Question

An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of

Answer 1

$\displaystyle\int_0^{60}r(t)\,\mathrm dt=\int_0^{60}80e^{-0.01t}\,\mathrm dt$

Let $u=-0.01t$ so that $-100\mathrm du=\mathrm dt$. Then the integral becomes

$\displaystyle-8000\int_0^{-0.6}e^u\,\mathrm du=8000\int_{-0.6}^0e^u\,\mathrm du=8000(e^0-e^{-0.6})\approx3610\text{ L}$