Chapter 8 - Further Applications of Integration - 8.4 Applications to Economics and Biology - 8.4 Exercises - Page 612: 3

$140,000$ dollar, $60,000$ dollar

$C(50) - C(0) = \int^{50}_{0} (0.6+0.008x)dx$ then $C(50) = 100+[0.6x + 0.004x^{2}]^{50}_{0} = 100 + (40 - 0) = 140$ or $140,000$ dollar Similarly, $C(100) - C(50) = [0.6x + 0.004x^{2}]^{100}_{50} = 60$ or $60,000$ dollar