Answer
The increase in the cost is $~\$58,000$.
Work Step by Step
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 $~\$58,000$