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Vi starter med å finne integrasjonsgrensene. Disse vil være nullpunktene til funksjonen $f$:

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\$f(x)=0\$

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\$\\frac{1}{\\var{a}}x^3-\\var{a}x=0\$

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\$\\frac{1}{\\var{a}}x(x^2-\\var{a}^2)=0\$

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\$x=-\\var{a} \\vee x=0 \\,\\,\\vee \\,\\,x=\\var{a}\$

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Arealet av flatestykket som ligger over $x$-aksen er:

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\$\\begin{eqnarray*} A_1 &=& \\int_{-\\var{a}}^0 \\left(\\frac{1}{\\var{a}}x^3-\\var{a}x\\right)dx \\\\&=& \\left[\\simplify{1/{4a}x^4-{a}/2x^2}\\right]_{-\\var{a}}^0 \\\\&=& 0-\\left(\\simplify{1/{4*a}}(-\\var{a})^4-\\simplify{{a}/2}(-\\var{a})^2\\right)\\\\ &=& \\simplify{{a^3}/4}\\end{eqnarray*}\$

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Arealet av flatestykket som ligger under $x$-aksen er:

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\$\\begin{eqnarray*} A_2 &=& -\\int_{0}^\\var{a} \\left(\\frac{1}{\\var{a}}x^3-\\var{a}x\\right)dx \\\\&=& -\\left[\\simplify{1/{4a}x^4-{a}/2x^2}\\right]_0^\\var{a} \\\\&=& -\\left(\\simplify{1/{4*a}}(\\var{a})^4-\\simplify{{a}/2}(\\var{a})^2\\right)+0\\\\ &=& \\simplify{{a^3}/4}\\end{eqnarray*}\$

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Arealet av hele flatestykket blir dermed

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\$A = A_1+A_2 = \\simplify{{a^3}/4}+\\simplify{{a^3}/4}=\\simplify{{a^3}/2}\$

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Finn arealet $A$ av det området som er avgrenset av grafen til $f$ og $x$-aksen ved regning

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$A$ = [[0]]

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Her er $A = A_1+A_2$, der $A_1$ er arealet over $x$-aksen og $A_2$ er arealet under $x$-aksen.

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Du må starte med å finne nullpunktene til funksjonen. Det gjør du ved å løse likningen $f(x)=0$

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Husk at:

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Hvis grafen til $f$ ligger over $x$-aksen i intervallet $[a, b]$ er arealet avgrenset av $x$-aksen, grafen til $f$ og linjene $x=a$, $x=b$ gitt ved $\\displaystyle{A=\\int_a^b f(x) dx}$.

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Hvis grafen til $f$ ligger under $x$-aksen i intervallet $[a, b]$ er arealet avgrenset av $x$-aksen, grafen til $f$ og linjene $x=a$, $x=b$ gitt ved $\\displaystyle{A=-\\int_a^b f(x) dx}$.

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{eqnline(a)}

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Figuren ovenfor viser grafen til funksjonen

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$\\displaystyle\\simplify[all, !Noleadingminus]{f(x)=1/{a} x^3-{a}x}$

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