// Numbas version: exam_results_page_options {"name": "Integration by Partial Fractions 1 - double root", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "c", "b", "nb", "s1", "a_", "c_", "b_", "nb_", "a1_", "a2_", "s1_", "new", "a1", "a2", "a3", "d", "d_", "b1", "b2", "b3", "p", "q", "c1", "c2", "c3", "p1", "q1", "i1", "i0"], "name": "Integration by Partial Fractions 1 - double root", "tags": ["algebra", "algebraic fractions", "algebraic manipulation", "combining algebraic fractions", "common denominator"], "preamble": {"css": "", "js": ""}, "advice": "

a)

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We use partial fractions to find $A$, $B$ and $C$ such that:
$\\simplify{({a1+a3}x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))} `\\;\\;\\;=\\simplify{A/(x+{a})+B/(x+{a})^2+C/(x+{b})}$

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Dividing both sides of the equation by $\\displaystyle \\simplify[std]{1/( (x+{a})^2(x+{b}) )}\\;\\;$ we obtain:

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$\\simplify{A(x+{a})(x+{b})+B(x+{b})+C(x+{a})^2 = {a1+a3}*x^2+{a1*a+a1*b+a2+2*a*a3}*x + {a1*a*b + a2*b + a3*a^2}}$

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$\\Rightarrow \\simplify[std]{(A+C)x^2+({a+b}A+B+{2a}C)x+({a*b}A+{b}B+{a*a}C)={a1+a3}*x^2+{a1*a+a1*b+a2+2*a*a3}*x + {a1*a*b + a2*b + a3*a^2}}$

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Identifying coefficients:

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Coefficient $x^2$: $\\simplify[std]{A+C={a1+a3} }$

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Coefficent $x$: $\\simplify[std]{ {a+b}A+B+{2a}C = {a1*a+a1*b+a2+2*a*a3} }$

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Constant term: $\\simplify{{a*b}A+{b}B+{a*a}C ={a1*a*b + a2*b + a3*a^2}}$

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On solving these equations we obtain $A = \\var{a1}$, $B=\\var{a2}$ and $C=\\var{a3}$

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Which gives:$\\simplify{({a1+a3}x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))} \\;\\;\\;=\\simplify{{a1}/(x+{a})+{a2}/(x+{a})^2+{a3}/(x+{b})}$

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Integration:

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For the integration, we use our answer for the partial fractions to help use and then recall that $\\int \\frac{1}{x+a}\\mathrm{d}x = \\ln(x+a)$ and $\\int \\frac{1}{(x+a)^2} \\mathrm{d}x = \\frac{-1}{x+a}$

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Hence we apply this to our problem to find $\\int \\simplify{(({a1+a3})x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))} \\mathrm{d}x = \\int \\simplify{{a1} / (x + {a}) + {a2}/(x + {a})^2 + {a3} / (x + {b})} \\mathrm{d}x = \\simplify{{a1}*ln(x + {a}) - {a2}/((x + {a}))+ {a3}*ln(x + {b})}$

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Apply same method to solve b), but substituting in the appropriate integral limits to get a numerical answer.

", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "

Split \$\\simplify{(({a1+a3})x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))}\$ into partial fractions.

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Input the partial fractions here: [[0]].

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Hence evaluate $\\int \\simplify{(({a1+a3})x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))} \\mathrm{d}x =$[[1]]

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(You may assume that the constant on integration is 0)

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Input as the sum of partial fractions.

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Split \$\\simplify{(({b1+b3})x^2+{b1*p+b1*q+b2+2*p*b3} * x + {b1*p*q + b2*q + b3*p^2})/ ((x + {p})^2 * (x + {q}))}\$ into partial fractions.

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Input the partial fractions here: [[0]].

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Hence evaluate $\\int^\\var{i1}_\\var{i0} \\simplify{(({b1+b3})x^2+{b1*p+b1*q+b2+2*p*b3} * x + {b1*p*q + b2*q + b3*p^2})/ ((x + {p})^2 * (x + {q}))} \\mathrm{d}x =$[[1]]

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Input as the sum of partial fractions.

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5/08/2012:

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Added tags.

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Added description.

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Changed to two questions, for the numerator and denomimator, rather than one as difficult to trap student input for this example. Still some ambiguity however.

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12/08/2012:

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Back to one input of a fraction and trapped input in Forbidden Strings.

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Used the except feature of ranges to get non-degenerate examples.

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Checked calculation.OK.

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Improved display in content areas.

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