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II. Integration – 2. Itereret integral 200 2.29. Opgave løst ☞ [S] 12.2 Iterated integrals Øvelse 16 xe xy 1 1 dA = xe xy dydx R 0 0 1 = [e 0 xy ] y=1 y=0 dx 1 = (e 0 x − e 0 )dx = [e x − x] 1 0 = e − 2 2.30. Test Fubini ☞ [S] 12.2 Iterated integrals Test Lad f(x,y) = xlny, R = [0,2] × [1,2]. Der gælder R xlny dA = 2 0 x dx 2 lny dy. 1 Afkryds: ja nej
II. Integration – 3. Generelle områder 201 Løsning 3. Generelle områder R xlny dA = = = 2 2 0 2 0 2 1 x 1 2 1 lny dy xlny dydx lny dydx 3.1. Oversigt ☞ [S] 12.1, 12.2, 12.3 Nøgleord og begreber ✌ Dobbelt integral ✌ Fubinis sætning ✌ Generelle områder ✌ Type I ✌ Type II ✌ Regneregler ✌ Nem ulighed 2 0 x dx
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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