An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of
r(t) = 70e−0.05t liters per minute. How much oil leaks out during the first hour?
How large is the the mess?
∫60070e−0.05t=∫70e−0.05tdt|Calculate the indefinite integral =70∫e−0.05tdt| (Move constant) ∫a⋅f(x)dx=a⋅∫f(x)dx
=70∫−20eudu|∫f(x(g))⋅g′(x)dx=∫f(u)du,u=g(x)⇒du=−0.05dt⇒dt=(−20)du=ddt(−0.05t)
=70(−20∫eudu)| Move constant =70(−20eu)| using∫eudu=eu=70(−20e(−0.05t)|for u=−0.05t=−1400e−t20
−1400e−3=−69.70| for t=60−1400e0=−1400| for t=0−1400−(−69.70)=−1330.31330 liters leak in the first hour
I’m glad I don’t have to mop this up.