problemset7s - MAT 137Y 20082009 Winter Session, Solutions...

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1 (SHE Section 5.2) 6 The function f ( x ) = x is an increasing function, so m i = f ( x i - 1 ) and M i = f ( x i ) for all i . Therefore L f ( P ) = f ( 0 ) · 1 25 + f ( 1 25 ) · 3 25 + f ( 4 25 ) · 5 25 + f ( 9 25 ) · 7 25 + f ( 16 25 ) · 9 25 = 0 + 1 5 · 3 25 + 2 5 · 1 5 + 3 5 · 7 25 + 4 5 · 9 25 = 3 125 + 10 125 + 21 125 + 36 125 = 70 125 = 14 25 . Likewise, U f ( P ) = f ( 1 25 ) · 1 25 + f ( 4 25 ) · 3 25 + f ( 9 25 ) · 5 25 + f ( 16 25 ) · 7 25 + f ( 1 ) · 9 25 = 1 5 · 1 25 + 2 5 · 3 25 + 3 5 · 1 5 + 4 5 · 7 25 + 1 · 9 25 = 1 125 + 6 125 + 15 125 + 28 125 = 45 125 = 19 25 . 12 For f ( x ) = x + 3 we have f is increasing on the interval [ a , b ] , so the lower sum and upper sums are L f ( P ) = ( x 0 + 3 )( x 1 - x 0 )+( x 1 + 3 )( x 2 - x 1 )+ ··· +( x n - 1 + 3 )( x n - x n - 1 ) , U f ( P ) = ( x 1 + 3 )( x 1 - x 0 )+( x 2 + 3 )( x 2 - x 1 )+ ··· +( x n + 3 )( x n - x n - 1 ) . For part (b), for each index i , x i - 1 + 3 1 2 ( x i - 1 + x i )+ 3 x i + 3 . Multiplying by Δ x i = x i - x i - 1 gives ( x i - 1 + 3 ) Δ x i 1 2 ( x 2 i - x 2 i - 1 )+ 3 ( x i - x i - 1 ) ( x i + 3 ) Δ x i . Summing from i = 1 to i = n , we find that L f ( P ) 1 2 ( x 2 1 - x 2 0 )+ 3 ( x 1 - x 0 )+ ··· + 1 2 ( x 2 n - x 2 n - 1 )+ 3 ( x n - x n - 1 ) U f ( P ) . The middle sum collapses to 1 2 ( x 2 n - x 2 0 )+ 3 ( x n - x 0 ) = 1 2 ( b 2 - a 2 )+ 3 ( b - a ) , thus Z b a ( x + 3 ) dx = 1 2 ( b 2 - a 2 )+ 3 ( b - a ) . 32 Let P = { x 0 , x 1 ,..., x n } be a regular partition of [ a , b ] and let Δ x = ( b - a ) / n . Since f is decreas- ing on [ a , b ] , U f ( P ) = f ( x 0 ) Δ x + f ( x 1 ) Δ x + ··· + f ( x n - 1 ) Δ x and L f ( P ) = f ( x 1 ) Δ x + f ( x 2 ) Δ x + ··· + f ( x n ) Δ x . Therefore,
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problemset7s - MAT 137Y 20082009 Winter Session, Solutions...

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