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II. Integration – 2. Itereret integral 196 z x 2.24. Fortsat Eksempel 3 - fortsat ☞ [S] 12.2 Iterated integrals π 2 y sin(xy)dA = y sin(xy)dxdy R 0 π 1 = 0 y [−cos(xy)] x=2 x=1 dy
II. Integration – 2. Itereret integral π 197 = (−cos(2y) + cos(y))dy 0 = − sin(2y) π + sin(y) 2 0 = 0 2.25. Test Fubini ☞ [S] 12.2 Iterated integrals Test Lad f(x,y) = xlny, R = [0,2] × [1,2]. Der gælder Løsning (x,y) ∈ [0,2] × [1,2] giver ved Fubinis sætning xlny dA = R 2 2 0 1 R xlny dA = 2 0 xlny dydx 2 xlny dxdy. 1 Afkryds: ja nej 2.26. Produktregel ☞ [S] 12.2 Iterated integrals Figur - Eksempel 5 R = [0, π 2 π ] × [0, ], f(x,y) = sin(x)cos(y) 2
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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Litteratur • Arnold, Ordinary dif
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