If a cup of coffee has temperature $95^{\circ} \mathrm{C}$ in a room where the temperature is $20^{\circ} \mathrm{C}$, then, according to Newton's Law of Cooling, the temperature of the coffee after $t$ minutes is
\[
T(t)=20+75 e^{-t / 50}
\]
What is the average temperature (in degrees Celsius) of the coffee during the first half hour?
Average temperature $=$ degrees Celsius

Step 1 ：To find the average temperature of the coffee during the first half hour, we need to integrate the temperature function from 0 to 30 minutes and then divide by the interval of time, which is 30 minutes.

Step 2 ：First, we integrate the function \(T(t)=20+75 e^{-t / 50}\) from 0 to 30. The integral of \(T(t)\) from 0 to 30 is \(\int_{0}^{30} (20+75 e^{-t / 50}) dt\).

Step 3 ：The integral can be split into two parts: \(\int_{0}^{30} 20 dt + \int_{0}^{30} 75 e^{-t / 50} dt\).

Step 4 ：The first integral is easy to compute: \(\int_{0}^{30} 20 dt = 20 * 30 = 600\).

Step 5 ：The second integral is a bit more complex. We can use the formula for the integral of \(e^{at}\), which is \(\frac{1}{a}e^{at}\). So, \(\int_{0}^{30} 75 e^{-t / 50} dt = -75*50(e^{-30 / 50}-e^{0})\).

Step 6 ：Computing the second integral gives us \(-75*50(e^{-30 / 50}-1) = -75*50*(-0.5493-1) = 1162.25\).

Step 7 ：Adding the two parts of the integral together gives us the total integral: \(600 + 1162.25 = 1762.25\).

Step 8 ：Finally, we divide the total integral by the interval of time to find the average temperature: \(\frac{1762.25}{30} = 58.7417\).

Step 9 ：So, the average temperature of the coffee during the first half hour is approximately \(\boxed{58.74^\circ C}\).