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Unformatted text preview: f ( x ) = x ( x 21) 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 ) = Z1 x ( x 21)xdx + Z √ 2 xx ( x 21) dx = Z1 x 32 xdx + Z √ 2 2 xx 3 dx = 7 / 4 • Question 4  Apostol 2.4.16 (pg 94) Consider f ( x ) = xx 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 1a ≤ x ≤ 0. If a ≤ 1, then 0 ≤ x ≤ (1a ). If a ≥ 1, then the area is R 1a x (1a )x 2 dx =(1a ) 3 / 6. Hence a = 4. If a ≤ 1, then the area is R 1a xx ( x 21) dx , implying a =2 • Question 5  Apostol 2.5.18 (pg 105) Z π ( x + sin x ) dx = x 2 2cos x  π = π 2 2 + 2...
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 Spring '08
 Borodin,A
 Math, Calculus, Linear Algebra, Algebra, Prime number, dx, LILIAN WONG*

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