Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 410: 67

The increase in the cost is $~\$58,000$.

We can find the increase in the manufacturing cost: $\int_{2000}^{4000}C'(x)~dx$ $=\int_{2000}^{4000}(3-0.01x+0.000006x^2)~dx$ $=(3x-0.005x^2+0.000002x^3)~\vert_{2000}^{4000}$ $=[3(4000)-0.005(4000)^2+0.000002(4000)^3]-[3(2000)-0.005(2000)^2+0.000002(2000)^3]$ $=(60,000)-(2000)$ $= \$58,000$ The increase in the cost is $~\$58,000$