DEL 1

Oppgave 1

$\int_{0}^{1} (2e^x+2x^2)dx$

$=[2e^x+\frac{2}{3}x^3]_{0}^{1}$

$=2e+\frac{2}{3}-2$

$=2e-\frac{4}{3}$

$\int\frac{2x-1}{x^2-x-6}dx$

$\frac{2x-1}{x^2-x-6}=\frac{A}{x-3}+\frac{B}{x+2}$

$2x-1=A(x+2)+B(x-3)$

$2x-1=Ax+2A+Bx-3B$

$-1=(A+B-2)x+2A-3B$

Vi får to likninger:

$A+B-2=0 \quad \wedge \quad 2A-3B=-1$

$A=-B+2 \quad \wedge \quad 2(-B+2)-3B=-1$

$A=-B+2 \quad \wedge \quad -5B=-5$

$A=1 \quad \wedge \quad B=1$

Setter verdiene for A og B inn i det opprinnelige integralet:

$\int\frac{2x-1}{x^2-x-6}dx=\int\frac{A}{x-3}+\frac{B}{x+2}dx$

$=\int\frac{1}{x-3}+\int\frac{1}{x+2}$

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