Oil leaks out of a tanker at a rate of r=f (t) r=f (t) liters per minute, where tt is in minutes. if f (t) = ae-ktf (t) = ae-kt, write a definite integral expressing the total quantity of oil which leaks out of the tanker in the first hour.

The rate the oil leaks out at is dV/dt = r = = > dV = f (t) dt = A e^ (-kt) dt; The integral is V = â" (t = a to b) A e^ (-kt) dt = = A â" (t = 0 to 60) e^ (-kt) dt = = - A/k e^ (-kt) | (t = 0 to 60) = = - A/k (e^ (-60k) - 1) = = A/k (1 - e^ (-60k)) That's the result you get solving the differential equation, as well : dV = A e^ (-kt) dt V = - A/k e^ (-kt) + C and put V (0) = 0 0 = - A/k + C = = > C = A/k V (t) = A/k - A/k e^ (-kt) = A/k (1 - e^ (-kt)) and plug in t = 60