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II. Integration – 2. Itereret integral 194 Det itererede integral tilnærmes af Riemann summen b b d A(x)dx = f(x,y)dydx a i a c A(x ∗ i )∆x som ved grænseovergang fører til volumenet, der netop er dobbeltintegralets værdi. 2.21. Fremgangsmåde ☞ [S] 12.2 Iterated integrals Eksempel 2 R = [0,2] × [1,2], f(x,y) = x − 3y 2 (x − 3y 2 2 2 )dA = (x − 3y 2 )dydx R = = 0 1 2 3 xy − y 0 y=2 y=1 dx 2 0 2 x = 2 = −12 (x − 7)dx 2 − 7x 0
II. Integration – 2. Itereret integral 195 2.22. Fortsat Eksempel 2 - fortsat ☞ [S] 12.2 Iterated integrals (x − 3y 2 2 2 )dA = (x − 3y 2 )dxdy R = = 1 2 1 2 1 0 x 2 2 − 3y2 x (2 − 6y 2 )dy x=2 x=0 = 2y − 2y 32 = −12 1 Bemærk, at midtpunktsreglen, [S] 12.1 Eksempel 3, gav tilnærmelsen − 95 8 dy = −11,875 2.23. Med sinus og cosinus ☞ [S] 12.2 Iterated integrals Figur - Eksempel 3 R = [1,2] × [0,π], f(x,y) = y sin(xy)
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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