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| (10 mellomliggende sideversjoner av samme bruker vises ikke) |
| Linje 9: |
Linje 9: |
| =DEL 1= | | =DEL 1= |
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| ==Oppgave 1== | | =DEL 2= |
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| ===a)===
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| $\int_{0}^{1} (2e^x+2x^2)dx$
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| $=[2e^x+\frac{2}{3}x^3]_{0}^{1}$
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| $=2e+\frac{2}{3}-2$
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| $=2e-\frac{4}{3}$
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| ===b)===
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| $\int\frac{2x-1}{x^2-x-6}dx$
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| Bruker delbrøkoppspalting:
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| $\frac{2x-1}{x^2-x-6}=\frac{A}{x-3}+\frac{B}{x+2}$
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| $2x-1=A(x+2)+B(x-3)$
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| $2x-1=Ax+2A+Bx-3B$
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| $-1=(A+B-2)x+2A-3B$
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| Vi får to likninger:
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| $A+B-2=0 \quad \wedge \quad 2A-3B=-1$
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| $A=-B+2 \quad \wedge \quad 2(-B+2)-3B=-1$
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| $A=-B+2 \quad \wedge \quad -5B=-5$
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| $A=1 \quad \wedge \quad B=1$
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| Setter verdiene for A og B inn i det opprinnelige integralet:
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| $\int\frac{2x-1}{x^2-x-6}dx=\int\frac{A}{x-3}+\frac{B}{x+2}dx$
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| $=\int\frac{1}{x-3}dx+\int\frac{1}{x+2}dx$
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| $=ln(x-3)+ln(x+2)+C$
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| ==Oppgave 2==
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| Vi har at
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| * $f'(x)=-\frac{2}{x^3}$
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| * $\int_{1}^{2}f(x)dx=\frac{11}{14}$
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| Integrerer f'(x) for å få et uttrykk for f(x):
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| $f(x)=\int -\frac{2}{x^3} dx = -2\int x^{-3}dx=\frac{-2}{-2}x^{-2} +C=\frac{1}{x^2}+C$
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| Bruker det bestemte integralet som likning for å finne C:
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| $\int_{1}^{2}(\frac{1}{x^2}+C )\, dx=\frac{11}{14}$
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| $[-\frac{1}{x}+Cx]_{1}^{2}=\frac{11}{14}$
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| $-\frac{1}{2}+2C-(-1+C)=\frac{11}{14}$
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| $C+\frac{1}{2}=\frac{11}{14}$
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| $C=\frac{11}{14}-\frac{7}{14}$
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| $C=\frac{4}{14}=\frac{2}{7}$
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| Vi har:
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| $f(x)=\frac{1}{x^2}+\frac{2}{7}$
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| ==Oppgave 3==
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| ===a)===
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| {| class="wikitable"
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| |-
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| ! k
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| ! 1
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| ! 2
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| ! 3
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| ! 6
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| |-
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| | $P(X=k)$
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| | $\frac{1}{6}$
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| | $\frac{1}{6}$
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| | $\frac{1}{6}$
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| | $\frac{1}{2}$
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| |}
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| $E(x)=\frac{1+2+3+6+6+6}{6}=\frac{24}{6}=4$
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| ===b)===
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