AP Calculus

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Special Focus: The Fundamental Theorem of Calculus At this point, I define a “new” notation for this limit of a Riemann sum. The new notation 10 is v ( t ) 10 ∫ dt, and we call v( t) dt 0 ∫ the definite integral of v(t) from t=0 to t=10. The 0 10 meaning of ∫ v ( t ) dt is exactly the same as lim ( v( t ) ( ) ( ) 0 1 ⋅ ∆t + v t2 ⋅ ∆t + ⋅⋅⋅ + v tn ⋅ ∆ t) . n→∞ 10 We discuss the form of the definite integral and emphasize that ∫ v ( t ) dt embodies 0 everything that appears in the limit of the corresponding Riemann sum. Evaluation Step 3: Understanding the statement of the evaluation part of the Fundamental Theorem Now it is time to pull everything together. We currently have ( ) = Change in position = lim v ( t1 ) ⋅ ∆t + v ( t2 ) ⋅ ∆t + ⋅⋅⋅ + v ( tn ) ⋅ ∆ t ∫ v ( t)dt. n→∞ Recall that v( t) = s′ ( t), so change in position = v( t) dt = s′ ( t) dt. Earlier we saw 0 0 10 that change in position = s( 10) − s( 0). Therefore ∫ s′ ( t) dt = s( 10) − s( 0 ). My goal is for the students to not only see ∫ s′ ( t) dt = s( 10) − s( 0 ) as a natural result, 0 b but also to generalize one more step to ∫ s ′( t ) dt = s ( b ) − s ( a ) . a Antiderivative Part of the Fundamental Theorem In an attempt to help students understand that the antiderivative part of the Fundamental Theorem guarantees that a continuous function on a open interval has an antiderivative, whether or not a closed form of the antiderivative can be found, I break the development into two steps. Antiderivative Step 1: Applying this part of the Fundamental Theorem to write the x derivative of a function that is defined by a definite integral of the form ∫ f ( t) dt , k where k is a constant 10 Generally, textbooks do a good job developing this concept, and I do not need to create supplementary problems. I do like to use some of the excellent AP free-response questions that are most easily completed by using this antiderivative part of the Fundamental Theorem. One example follows. ∫ 10 0 ∫ 10 10 0 AP® Calculus: 2006–2007 Workshop Materials 11
Special Focus: The Fundamental Theorem of Calculus 2002 AB4/BC4 Free-response question AB4/BC4 from the 2002 AP Exam gives an example of this application of the antiderivative part of the FTC. Students were not allowed to use a calculator on this question. The graph of the function f shown above consists of two line segments. Let g be the function given by g( x) = ∫ f ( t) dt . 0 (a) Find g( −1 ), g ′( − ), x 1 and g ′′( −1 ). (b) For what values of x in the open interval (–2, 2) is g increasing? Explain your reasoning. (c) For what values of x in the open interval (–2, 2) is the graph of g concave down? Explain your reasoning. (d) Sketch the graph of g on the closed interval [–2, 2]. We will focus on parts (a) and (b) here. Students needed to use the Fundamental Theorem to determine that g′ ( x) = f ( x) in order to answer part (a). For part (b), some students attacked the problem by trying to make an area argument. Others used the fact that g′ ( x) = f ( x) and that from the given graph f ( x) > 0 on − 1 < x < 1, so f is increasing on the interval −1 ≤ x ≤ 1. The students who used this last argument were much more successful than those who tried an area argument. 12 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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