[tex]\displaystyle\int_0^{60}r(t)\,\mathrm dt=\int_0^{60}80e^{-0.01t}\,\mathrm dt[/tex]
Let [tex]u=-0.01t[/tex] so that [tex]-100\mathrm du=\mathrm dt[/tex]. Then the integral becomes
[tex]\displaystyle-8000\int_0^{-0.6}e^u\,\mathrm du=8000\int_{-0.6}^0e^u\,\mathrm du=8000(e^0-e^{-0.6})\approx3610\text{ L}[/tex]
