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The Definite Integral as an Accumulator Bob Arrigo Scarsdale High School Scarsdale, NY

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Presentation on theme: "The Definite Integral as an Accumulator Bob Arrigo Scarsdale High School Scarsdale, NY"— Presentation transcript:

1 The Definite Integral as an Accumulator Bob Arrigo Scarsdale High School Scarsdale, NY r1arrigo@gmail.com www.BCCalculus.com

2 Traditional applications of the Definite Integral prior to the Calculus reform movement Area, volume, total distance traveled.. (AB) Arc length, work.. (BC) Mass, fluid pressure.. (Some college Calculus courses)

3 Calculus Reform in the early 90’s brought in “broader”, more robust applications of the definite integral…… most prominently, use of the definite integral to calculate “net change”, or “accumulated change.”

4 Types of Integrals Definite Integrals…limits of Riemann sums …”summing up infinitely many infinitesimally small products” Indefinite Integrals….a family of functions Integral functions….functions defined by an integral

5 The definite integral provides net change in a quantity over time. The definite integral of a rate function yields accumulated change of the associated function over some interval.

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7 Motivate with a water flow problem: The rate at which water flows into a tank, in gallons per hour, is given by a differentiable function R of time t. Values of R are given at various times t during a 24 hour period. Approximate the number of gallons of water that flowed into the tank over the 24 hour period. tR(t) 013 615 1218 14 2410

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12 This is an approximation for the total flow in gallons of water from the pipe in the 24-hour period.

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14 Summing up lots of distances, each of which equals the product (rate)(time)

15 Method I to get the total distance traveled: Break up the interval [0,6] into smaller and smaller subintervals. To get the actual distance traveled, use more, smaller subintervals.

16 tv(t) 00 27.2 412.8 616.8 Method II

17 Since method I and method II, both yield total distance, We get: Answer method I = Answer method II

18 Since method I and method II both yield total distance, We get: Answer method I = Answer method II

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23 End Amt = Start Amt + NET CHANGE

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32 Since is positive for and is negative for, the maximum value for occurs at time.

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