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II. Integration – 4. Koordinatskift 246 z x R = {(x,y)|0 ≤ x ≤ 2,0 ≤ y ≤ 4 − x 2 } 4.36. Opgave ☞ Matematik Alfa 1, August 2002 Opgave 1 - alternativt er et Type I område. R = {(x,y)|0 ≤ x ≤ 2,0 ≤ y ≤ 4 − x 2 } y
II. Integration – 4. Koordinatskift 247 Integralet er R x 2 y dA = 2 √ 4−x2 0 0 x 2 y dy dx 4.37. Opgave Opgave 1 - alternativt - fortsat ☞ Matematik Alfa 1, August 2002 x 2 2 y dA = √ 4−x2 x 2 y dy dx R = 0 0 2 1 0 2 x2y 2 √ y= 4−x2 y=0 2 1 = 0 2 (4x2 − x 4 )dx 2 = 3 x3 − 1 10 x5 2 = 32 15 4.38. Populært ☞ [S] 12.5 Applications of double integrals Anvendelser 0 dx
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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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1. Grafiske/numeriske metoder IV Di
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V. Matricer - 4. Determinanter = (
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1. Egenvektorer VI Egenvektorer og
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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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