alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# Dx example 2 let fx x for x 1 2 3 we may compute

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Unformatted text preview: contin uous summation of UNCOUNTABLY many continuous summation c ontin summa terms from the discrete summation of a finite number of terms using + or discrete summation discr summa COUNTABLY many terms using Σ . The dx in dy/dx of differentiation refers to the calculation at a point or instant. differentia dx dif entiation point instant The ∫ f (x) . dx of integr ation refers to the contin uous summation over an integ continuous summation inte c ontin summa interv inter val of contiguous instants dx . contiguous dx 132 pez method Tr a pe z oidal m ethod f(x) B f(x) A x- 0 x- 1 x- 2 (0 , 0 ) x 0= a x1 x2 x3 . . b = x n . x area Instead of approximating the ar ea under the cur v e f(x) by rectangular ar strips we could use trapezoidal strips. 1 Area of a trapezoid is: ( -- s um of the parallel sides) * ( perpendicular distance ) 2 APPROXIMATE step discrete summation discr 1. APPR OXIMATE AREA step : discr ete summation of FINITELY many terms. f(x ) + f(x 1 ) . f( x ) + f(a ) . ( x 1 -- a ) + 2 ( x 2 -- x 1 ) + . . . F( x ) = 1 --2 2 f( b ) + f( x n -- 1 ) . + ( b -- x n -- 1 ) 2 It is not difficult to see that in sub-interval ∆x 0 = [ a, x 1 ] there is some MEAN point (see note on next page) -x 0 s uch that: f( x1 ) + f( a ) f ( -x 0 ) = 2 Similarly, in sub-inter val ∆x 1 = [ x 1, x 2] there some MEAN point -x 1 s uch that: f( x 2) + f( x1 ) f ( -x 1) = 2 and so on. So we can write: n−1 -- ) (x -- a) + f(x ) (x -- x ) + ... + f( -x ) (b -- x ) = Σ f (x- ). ∆x --. F(x) = f(x 0 1 --- 1 1 2 n -- 1 n -- 1 i i i=0 133 step discrete summation 2. TENDS TO step : discrete summation as n → ∞ of COUNTABLY many terms. discr All we have to do now is let n get larger and larger by letting n → ∞ . T he sub-intervals ∆x i will get smaller and smaller. We can denote the sub-intervals ∆x i b y δ xi for i = 0, 1, 2, . . . , n → ∞ : δ x 0 = ( x 1 -- a), δ x 1 = ( x 2 -- x 1), . . . , δ x n -- 1 = ( b -- x n -- 1) --F(x) = f( -x 0) δ x 0 + f ( -x 1) δ x...
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