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?
+2511 How large is the the mess? 
\(\large \int_{0}^{60} 70e^{-0.05t} = \\ \int \:70e^{-0.05t}dt \hspace {14pt} |\text {Calculate the indefinite integral }\\ =70\int \:e^{-0.05t}dt \hspace {14pt} | \text { (Move constant) } \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\ \)
\(=70\int \:-20e^udu \hspace {14pt}|\int f (x (g)) \cdot {g}'(x)dx =\int f(u)du, u=g(x)\\ \hspace {84pt} \Rightarrow \:du=-0.05dt\\ \hspace {84pt} \Rightarrow \:dt=\left(-20\right)du\\ \hspace {84pt}=\frac{d}{dt}\left(-0.05t\right) \)
\(=70(-20\int e^udu) |\text { Move constant }\\ =70(-20e^u) |\text { using} \int e^udu=e^u\\ =70(-20e^{(-0.05t)}| \:\text{for }u=-0.05t\\ =-1400e^{-\frac{t}{20}}\\ \)
\(-1400e^{-3} = -69.70 \hspace {14pt}| \text{ for t=60}\\ -1400e^{0} = -1400 \hspace {20pt}| \text{ for t=0}\\\\ -1400-(-69.70) = -1330.3\\ \hspace {1pt}\\ \text {1330 liters leak in the first hour}\)
I’m glad I don’t have to mop this up. 
