APcal 2004-Solutions

# that is k sign 0 1 2 3 20 21 22 2004

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Unformatted text preview: 1 f − P = < = < . 4 4! 4!⋅10 24 ⋅16 100 10 10 10 This polynomial can be obtained by integrating the first three terms in P(x), and 2 5 2 2 25 2 3 taking into account that G(0) = 0. We get x+ x− x. 2 2⋅2 2⋅6 Notes: 1. The sign for the k-th term goes +, +, –, –, etc., that is k sign 0 + 1 + 2 – 3 – ... ... 20 + 21 + 22 – 2004 AB (Form B) AP Calculus Free-Response Solutions and Notes Question 1 (a) Area = ∫ 10 x − 1 dx 1 (b) = 18 . y 3 1 10 ( ) 2 10 Volume = π ∫ 32 − 3 − x − 1 dx 1 (c) x ≈ 212.058 . y 10–x 3 1 10 Volume = π ∫ ( 9 − y 2 ) dy 3 0 2 1 x ≈ 407.150 . Notes: 1. y = x − 1 ⇒ x = y 2 + 1 ⇒ 10 − x = 9 − y 2 11 12 FREE-RESPONSE SOLUTIONS ~ 2004 AB (FORM B) Question 2 M (a) 6 Increasing, because R(6) = 5 6 cos 5 (b) Increasing at a decreasing rate, because R′(6) (c) 31 t Number of mosquitoes = 1000 + ∫ 5 t cos dt 0 5 (d) ≈ 4.438 > 0 . R (t ) ≈ −1.913 < 0 . ≈ 964 . 0 6 31 1 5π 5π 15π 15π , R(t) < 0 for <t < , and R(t) > 0 for < t ≤ 31 . 2 2 2 2 5π Therefore the maximum could occur either at t = or at t = 31. The number of 2 5π 5π t mosquitoes at t = is 1000 + ∫ 2 5 t cos dt ≈ 1039 . This is greater than 0...
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## This note was uploaded on 07/07/2013 for the course MATH AP taught by Professor Mr.snickles during the Fall '11 term at Benjamin N Cardozo High School.

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