Answer
$V(t)=73,584t^{3}+709,560t^{2}-525,600t+76,212$
Work Step by Step
$V(t) $ is an antiderivative of the given instantaneous rate of change $m(t).$
$V(t)=\displaystyle \int(525,600(0.42t^{2}+2.7t-1) )dt$
$=525,600(0.42\displaystyle \cdot\frac{t^{3}}{3}+2.7\cdot\frac{t^{2}}{2}-t)+C$
$=73,584t^{3}+709,560t^{2}-525,600t+C,$
which is a collection of functions. To find the exact function, we find $C.$
The middle of the year 2005 is $t=0.5$ years after the start of 2005. We are given
$V(0.5)=0\qquad $hours
$0=73584(0.5)^{3}+709560(0.5)^{2}-525,600(0.5)+C,$
$0=-76212+C$
$C=76212$
Thus,
$V(t)=73,584t^{3}+709,560t^{2}-525,600t+76,212$
