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II. Integration – 2. Itereret integral 182 Lad R = [−1,1] × [−1,1]. R ( 1 − x2 + 1 − y2 )dA = 1 − x2 dA + R R 1 − y 2 dA De to dobbelt integraler findes som volumener af halvcylindre: I alt ( 1 − x2 + 1 − y2 )dA = 1 2 π12 · 2 · 2 = 2π 2. Itereret integral R 2.1. Oversigt ☞ [S] 5.2, 5.3, 5.4, 12.1, 12.2 Nøgleord og begreber ✌ Analysens hovedsætning ✌ Stamfunktioner ✌ Itereret integral ✌ Test itereret integral ✌ Fubinis sætning ✌ Test Fubini
II. Integration – 2. Itereret integral 183 ✌ Eksempler ✌ Test produkt 2.2. Analysens hovedsætning ☞ [S] 5.4 The fundamental theorem of calc. Sætning (Analysens fundamentalsætning) Antag, at f : [a,b] → R er kontinuert. Så er x g(x) = f(t)dt differentiabel og a g ′ (x) = f(x) For en stamfunktion F(x), F ′ (x) = f(x), er b f(x)dx = F(b) − F(a) a 2.3. Overbevis ☞ [S] 5.4 The fundamental theorem of calculus Sætning (Analysens fundamentalsætning) Bevis g ′ (x) = lim h→0 g(x + h) − g(x) h
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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