// Numbas version: finer_feedback_settings {"name": "Testing section 8g - Integration by partial fractions", "metadata": {"description": "", "licence": "None specified"}, "duration": 1800, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "variable_overrides": [[], [], []], "questions": [{"name": "Integration by partial fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(b1=a,b1+s3,b1)", "description": "", "name": "b"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "c"}, "s3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s3"}, "b1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s2*random(1..9)", "description": "", "name": "b1"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "a"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s2"}}, "ungrouped_variables": ["a", "c", "b", "s3", "s2", "s1", "b1"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 1, "scripts": {}, "gaps": [{"answer": "({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C", "vsetrange": [11, 12], "checkingaccuracy": 0.0001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "
Input all numbers as fractions or integers and not decimals.
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t$I=\\;$[[0]]
\n\t\t\tInput all numbers as fractions or integers and not decimals.
\n\t\t\tInput the constant of integration as $C$.
\n\t\t\tClick on Show steps for help if you need it. You will lose 1 mark if you do so.
\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\tFirst of all factorise the denominator.
\n\t\t\t\t\tYou have to find $a$ and $b$ such that $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+a)*(x+b)}$.
\n\t\t\t\t\tThen use partial fractions to write:
\\[\\simplify[std]{{c}/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]
for suitable integers or fractions $A$ and $B$.
\n\t\t\t\t\t", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\n\tFind the following integral.
\n\t\\[I = \\simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\]
\n\tInput all numbers as fractions or integers and not decimals.
\n\tInput the constant of integration as $C$.
\n\t", "tags": ["2 distinct linear factors", "Calculus", "MAS1601", "Steps", "checked2015", "completing the square", "constant of integration", "factorising a quadratic", "factorizing a quadratic", "indefinite integration", "integration", "partial fractions", "two distinct linear factors"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t5/08/2012:
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\n\t\tNote the checking range is chosen so that the arguments of the log terms are always positive - could have used abs - might be better?
\n\t\tImproved display of Advice.
\n\t\tAdded information about Show steps, also introduced penalty of 1 mark.
\n\t\tAdded !noLeadingMinus to ruleset std for display purposes.
\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "\n\t\tFactorise $x^2+bx+c$ into 2 distinct linear factors and then find $\\displaystyle \\int \\frac{a}{x^2+bx+c }\\;dx$ using partial fractions or otherwise.
\n\t\t\n\t\t"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t
First we factorise $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+{a})*(x+{b})}$.
\n\tYou can do this by spotting the factors or by completing the square.
\n\tNext we use partial fractions to find $A$ and $B$ such that:
\\[\\simplify[std]{{c}/((x +{a})*(x+{b})) = A/(x+{a})+B/(x+{b})}\\]
Multiplying both sides of the equation by $\\displaystyle \\simplify[std]{1/((x +{a})*(x+{b}))}$ we obtain:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}}\\Rightarrow \\simplify[std]{(A+B)x+{b}A+{a}B={c}}$
\n\tIdentifying coefficients:
\n\tConstant term: $\\simplify[std]{{b}*A+{a}*B = {c}}$
\n\tCoefficent $x$: $ \\simplify[std]{A + B = 0}$ which gives $A = -B$.
\n\tHence we obtain $\\displaystyle \\simplify[std]{A = {c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={-c}/{b-a}}$
\n\tWhich gives: \\[\\simplify[std]{{c}/((x +{a})*(x+{b})) = ({c}/{b-a})*(1/(x+{a}) -1/(x+{b}))}\\]
\n\tSo \\[\\begin{eqnarray*} I &=& \\simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\\\ &=& \\simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}\\\\ &=& \\simplify[std]{({c}/{b-a})*(Int(1/(x+{a}),x) -Int(1/(x+{b}),x))}\\\\ &=& \\simplify[std]{({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C} \\end{eqnarray*}\\]
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", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n$I=\\;$[[0]]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\nClick on Show steps for help if you need it. You will lose 1 mark if you do so.
\n ", "steps": [{"type": "information", "prompt": "\nFirst of all factorise the denominator.
\nYou have to find $a$ and $b$ such that $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+a)*(x+b)}$
\nThen use partial fractions to write:
\\[\\simplify[std]{({c}*x+{d})/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]
for suitable integers or fractions $A$ and $B$.
\n ", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\nFind the following integral.
\n\\[I = \\simplify[std]{Int(({c}*x+{d})/(x^2+{a+b}*x+{a*b}),x )}\\]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\n ", "tags": ["2 distinct linear factors", "Calculus", "MAS1601", "Steps", "checked2015", "completing the square", "constant of integration", "factorising a quadratic", "indefinite integration", "integration", "logarithms", "partial fractions", "two distinct linear factors"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t5/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tAdded decimal point as forbidden string.
\n \t\tNote the checking range is chosen so that the arguments of the log terms are always positive - could have used abs - might be better?
\n \t\tImproved display of Advice.
\n \t\tAdded information about Show steps, also introduced penalty of 1 mark.
\n \t\tAdded !noLeadingMinus to ruleset std for display purposes.
