// Numbas version: exam_results_page_options {"name": "Numbastest uke 42: Integrasjonsteknikker", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "variable_overrides": [[], [], []], "questions": [{"name": "Integrasjon ved substitusjon", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Elena Malyutina", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7213/"}], "tags": [], "metadata": {"description": "", "licence": "All rights reserved"}, "statement": "

I denne oppgaven regner vi ut det ubestemte integralet

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\\[ \\int (\\simplify{2{a}x+{b}})\\cdot e^{\\simplify{{a}x^2+{b}x+{c}}}dx\\]

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steg for steg ved bruk av substitusjon

", "advice": "

Først, bestemmer vi den kjernefunksjonen (indrefunksjonen) i integralet $y(x)=\\simplify{{a}x^2+{b}x+{c}}$

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Videre regner ut $y'(x)=\\dfrac{dy}{dx}=\\simplify{2{a}x+{b}}$

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Vi får en likning $\\dfrac{dy}{dx}=\\simplify{2{a}x+{b}}$ og fra denne likningen finner $dx=\\dfrac{dy}{\\simplify{2{a}x+{b}}}$

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Setter uttrykket for $y$ og $dx$ inn i integralet, forenkler og skriver integralet hvor integranden blir funksjon av $y$:

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\\[ \\int (\\simplify{2{a}x+{b}})\\cdot e^{\\simplify{{a}x^2+{b}x+{c}}}dx= \\int {(\\simplify{2{a}x+{b}})}\\cdot e^{y}\\dfrac{dy}{(\\simplify{2{a}x+{b}})}=\\int e^y dy\\]  

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Regner ut integralet

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\\[\\int e^y dy=e^y+C\\]  

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Og skriver inn tilbake uttrykk for $y(x)$:  

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\\[\\int e^y dy=e^y+C=e^{\\simplify{{a}x^2+{b}x+{c}}}+C\\]  

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Først, bestem kjernefunksjonen (indrefunksjonen) i integralet $y(x)=$ [[0]]

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Videre regn ut $y'(x)=\\dfrac{dy}{dx}=$ [[1]]. 

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Vi får en likning $\\dfrac{dy}{dx}=$ [[1]] og fra denne likningen finn $dx=$[[2]]

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Sett uttrykket for $y$ og $dx$ inn i integralet, forenkl og skriv integralet hvor integranden blir funksjon av $y$:

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$\\int$[[3]]$dy$    

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Regn ut integralet

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$\\int$[[3]]$dy$=[[4]]

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Og skriv inn tilbake uttrykk for $y(x)$:  [[5]]

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"showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Elena Malyutina", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7213/"}], "tags": [], "metadata": {"description": "", "licence": "All rights reserved"}, "statement": "

I denne oppgaven regner vi ut det ubestemte integralet 

\n

\\[ \\int (\\simplify{{b}x+{c}})\\cdot e^{x}dx\\]

\n

steg for steg ved bruk av delvis integrasjon. 

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Setningen som vi skal bruke sier at for funksjonene $u(x)$ og $v(x)$ som har kontinuerlige deriverte gjelder formelen

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\\[\\int u(x)\\cdot v'(x)dx=u(x)\\cdot v(x) -\\int u'(x)\\cdot v(x) dx  \\]

", "advice": "

Først, er funksjonene $u(x)=\\simplify{{b}x+{c}}$ og $v'(x)=e^x$

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Videre regner ut $u'(x)=\\var{b}$ og $v(x)=e^x$

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Nå bruker setningen og skriver det opprinnelige integralet som differansen og regner ut

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\\[\\int (\\simplify{{b}x+{c}})\\cdot e^{x}dx=(\\simplify{{b}x+{c}})\\cdot e^{x}-\\int \\var{b} e^x dx=(\\simplify{{b}x+{c-b}})e^x+C\\]

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Først, bestem funksjonene $u(x)=$ [[0]] og $v'(x)=$[[1]]

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Videre regn ut $u'(x)=$ [[2]] og $v(x)=$[[3]] 

\n

Nå bruk setningen ovenfor og skriv det opprinnelige integralet som differansen og regn ut

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$\\int (\\simplify{{b}x+{c}})\\cdot e^{x}dx=$ [[4]] $-\\int$ [[5]]$dx=$[[6]]

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Velg passende integrasjonsmetode for hvert integral

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