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# Worked examples: interpreting definite integrals in context

Interpreting expressions involving definite integrals in a real-world context.

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• Why would we use integrals to represent Julia's revenue? Isn't that unnecessarily complicated?
• I think integrals allow us to see the accumulation of her revenue at a certain interval of time.
• what would the definite integral of just k(t) from 0-4 be measuring? Its units would be kg*s.
• The owner of the sauce factory might charge a potential competitor who doesn't have a factory—perhaps because they are just starting out in the business—a usage fee of \$1 per hour per kilogram of ketchup produced. The definite integral of k(t) from 0 to 4 would then measure the total fee for those four hours.
• For the ketchap problem, what happens if we take the integral to K(t) rather than K'(t). We will get something Kg hours. What does this means ?
• The value in kg/hour mean how many kg are produced per hour. Since it is the derivative, it just shows how much the amount of ketchup is growing, at that instant, with respect to time.
(1 vote)
• Are the domains of k'(t) discrete in the second problem? Since t is in hours, would t have to be integers only or does it take any real number as its domain?
• Wouldn't the 3 represent the amount made from time = 0 to time = 1, if we put time as the horizontal axis? The month has to elapse; it doesn't just exist. The 3 and the integral from 1 to 5 are separate values/entities, yes? Could we say that 3 = the definite integral from 0 to 1 r(t) dt and we add to that the definite integral from 1 to 5 r(t) dt? We then add those together and get our \$19,000. Any context or perspective is appreciated.
(1 vote)
• 3 is indeed the amount of money made from t = 0 to t = 1. Essentially, t = 0 is when no months have passed, and t = 1 is when 1 month has passed. So, going from t = 0 to t = 1 means one month has passed.

We could write 3 as the integral of r(t) from 0 to 1, but here's the issue: we don't know if the rate of her getting revenue between t = 0 to t = 1 is the same as the rate between t = 1 and t = 5. So, with the information given, we can't write 3 as an integral.