111_quiz7_spr08_key

# 111_quiz7_spr08_key - APMA 111 Quiz 7 March 18, 2008 NAME:...

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Unformatted text preview: APMA 111 Quiz 7 March 18, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. [7 pts] Find the centroid of the region bounded by the curves y = x + 2 and y = x 2 . Your answer should include a sketch of the region. Solution: First find the points of intersection: x 2 = x + 2 ⇒ x 2- x- 2 = 0 ⇒ ( x + 1)( x- 2) = 0 ⇒ x =- 1 , 2 . Since x + 2 > x 2 for- 1 < x < 2, the area of the region is A = Z 2- 1 ( x + 2- x 2 ) dx = ( x 2 / 2 + 2 x- x 3 / 3) fl fl fl fl 2- 1 = 2 + 4- 8 / 3- (1 / 2- 2 + 1 / 3) = 8- 3- 1 / 2 = 9 / 2 , while Z 2- 1 x ( x + 2- x 2 ) dx = Z 2- 1 ( x 2 + 2 x- x 3 ) dx = ( x 3 / 3 + x 2- x 4 / 4) fl fl fl fl 2- 1 = 8 / 3 + 4- 4- (- 1 / 3 + 1- 1 / 4) = 3- 1 + 1 / 4 = 9 / 4 , and Z 2- 1 £ ( x + 2) 2- ( x 2 ) 2 / dx = Z 2- 1 ( x 2 + 4 x + 4- x 4 ) dx = ( x 3 / 3 + 2 x 2 + 4 x- x 5 / 5) fl fl 2- 1 = 8 / 3 + 8 + 8- 32 / 5- (- 1 / 3 + 2- 4 + 1 / 5) = 3 + 16 + 2- 33 / 5 = 15- 3 / 5 = 72 / 5 ....
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## This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.

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111_quiz7_spr08_key - APMA 111 Quiz 7 March 18, 2008 NAME:...

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