AP Calculus

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Special Focus: The Fundamental Theorem of Calculus 25. 2003 BC18 The graph of the function f shown in the figure above has horizontal tangents at x = 3 2x and x = 6. If g( x) = ∫ f ( t) dt, what is the value of g ′( 3 )? 0 (A) 0 (B) –1 (C) –2 (D) –3 (E) –6 26. 2003 BC27 3 d ⎛ x ln 2 ⎞ t + 1 dt dx 0 ( ) ⎝ ⎜ ∫ ⎠ ⎟ = (A) 2x 6 3 x + 1 (B) ( ) 2 6 (E) 3x ln x + 1 3x 6 2 x + 1 ( ) (D) 2 x 3 ln ( x 6 + 1) (C) ln x 6 + 1 AP® Calculus: 2006–2007 Workshop Materials 91
Special Focus: The Fundamental Theorem of Calculus 27. 2003 BC80 Insects destroyed a crop at the rate of 100 −0. 1t e −3t tons per day, where time t is 2 − e measured in days. To the nearest ton, how many tons did the insects destroy during the time interval 7 ≤ t ≤ 14? (A) 125 (B) 100 (C) 88 (D) 50 (E) 12 28. 2003 BC87 A particle moves along the x-axis so that at any time t ≥ 0, its velocity is given by 2 ( ) v( t) = cos 2 − t . The position of the particle is 3 at time t = 0. What is the position of the particle when its velocity is first equal to 0? (A) 0.411 (B) 1.310 (C) 2.816 (D) 3.091 (E) 3.411 Multiple-Choice Question Solutions 1. (1969 BC12) x 2 −t (E) By the Fundamental Theorem of Calculus, if F( x) = ∫ e dt , then 0 F′ ( x) = e −x 2 . 2. (1969 BC22) x 1 (E) Since f ( x) = ∫ dt , f ( − ) = ∫ − 1 1 0 1 1 dt = − dt 0 3 0 ∫ < 0 3 −1 3 t + 2 t + 2 t + 2 because the integrand is positive on the interval [ −1, 0 ]. Since f ( −1) < 0 , E is 0 1 false. A is true because f ( 0) = ∫ dt = 0. B is true because f is 0 3 t + 2 differentiable for all x ≥ 0. C is true because the integrand is positive on the 1 interval [ 0, 1 ]. D is true because f ′( x) = by the FTC. 3 x + 2 3. (1973 AB20) b (D) By the Fundamental Theorem of Calculus, ∫ f ( x) dx = F( b) − F( a) since a F′ ( x) = f ( x). 92 AP® Calculus: 2006–2007 Workshop Materials
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