\n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Factorise $x^2+cx+d$ into 2 distinct linear factors and then find $\\displaystyle \\int \\frac{ax+b}{x^2+cx+d}\\;dx,\\;a \\neq 0$ using partial fractions or otherwise.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "First we factorise $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+{a})*(x+{b})}$. You can do this by spotting the factors or by completing the square.
Next we use partial fractions to find $A$ and $B$ such that:
\\[\\displaystyle \\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b})) = A/(x+{a})+B/(x+{b})}\\]
Multiplying both sides of the equation by $\\displaystyle \\simplify[std]{1/((x +{a})*(x+{b}))}$ we obtain:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$
\nIdentifying coefficients:
\nConstant term: $\\simplify[std]{{b}*A+{a}*B = {d}}$
\nCoefficent $x$: $ \\simplify[std]{A+B={c}}$ which gives $A =\\var{c} -B$
\nOn solving these equations we obtain $\\displaystyle \\simplify[std]{A = {d-a*c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={d-b*c}/{a-b}}$
\nWhich gives: \\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b})) = ({d-a*c}/{b-a})*(1/(x+{a}) )+({d-b*c}/{a-b})*(1/(x+{b}))}\\]
\nSo \\[\\begin{eqnarray*} I &=& \\simplify[std]{Int(({c}*x+{d})/(x^2+{a+b}*x+{a*b}),x )}\\\\ &=&\\simplify[std]{Int(({c}*x+{d})/((x +{a})*(x+{b})),x )}\\\\ &=& \\simplify[std]{({d-a*c}/{b-a})*(Int(1/(x+{a}),x)) +({d-b*c}/{a-b})Int(1/(x+{b}),x)}\\\\ &=& \\simplify[std]{({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C} \\end{eqnarray*}\\]
"}, {"name": "Integration by partial fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(b1=a,b1+s3,b1)", "description": "", "name": "b"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "c"}, "s3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s3"}, "b1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s2*random(1..9)", "description": "", "name": "b1"}, "d": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(d1=a*c,if(d1+1=b*c,d1+2,d1+1),if(d1=b*c,if(d1+1=a*c,d1+2,d1+1),d1))", "description": "", "name": "d"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s2"}, "d1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s3*random(1..9)", "description": "", "name": "d1"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "a"}}, "ungrouped_variables": ["a", "c", "b", "d", "s3", "s2", "s1", "b1", "d1"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 1, "scripts": {}, "gaps": [{"answer": "({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C", "vsetrange": [11, 12], "checkingaccuracy": 0.0001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "Input all numbers as fractions or integers and not decimals.
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n$I=\\;$[[0]]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\nClick on Show steps for help if you need it. You will lose 1 mark if you do so.
\n \n ", "steps": [{"type": "information", "prompt": "\nUse partial fractions in order to write:
\\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\\; = \\simplify[std]{A/(x+{a})+B/(x+{b})}\\]
for suitable integers or fractions $A$ and $B$.
\n ", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\nFind the following integral.
\n\\[I = \\simplify[std]{Int(({c}*x+{d})/((x +{a})*(x+{b})),x )}\\]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\n \n ", "tags": ["2 distinct linear factors", "Calculus", "calculus", "checked2015", "compare coefficients", "identify coefficients", "integration", "logarithms", "MAS1601", "mas1601", "partial fractions", "Steps", "steps", "two distinct linear factors"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t \t\t5/08/2012:
\n \t\t \t\tAdded tags.
\n \t\t \t\tAdded description.
\n \t\t \t\tAdded decimal point as forbidden string.
\n \t\t \t\tNote the checking range is chosen so that the arguments of the log terms are always positive - could have used abs - might be better?
\n \t\t \t\tImproved display of Advice.
\n \t\t \t\tAdded information about Show steps, also introduced penalty of 1 mark.
\n \t\t \t\tAdded !noLeadingMinus to ruleset std for display purposes.
\n \t\t \n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Find $\\displaystyle\\int \\frac{ax+b}{(x+c)(x+d)}\\;dx,\\;a\\neq 0,\\;c \\neq d $.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\nUsing partial fractions we have to find $A$ and $B$ such that:
\\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\\;= \\simplify[std]{A/(x+{a})+B/(x+{b})}\\]
Multiplying both sides of the equation by $\\displaystyle \\simplify[std]{1/((x +{a})*(x+{b}))}$ we obtain:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$
\nIdentifying coefficients:
\nConstant term: $\\simplify[std]{{b}*A+{a}*B = {d}}$
\nCoefficent $x$: $ \\simplify[std]{A+B={c}}$ which gives $A =\\var{c} -B$
\nOn solving these equations we obtain $\\displaystyle \\simplify[std]{A = {d-a*c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={d-b*c}/{a-b}}$
\nWhich gives: \\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\\; =\\simplify[std]{ ({d-a*c}/{b-a})*(1/(x+{a}) )+({d-b*c}/{a-b})*(1/(x+{b}))}\\]
\nSo \\[\\begin{eqnarray*} I &=& \\simplify[std]{Int(({c}*x+{d})/((x +{a})*(x+{b})),x )}\\\\ &=& \\simplify[std]{({d-a*c}/{b-a})*(Int(1/(x+{a}),x)) +({d-b*c}/{a-b})Int(1/(x+{b}),x)}\\\\ &=& \\simplify[std]{({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C} \\end{eqnarray*}\\]
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