// 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.

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$I=\\;$[[0]]

\n\t\t\t

Input all numbers as fractions or integers and not decimals.

\n\t\t\t

Input the constant of integration as $C$.

\n\t\t\t

Click 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\t

First of all factorise the denominator.

\n\t\t\t\t\t

You 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\t

Then use partial fractions to write:
\\[\\simplify[std]{{c}/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]

\n\t\t\t\t\t

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\t

Find the following integral.

\n\t

\\[I = \\simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\]

\n\t

Input all numbers as fractions or integers and not decimals.

\n\t

Input 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\t

5/08/2012:

\n\t\t

Added tags.

\n\t\t

Added description.

\n\t\t

Added decimal point as forbidden string.

\n\t\t

Note 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

Improved display of Advice. 

\n\t\t

Added information about Show steps, also introduced penalty of 1 mark.

\n\t\t

Added !noLeadingMinus to ruleset std for display purposes.

\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "\n\t\t

Factorise $x^2+bx+c$ 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\t

You can do this by spotting the factors or by completing the square.

\n\t

Next 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:

\n\t

$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}}\\Rightarrow \\simplify[std]{(A+B)x+{b}A+{a}B={c}}$

\n\t

Identifying coefficients:

\n\t

Constant term: $\\simplify[std]{{b}*A+{a}*B = {c}}$

\n\t

Coefficent $x$: $ \\simplify[std]{A + B = 0}$ which gives $A = -B$.

\n\t

Hence we obtain $\\displaystyle \\simplify[std]{A = {c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={-c}/{b-a}}$

\n\t

Which gives: \\[\\simplify[std]{{c}/((x +{a})*(x+{b})) = ({c}/{b-a})*(1/(x+{a}) -1/(x+{b}))}\\]

\n\t

So \\[\\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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Input all numbers as fractions or integers and not decimals.

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$I=\\;$[[0]]

\n

Input all numbers as fractions or integers and not decimals.

\n

Input the constant of integration as $C$.

\n

Click on Show steps for help if you need it. You will lose 1 mark if you do so.

\n ", "steps": [{"type": "information", "prompt": "\n

First of all factorise the denominator.

\n

You have to find $a$ and $b$ such that $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+a)*(x+b)}$

\n

Then use partial fractions to write:
\\[\\simplify[std]{({c}*x+{d})/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]

\n

for suitable integers or fractions $A$ and $B$.

\n ", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\n

Find the following integral.

\n

\\[I = \\simplify[std]{Int(({c}*x+{d})/(x^2+{a+b}*x+{a*b}),x )}\\]

\n

Input all numbers as fractions or integers and not decimals.

\n

Input 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\t

5/08/2012:

\n \t\t

Added tags.

\n \t\t

Added description.

\n \t\t

Added decimal point as forbidden string.

\n \t\t

Note 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

Improved display of Advice. 

\n \t\t

Added information about Show steps, also introduced penalty of 1 mark.

\n \t\t

Added !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:

\n

$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$

\n

Identifying coefficients:

\n

Constant term: $\\simplify[std]{{b}*A+{a}*B = {d}}$

\n

Coefficent $x$: $ \\simplify[std]{A+B={c}}$ which gives $A =\\var{c} -B$

\n

On solving these equations we obtain $\\displaystyle \\simplify[std]{A = {d-a*c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={d-b*c}/{a-b}}$

\n

Which 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}))}\\]

\n

So \\[\\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*}\\]

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Input all numbers as fractions or integers and not decimals.

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$I=\\;$[[0]]

\n

Input all numbers as fractions or integers and not decimals.

\n

Input the constant of integration as $C$.

\n

Click on Show steps for help if you need it. You will lose 1 mark if you do so.

\n \n ", "steps": [{"type": "information", "prompt": "\n

Use partial fractions in order to write:
\\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\\; = \\simplify[std]{A/(x+{a})+B/(x+{b})}\\]

\n

for suitable integers or fractions $A$ and $B$.

\n ", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\n

Find the following integral.

\n

\\[I = \\simplify[std]{Int(({c}*x+{d})/((x +{a})*(x+{b})),x )}\\]

\n

Input all numbers as fractions or integers and not decimals.

\n

Input 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\t

5/08/2012:

\n \t\t \t\t

Added tags.

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Added description.

\n \t\t \t\t

Added decimal point as forbidden string.

\n \t\t \t\t

Note 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\t

Improved display of Advice. 

\n \t\t \t\t

Added information about Show steps, also introduced penalty of 1 mark.

\n \t\t \t\t

Added !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": "\n

Using 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:

\n

$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$

\n

Identifying coefficients:

\n

Constant term: $\\simplify[std]{{b}*A+{a}*B = {d}}$

\n

Coefficent $x$: $ \\simplify[std]{A+B={c}}$ which gives $A =\\var{c} -B$

\n

On solving these equations we obtain $\\displaystyle \\simplify[std]{A = {d-a*c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={d-b*c}/{a-b}}$

\n

Which 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}))}\\]

\n

So \\[\\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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You have 5  minutes left.

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This is only for practice. It is not a summative test.

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