I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.
A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.
I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?
Best Answer
Hint :
The average value of a function $f(x)$ on the intervall $[a,b]$ is given by : $$\frac{1}{b-a}\int_a^bf(x)dx$$
Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.
$T(0)=195$, so $T(t)=\int_0^tT'(s)ds+T(0)=\int_0^t−4.5e^{−0.07s}ds+195=−4.5\int_0^te^{−0.07s}ds+195=\frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=\frac{4,5}{0.07}(e^{-0.07t}-1)+195$
$$\frac{1}{5-0}\int_0^5T(x)dx=\frac{1}{5}\int_0^5\frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10\int_0^5e^{-0.07t}dt-50+195=-\frac{10}{0.07}( e^{-0.35}-1)+145\approx187.187$$