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# Chapter1 - DANANG UNIVERSITY OF TECHNOLOGY DANANG...

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DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II DANANG UNIVERSITY OF TECHNOLOGY LECTURE ON CALCULUS WITH ANALYTIC GEOMETRY II Dr. Nguyen Chanh Dinh

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DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Chapter 1     INTEGRALS I. AREAS AND DISTANCES 1. The Area Problem: Find the area of the region S that lies under the  curve  y = f ( x ) on [ a,b ]?.  Here   f ≥ 0 is assumed to be continuous. Idea:   Subdividing  S into n strips of equal width  1 2 , , , n S S S L
DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II The width of each of the n strips is These strips divide the intervals [ a,b ] into n  subinterval b a x n - ∆ = 0 1 1 2 1 [ , ],[ , ], ,[ , ] n n x x x x x x - L

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DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II where  The right endpoints of the subintervals are The area of S is approximated by the sum: 0 , . n x a x b = = 1 2 3 , 2 , 3 , x a x x a x x a x = + ∆ = + ∆ = + ∆ L 1 2 ( ) ( ) ( ) . n n R f x x f x x f x x = ∆ + ∆ + + L
DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Definition:   The  area  A of the region S that lies  under the graph of the continuous function  f   is  the limit: [ ] 1 2 lim lim ( ) ( ) ( ) n n n n A R f x x f x x f x x f = = ∆ + ∆ + + L

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DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II * i x In fact, instead of using left endpoints or right  endpoints, we could take the height of the i-th  rectangle to be the value of   f   at any   number      in the i-th subinterval           . We call the numbers                   the  sample  points.   1 [ , ] i i x x - * i x * * * 1 2 , , , n x x x L
DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II A more general expression for the area of S  is * * * 1 2 * 1 lim ( ) ( ) ( ) lim ( ) . n n n i n i A f x x f x x f x x f x x f f = = ∆ + ∆ + + = L

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DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Example Let A  be the area of the region that lies  under the graph of   ( ) x f x e - = between x = 0 and x = 2.
DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Solution:  Since a = 0 and b = 2, the  width of a subinterval is so 2 0 2 x n n - ∆ = = 1 2 2 4 2 2 , , , , , , i n i n x x x x n n n n = = = = L L

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DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II 1 2 1 2 2/ 4/ 2 / ( ) ( ) ( ) 2 2 2 . n n n x x x n n n n R f x x f x x f x x e x e x e x e e e n n n - - - - - - = ∆ + ∆ + + = ∆ + ∆ + + = + + + L L L
DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II According to the definition, the area is 2 / 1 2 lim . n

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Chapter1 - DANANG UNIVERSITY OF TECHNOLOGY DANANG...

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