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II. Integration – 2. Itereret integral 192 Partial integral A(y) = 1 0 (2x + 2ye x )dx = x 2 + 2ye x x=1 x=0 = (1 + 2ye) − (0 + 2y) = 2(e − 1)y + 1 2.17. Igen ens ☞ [S] 12.2 Iterated integrals Eksempel - fortsat Itereret integral 1 1 0 0 A(y) = 2(e − 1)y + 1 (2x + 2ye x )dxdy = 1 0 (2(e − 1)y + 1)dy = (e − 1)y 2 + y 1 0 = ((e − 1) + 1) − (0 + 0) 2.18. Fubini: Alle veje fører til Rom ☞ [S] 12.2 Iterated integrals = e
II. Integration – 2. Itereret integral 193 4 Sætning (Fubinis sætning) Lad R = [a,b] × [c,d] og antag f : R → R er kontinuert. Så er dobbeltintegralet lig de itererede integraler b d f(x,y)dA = f(x,y)dydx og R R a d b f(x,y)dA = f(x,y)dxdy 2.19. Overbevis ☞ [S] 12.2 Iterated integrals Fubinis sætning - begrundelse Lad f(x,y) ≥ 0. Volumenet under grafen er dobbeltintegralet f(x,y)dA Det partielle integral R d A(x) = f(x,y)dy er arealet af et snit gennem området under grafen for x fast. c 2.20. Mere overbevis ☞ [S] 12.2 Iterated integrals Fubinis sætning - begrundelse c c a
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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