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The number of people waiting in line to buy a new piece of technology is measured by the differentiable function 
f, where 
f(t) is measured in people and 
t is measured in hours after the store opened. What are the units of 
int_(1)^(5)f^(')(t)dt ?
people
hours
people / hour
hours / person
people / hour 
^(2)
hours / person 
^(2)

The number of people waiting in line to buy a new piece of technology is measured by the differentiable function f f , where f(t) f(t) is measured in people and t t is measured in hours after the store opened. What are the units of 15f(t)dt \int_{1}^{5} f^{\prime}(t) d t ?\newlinepeople\newlinehours\newlinepeople / hour\newlinehours / person\newlinepeople / hour 2 { }^{2} \newlinehours / person 2 { }^{2}

Full solution

Q. The number of people waiting in line to buy a new piece of technology is measured by the differentiable function f f , where f(t) f(t) is measured in people and t t is measured in hours after the store opened. What are the units of 15f(t)dt \int_{1}^{5} f^{\prime}(t) d t ?\newlinepeople\newlinehours\newlinepeople / hour\newlinehours / person\newlinepeople / hour 2 { }^{2} \newlinehours / person 2 { }^{2}
  1. Understand rate of change integral: Understand the integral of a rate of change.\newlineThe integral of a rate of change gives us the net change in the quantity over the interval. In this case, f(t)f'(t) represents the rate of change of the number of people with respect to time. The integral of f(t)f'(t) from 11 to 55 will give us the net change in the number of people from hour 11 to hour 55.
  2. Determine integral units: Determine the units of the integral.\newlineSince f(t)f'(t) is the derivative of the number of people with respect to time, its units are "people per hour". When we integrate f(t)f'(t) with respect to time, we multiply "people per hour" by "hours" (the units of tt). This will give us the units of the integral.
  3. Calculate integral units: Calculate the units of the integral.\newlineMultiplying "people per hour" by "hours" cancels out the "hours" in the denominator, leaving us with just "people". Therefore, the units of the integral 15f(t)dt\int_{1}^{5}f'(t)\,dt are "people".

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