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?

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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:

$$ 200-29=\boxed{171} $$