I need some help with a question involving Newtons law of cooling
Formula: $$T(t) = \frac{\int T_ske^{-kt} \mathrm dt+c}{e^{-kt}}$$
$t =$ time in minutes
$T(t) =$ temperature of the object at time t $$T_s = \text{surrounding constant temperature}=21$$
$k =$ constant
Question: Using this formula, determine the temperature of a cold drink you have removed from the refrigerator, given the following information: The ambient temperature of the room is 21 degrees celsius. You take a very cold drink from the refrigerator, but the phone rings and you leave the drink on the bench. 5 minutes into your phone call you find that the temperature of your drink is 7.3 degrees celsius. knowing that the drink was 3.5 degrees celsius when you removed it from the refrigerator, what is the temperature when you hang up the phone and retrieve your drink 10 minutes after taking it out of the refrigerator?
We have that$$\frac{d}{dt} e^{-kt}=-k e^{-kt}$$ This is a fact about differentiating exponentials which you should try to remember. Next we can use the fact that integration is the "opposite" of differentiation to write $$\int -ke^{-kt}\,dt=e^{-kt}+\alpha\\\implies \int e^{-kt}\,dt=-\frac1ke^{-kt}+\alpha$$In the second line, we divided both sides by $k$. Also, we have a constant of integration $\alpha$.
In your question you have $\int T_s ke^{-kt}\,dt$. Since $T_s$ and $k$ are constants, you can take them outside the integral, which will give you the integral I have shown above. With this, I believe you should be able to do the rest of the question. Let me know if you get stuck again.