John's temperature is rising at a rate of r(t) = 0.3/1+t^2 degrees Fahrenheit per minute at t minutes since the thermometer is inserted. What are the units of the integral 0 to 5 r(t)dt? What does is represent in practical terms? Evaluate the intergral
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If r(t) has units °F/min and t (and hence dt) has units of minutes, then r(t) dt has units of °F, and integration (which is a sort of generalized sum) does not change this. What the integral gives is the difference (in F°) between his temperature at time t=5 and his temperature at time t=0. There is a subtle distinction between the difference of two temperatures, and a single temperature. But that is physics, not math, so perhaps the distinction doesn't matter here, and °F will turn out to be the correct answer. As for the integral itself, it is elementary:
5
∫ [0.3/(1 + t²)] dt = 0.3 arctan(t) evaluated from 0 to 5
0
= 0.3 arctan(5) - 0.3 arctan(0)
= 0.3 arctan(5)
≈ 0.412
5
∫ [0.3/(1 + t²)] dt = 0.3 arctan(t) evaluated from 0 to 5
0
= 0.3 arctan(5) - 0.3 arctan(0)
= 0.3 arctan(5)
≈ 0.412