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II. Integration – 3. Generelle områder 212 3.17. Type II ☞ [S] 12.3 Double integrals over general regions Eksempel 2 - fortsat (x D 2 + y 2 4 )dA = 0 √ y 1 2 y (x 2 + y 2 )dx dy 4 x= √ y = = = 0 3 x + xy2 3 dy 0 x= y 2 4 ( 1 3 y3/2 + y 5/2 − 1 24 y3 − 1 2 15 y5/2 + 2 7 y7/2 − 13 96 y4 = 216 35 4 0 2 y3 )dy 3.18. Type I ☞ [S] 12.3 Double integrals over general regions Eksempel 2 Givet funktionen f(x,y) = x 2 + y 2 på Type I mængden D = {(x,y)|0 ≤ x ≤ 2,x 2 ≤ y ≤ 2x}
II. Integration – 3. Generelle områder 213 Dobbelt integralet beregnes itereret f(x,y)dA = D 2 2x 0 x2 (x 2 + y 2 )dy dx 3.19. Type I ☞ [S] 12.3 Double integrals over general regions Eksempel 2 - figur y 4 y = 2x y = x 2 D = {(x,y)|0 ≤ x ≤ 2,x 2 ≤ y ≤ 2x} 3.20. Type I ☞ [S] 12.3 Double integrals over general regions 2 x
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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I. Differentiation - 1. Kontinuitet
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V. Matricer - 4. Determinanter = (
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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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Identitetsmatricen, 383 Identitetsm
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