Chapter 8 - Further Applications of Integration - 8.4 Applications to Economics and Biology - 8.4 Exercises - Page 612: 1

$21,104$

$C(4000) - C(0) = \int^{4000}_{0} C'(x) dx$ So, $C(4000) = 18,000 + \int^{4000}_{0} (0.82 - 0.000 03 x + 0.000000003x^{2})dx$ $= 18,000 + [0.82x - 0.000015x^{2} + 0.000000001x^{3}]^{4000}_{0} = 18,000+3,104 = 21,104$