Evaluate the definite integral to two decimal places.

${\displaystyle \underset{0}{\overset{25}{}}}800e^{0\mathrm{.}02t}{e}^{0\mathrm{.}06(25-t)}dt$

${\displaystyle \underset{0}{\overset{25}{}}}800e^{0\mathrm{.}02t}{e}^{0\mathrm{.}06(25-t)}dt=$

$($Round to two decimal places as needed.$)$