Sol-162E2-S2004

# Sol-162E2-S2004 - MATH 162 – SPRING 2004 – SECOND EXAM...

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Unformatted text preview: MATH 162 – SPRING 2004 – SECOND EXAM SOLUTIONS The following formulas were given on the exam: Moments and center of mass M x = Z b a 1 2 ( ( f ( x )) 2- ( g ( x )) 2 ) dx, M y = Z b a x ( f ( x )- g ( x )) dx x = M y M , y = M x M , Arc length L = Z b a p 1 + ( f ( x )) 2 dx Area of a surface of revolution S = Z b a 2 πf ( x ) p 1 + ( f ( x )) 2 dx 1) The mass of the region bounded by f ( x ) = 1 2 √ 4- 2 x 2 , g ( x ) =- 1 2 √ 4- 2 x 2 and the y-axis is π √ 2. Its center of mass is A) ( 8 3 π √ 2 , 0) B) (0 , 8 3 π √ 2 ) C) ( 1 2 , 8 3 π √ 2 ) D) (0 , 4 3 π √ 2 ) E) ( 4 3 π √ 2 , 0) Solution: According to the formulas above, the center of mass is x = 1 π √ 2 Z √ 2 x √ 4- 2 x 2 dx, y = 0 this is because f ( x ) 2 = g ( x ) 2 . 1 2 To compute this integral we just set u = 4- 2 x 2 . Then du =- 4 x dx and the integral becomes x = 1 π √ 2 Z 4 √ u du 4 = 1 4 π √ 2 2 3 u 3 2 4 = 1 4 π √ 2 2 3 8 = 4 3 π √ 2 . The correct answer is E. 2) The improper integral Z 1 ln x dx = A) 2ln2 B)- 4ln2 C) 2ln2 D)- 1 E)- 1 9 Solution: By definition of improper integrals Z 1 ln x dx = lim a → Z 1 a ln x dx. Integration by parts gives Z 1 a ln x dx = ( x ln x- x ) | 1 a =- 1- ( a ln a- a ) . Now we have to compute lim a → a ln a- a = lim a → a ln a To do this we use L’Hosptial’s rule and write lim a → a ln a = lim a → ln a 1 a = lim a → 1 a- 1 a 2 = lim a →- a = 0 . So Z 1 ln x dx =- 1 The correct answer is D. Unfortunately this question had two identical alternatives, but but both are incorrect. 3 3) The improper integral Z ∞ x e- x 2 dx is equal to A) 1 3 B) 1 4 C) 1 2 D) 1 E) 2 Solution: By definition Z ∞ x e- x 2 dx = lim...
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Sol-162E2-S2004 - MATH 162 – SPRING 2004 – SECOND EXAM...

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