A rain gauge in a hill town records the rainfall intensity R(t) = 12te^(-t/2) millimetres per hour…

Step 1. An antiderivative. By parts, $\displaystyle\int te^{-t/2}\,dt = -2te^{-t/2} - 4e^{-t/2}$. So $\displaystyle\int 12te^{-t/2}\,dt = -24te^{-t/2} - 48e^{-t/2}$.

Step 2. The total. At $t = 4$ the value is $-96e^{-2} - 48e^{-2} = -144e^{-2}$. At $t = 0$ it is $-48$. The difference is $48 - 144e^{-2} \approx 28.51$ mm.

Step 3. The mean. $\dfrac{28.51\ \text{mm}}{4\ \text{h}} \approx 7.13$ mm/h.

Step 4. The peak. $R'(t) = 12e^{-t/2}\left(1 - \dfrac{t}{2}\right) = 0$ at $t = 2$ h, where $R = 24e^{-1} \approx 8.83$ mm/h.