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II. Integration – 3. Generelle områder 214 Eksempel 2 - fortsat (x D 2 + y 2 2 2x )dA = 0 x2 (x 2 + y 2 )dy dx 2 = x 0 2 y + 1 3 y3 y=2x y=x2 dx 2 = (x 0 2 (2x) + 1 3 (2x)3 − x 2 (x 2 ) − 1 2 = − 1 21 x7 − 1 5 x5 + 7 6 x4 = 216 35 0 3 (x2 ) 3 )dx 3.21. Type II ☞ [S] 12.3 Double integrals over general regions Eksempel 3 Givet funktionen f(x,y) = xy på Type II mængden D = {(x,y)| − 2 ≤ y ≤ 4, 1 2 y2 − 3 ≤ x ≤ y + 1}
II. Integration – 3. Generelle områder 215 Dobbelt integralet beregnes itereret f(x,y)dA = D 4 −2 y+1 1 2 y2 −3 xy dx dy 3.22. Type II ☞ [S] 12.3 Double integrals over general regions Eksempel 3 - figur y x = 1 2 y3 3 4 x = y + 1 D = {(x,y)| − 2 ≤ y ≤ 4, 1 2 y2 − 3 ≤ x ≤ y + 1} 5 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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V. Matricer - 4. Determinanter 419
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V. Matricer - 4. Determinanter = (
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