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# 16bf10lec11 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

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Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon Math 16b Review of Riemann Sums If a < b , f ( x ) is a function on [ a , b ] , and a = x ≤ a 1 ≤ x 1 ≤ a 2 ≤ ··· a n ≤ x n = b , then the Riemann sum associated to these data is n X i = 1 f ( a i )( x i- x i- 1 ) By de nition , the integral, R b a f ( x ) dx is the limit (if it exists) of these Riemann sums as maximum of | x i + 1- x i | tends to zero. Thomas Scanlon Math 16b An example Let us approximate R 1 4 1 + x 2 dx . We shall decompose the interval with two subintervals: [ , . 25 ] and [ . 25 , 1 ] with sample points 0 . 1 and 0 . 5. Thomas Scanlon Math 16b The calculation With f ( x ) = 4 1 + x 2 and the interval [ , 1 ] decomposed into [ , . 25 ] and [ . 25 , 1 ] with sample points 0 . 1 and 0 . 5, we have the Riemann sum 2 X i = 1 f ( a i )( x i- x i- 1 ) = f ( . 1 )( . 25- 1 ) + f ( . 5 )( 1- . 25 ) = 4 / ( 1 + ( . 1 ) 2 )( . 25 ) + 4 / ( 1 + ( . 5 ) 2 )( . 75 ) = 3 . 96 ( . 25 ) + 3 . 2 ( . 75 ) = 3 . 39 Thomas Scanlon Math 16b Uniform divisions For the sake of convenience, we often assume that the interval [ a , b ] has been decomposed into N pieces of equal length, for some positive integer N . The length of each piece is then Δ := b- a N . So, x i = a + i Δ and x i- 1 ≤ a i ≤ x i and for such a uniform decomposition, the Riemann sum is N X i = 1 ( f ( a i )( x i- x i- 1 )) = N X i = 1 ( f ( a i )( a + i Δ- a- i- 1 Δ)) = N X i = 1 f ( a i )Δ Thomas Scanlon Math 16b Left- and Right-hand rules Let a ≤ b , N , and f ( x ) be given.be given....
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16bf10lec11 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

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