 \cr
& \left( {\text{b}} \right)\$ 8687.5 \cr} $$
Work Step by Step
$$\eqalign{
& {\text{The marginal cost is given by }} \cr
& C'\left( x \right) = 200 - 0.05x \cr
& \cr
& {\text{From the definition of the marginal cost }}\left( {{\text{page 177}}} \right){\text{ we have}} \cr
& {\text{that the cost function }}C\left( x \right){\text{ is the cost required to produce }}x \cr
& {\text{units of a product}}{\text{.}} \cr
& \cr
& {\text{Using the Theorem 6}}{\text{.3}}{\text{, }}\left( {{\text{Net change and future value}}} \right) \cr
& Q\left( b \right) - Q\left( a \right) = \int_a^b {Q'\left( t \right)} dt \cr
& \cr
& {\text{Then applying the concept theorem}}{\text{, the cost in dollars of}} \cr
& {\text{producion is given by}} \cr
& C\left( b \right) - C\left( a \right) = \int_a^b {C'\left( t \right)} dt \cr
& \cr
& \left( {\bf{a}} \right){\text{ }}a = 100{\text{ units}}{\text{, b}} = 150{\text{ units}} \cr
& C\left( {150} \right) - C\left( {100} \right) = \int_{100}^{150} {\left( {200 - 0.05x} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {200x - \frac{{0.05}}{2}{x^2}} \right]_{100}^{150} \cr
& = \left[ {200\left( {150} \right) - \frac{{0.05}}{2}{{\left( {150} \right)}^2}} \right] - \left[ {200\left( {100} \right) - \frac{{0.05}}{2}{{\left( {100} \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& = 29437.5 - 19750 \cr
& = \$ 9687.5 \cr
& \cr
& \left( {\bf{b}} \right){\text{ }}a = 500{\text{ units}}{\text{, b}} = 550{\text{ units}} \cr
& C\left( {550} \right) - C\left( {500} \right) = \int_{500}^{550} {\left( {2000 - 0.5x} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {200x - \frac{{0.05}}{2}{x^2}} \right]_{500}^{550} \cr
& = \left[ {200\left( {550} \right) - \frac{{0.05}}{2}{{\left( {550} \right)}^2}} \right] - \left[ {200\left( {500} \right) - \frac{{0.05}}{2}{{\left( {500} \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& = 102437.5 - 93750 \cr
& = \$ 8687.5 \cr} $$
