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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online