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
$$T(t)=20+75e^{-t/50}$$
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 $T(t)=20+75e^{-t/50}$ from 0 to 30. The integral of $T(t)$ from 0 to 30 is ${\int }_0^{30}(20+75e^{-t/50})dt$.

Step 3 ：The integral can be split into two parts: ${\int }_0^{30}20dt+{\int }_0^{30}75e^{-t/50}dt$.

Step 4 ：The first integral is easy to compute: ${\int }_0^{30}20dt=20\ast 30=600$.

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, ${\int }_0^{30}75e^{-t/50}dt=-75\ast 50(e^{-30/50}-e^0)$.

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

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 $\boxed{{\displaystyle {58.74}^{\circ }C}}$.