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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 kth term goes +, +, –, –, etc., that is
k
sign 0
+ 1
+ 2
– 3
– ...
... 20
+ 21
+ 22
– 2004 AB (Form B)
AP Calculus FreeResponse
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 FREERESPONSE 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.
 Fall '11
 Mr.Snickles
 Calculus

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