Answer
$\$65,230.48$
Work Step by Step
We can compute the future value after 6 years:
$\int_{0}^{T}f(t)~e^{r(T-t)}~dt$
$= \int_{0}^{T}8000~e^{0.04t}~e^{r(T-t)}~dt$
$= \int_{0}^{T}8000~e^{rT}~e^{(0.04-r)t}~dt$
$= \frac{1}{0.04-r}\cdot 8000~e^{rT}~e^{(0.04-r)t}~\vert_{0}^{T}$
$= \frac{1}{0.04-r}\cdot 8000~[e^{rT}~e^{(0.04-r)T}- e^{rT}~e^{(0.04-r)(0)}~]$
$= \frac{1}{0.04-r}\cdot 8000~[e^{0.04T}- e^{rT}~]$
$= \frac{1}{0.04-0.062}\cdot 8000~[e^{(0.04)(6)}- e^{(0.062)(6)}~]$
$= -\frac{1}{0.022}\cdot 8000~[e^{0.24}- e^{0.372}~]$
$= \$65,230.48$
