FA06Ma.9Sec1Sol3

# FA06Ma.9Sec1Sol3 - f x = x x 2-1 and g x = x on the...

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SOLUTIONS TO HOMEWORK 3 MATH 1A (SECTION 1) FALL 2006 MANWAH LILIAN WONG* Question 1(a) - Apostol 1.26.10 (pg 83) 3 2 (3 x 2 - 4 x + 2) dx = (3 x 3 3 - 4 x 2 2 + 2 x ) | 3 2 = ( x 3 - 2 x 2 + 2 x ) | 3 2 = (3 3 - 2(3 2 ) + 2(3) - 2 3 - 2(2 2 ) + 2(2) = 11 Question 1(b) - Apostol 1.26.15 (pg 83) 2 0 ( x - 1)(3 x - 1) dx = 2 0 3 x 2 - 4 x + 1 dx = 3 x 3 3 - 4 x 2 2 + xdx | 2 0 = 2 Question 2 - Apostol 1.26.21 (pg 83) ( a ) c 0 x (1 - x ) dx = c 0 ( x - x 2 ) dx = ( x 2 - x 3 3 ) | c 0 = c 2 - c 3 3 Hence, c = 0 , 3 2 . (b) The clumsy way: note that x (1 - x ) 0 when 0 x 1 and x (1 - x ) 0 when x 0 or x 1. Suppose c 1. Then the argument is the same as in part (a), and c = 0 , 3 2 . If c > 1, then the Date : October 11th, 2006. * MC 253-37, Mathematics Department, California Institute of Technology, Pasadena, CA 91125, USA. E-mail: [email protected] 1

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2 M.-W. L. WONG problem becomes solving for 1 0 x (1 - x ) dx + c 1 - x (1 - x ) dx = 0, which is equivalent to solving 1 0 x (1 - x ) dx = c 1 x (1 - x ) dx = 0. This implies 1 / 3 = c 2 / 2 - c 3 / 3 and by plotting it we know that this happens when c < 0. Contradiction. The clever way: Note that | x (1 - x ) | > 0 x = 0 , 1, hence unless c = 0, c 0 | x (1 - x ) | dx > 0 . Question 3 - Apostol 2.4.10 (pg 94)
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Unformatted text preview: f ( x ) = x ( x 2-1) and g ( x ) = x on the interval [-1 , √ 2]. Note that g ≥ f on [0 , √ 2] and f ≥ g on [-1 , 0]. Hence the area of S is: Area ( S ) = Z-1 x ( x 2-1)-xdx + Z √ 2 x-x ( x 2-1) dx = Z-1 x 3-2 xdx + Z √ 2 2 x-x 3 dx = 7 / 4 • Question 4 - Apostol 2.4.16 (pg 94) Consider f ( x ) = x-x 2 and g ( x ) = ax . We need to decide on the conditions on a such that f ( x ) ≥ g ( x ). If a ≥ 1, then we must have 1-a ≤ x ≤ 0. If a ≤ 1, then 0 ≤ x ≤ (1-a ). If a ≥ 1, then the area is R 1-a x (1-a )-x 2 dx =-(1-a ) 3 / 6. Hence a = 4. If a ≤ 1, then the area is R 1-a x-x ( x 2-1) dx , implying a =-2 • Question 5 - Apostol 2.5.18 (pg 105) Z π ( x + sin x ) dx = x 2 2-cos x | π = π 2 2 + 2...
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