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# Worked examples: interpreting definite integrals in context

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.D (LO)
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CHA‑4.D.1 (EK)
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CHA‑4.D.2 (EK)
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CHA‑4.E (LO)
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CHA‑4.E.1 (EK)

## Video transcript

Julia's revenue is R of T thousand dollars per month where t is the month of the year julia had made three thousand dollars in the first month of the year what does three plus the definite integral from one to five of our F T DT equals 19 mean and we have some choices so like always pause the video and see if you can work through it alright now let's work through this together so they tell us that she made three thousand dollars in the first month and we also see this three here so that's interesting maybe there this they represent the same thing we don't know for sure yet but let's look at this definite integral the definite integral from 1 to 5 of R of T DT this is the area under this rate curve R of T is the rate at which Julia makes revenue on a monthly basis so if you take the area under that rate curve that's going to give you the net change in revenue from month 1 to month 5 how much that increased and so if you add that to the amount that she made in month 1 well that tells you the total she makes from essentially time zero all the way to month five and they're saying that is equal to 19 so let's see which of these choices are consistent with that Julia made an additional 19 thousand dollars between months 1 and 5 choice a would be correct if you didn't see this 3 over here because just the definite integral is the additional between months 1 and 5 but that's not what this expression says it says 3 plus this is equal to 19 if it said Julia made an additional \$16,000 well then that would make sense cuz you could subtract 3 from both sides and you'd get that result but that's not what they're saying Julia made an average of 19 thousand dollars per month well once again that's also not right because we just said from from the beginning from time zero all the way until the fifth month she made a total of \$19,000 not the average per month is 19,000 Julia made \$19,000 in the fifth month once again this is not just saying what happened in the fifth month this is saying we have the three thousand dollars from the first month and then we have the additional from month between months one and months so that's not that so this better be our choice by the end of the fifth month julia had made a total of \$19,000 yes that is correct she made three thousand and month one and then as we go between month one to the end of month five to the end of the fifth month she has made a total of \$19,000 let's do another one of these so here we're told the function K of T gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day so this is really quantity as a function of time it isn't rate what does the definitely go from zero to four of K prime of T DT represented once again pause the video and see if you can work through it well K if T is the amount of ketchup as a function of time so K prime of T that's going to be the rate at which our amount of ketchup is changing as a function of time but once again when you're taking the area under the rate curve that tells you the net change in the original quantity in the amount of ketchup and it's the net change between time 0 and time 4 so let's see which of these choices match up to that the average rate of change of the ketchup production over the first 4 hours no that does not tell us the average rate of change there's other ways to calculate that the time it takes to produce four kilograms of ketchup so does this represent the time it takes to represent four kilograms of ketchup no this four is the time right over here this is gonna tell you how much ketchup gets produced from time 0 to time for the instantaneous rate of production at t equals four now this would be K prime of four that's not what this integral represents the amount of ketchup produced over the first four hours yep that is exactly right
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