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Vi starter med å finne integrasjonsgrensene. Disse vil være nullpunktene til funksjonen $f$:
\n\\[f(x)=0\\]
\n\\[\\frac{1}{\\var{a}}x^3-\\var{a}x=0\\]
\n\\[\\frac{1}{\\var{a}}x(x^2-\\var{a}^2)=0\\]
\n\\[x=-\\var{a} \\vee x=0 \\,\\,\\vee \\,\\,x=\\var{a}\\]
\nArealet av flatestykket som ligger over $x$-aksen er:
\n\\[\\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*}\\]
\nArealet av flatestykket som ligger under $x$-aksen er:
\n\\[\\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*}\\]
\nArealet av hele flatestykket blir dermed
\n\\[A = A_1+A_2 = \\simplify{{a^3}/4}+\\simplify{{a^3}/4}=\\simplify{{a^3}/2}\\]
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\nFiguren ovenfor viser grafen til funksjonen
\n$\\displaystyle\\simplify[all, !Noleadingminus]{f(x)=1/{a} x^3-{a}x}$
\n", "parts": [{"gaps": [{"minValue": "a^3/2", "allowFractions": true, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "type": "numberentry", "correctAnswerFraction": true, "correctAnswerStyle": "plain", "showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "maxValue": "a^3/2", "scripts": {}}], "steps": [{"variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "showFeedbackIcon": true, "type": "extension", "scripts": {}, "prompt": "Her er $A = A_1+A_2$, der $A_1$ er arealet over $x$-aksen og $A_2$ er arealet under $x$-aksen.
\nHusk at:
\nHvis 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 $\\displaystyle{A=\\int_a^b f(x) dx}$.
\nHvis 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 $\\displaystyle{A=-\\int_a^b f(x) dx}$.
"}], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 0, "stepsPenalty": 0, "variableReplacements": [], "showFeedbackIcon": true, "type": "gapfill", "scripts": {}, "prompt": "Finn arealet $A$ av det området som er avgrenset av grafen til $f$ og $x$-aksen ved regning
\n$A$ = [[0]]
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