Definite Integrals Question 16

calculator allowed A cup of tea is cooling in a room that has a constant temperature of $70$ degrees Fahrenheit ( $^\circ \text{F}$ ). If the initial temperature of the tea, at time $t=0$ , is $200^\circ \text{F}$ and the temperature of the tea changes at the rate $R(t)=-6.55e^{-0.045t}$ degrees Fahrenheit per minute, what is the temperature, to the nearest degree, of the tea after $5$ minutes?

171^\circ \text{F}

130^\circ \text{F}

41^\circ \text{F}

29^\circ \text{F}

Approach

The change in temperature is found with the definite integral:

\int\limits_0^5 -6.55e^{-0.045t}   dt

Evaluating the integral using the calculator:

\int\limits_0^5 -6.55e^{-0.045t}   dt \approx -29

Subtracting from the initial temperature:

200-29=\boxed{171}