AP Calculus

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Special Focus: The Fundamental Theorem of Calculus Graph of f(t) Question 1 To answer question 1 correctly, students must understand that ∫ 3 ∫ 4 4 f ( t) dt = f ( t) dt = − . This level of understanding was indicated as a prerequisite 2 3 3 b above (i.e., ∫ f ( x) dx takes on a negative value that is the additive inverse of area when a f ( x) ≤ 0 on the interval [a,b]), but teachers should be alert for students who have not yet achieved this level of understanding. If a student needs additional help with this concept, teachers have several options. Among them are: • If the class has investigated a velocity context such as the one Ray Cannon introduced, f could be identified as a velocity function, and students could rethink why these integrals have a negative value. For example: During the first 2 seconds, a car moves away from a point of reference, thus increasing its distance from that point. In these last 2 seconds, the car moves back toward the point of reference, and therefore its distance from that reference point is decreasing. • If the class has investigated Riemann sums, the students could partition the interval [0,4] into eight subintervals and sketch a left- or right-hand Riemann sum. It should then be clear that the products of the form f ( x) × ∆ x are positive on [0,2] and negative on [2,4]. AP® Calculus: 2006–2007 Workshop Materials 43
Special Focus: The Fundamental Theorem of Calculus (a) Given that F x x ( ) = ∫ ( ) 0 f t dt , x 0 1 2 3 4 F(x) 0 4/3 8/3 4/3 0 (b) Teachers should only expect the five specific points from the above table to be graphed carefully. Celebrate graphs that show appropriate increasing and decreasing tendencies. A graph of F based on knowing the rule for f is provided below. The graph of F on [0,4] Question 2 0 Many students will have trouble evaluating G at x=0 and x=1. Since G( 0) = ∫ f ( t) dt 2 1 and G( 1) = ∫ f ( t) dt, both integrals have an upper limit that is smaller than the lower 2 limit. Some students will believe that there is no such expression and will leave those entries in the table blank. Usually these students believe that the domain of the function defined by the integral has the lower limit as its minimum domain value. Others may merely ignore the difficulty and write the answers that would result if the limits were reversed. In any case, this question usually generates considerable debate among students. If students have trouble resolving this issue, teachers might return to the velocity context introduced in “From Riemann Sums to Net Change” and think about running time backward, as if playing a video in reverse. 44 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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