CF Maths In class test three question 8 Integration by partial fractions with limits

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Created this as a copy of Integration by partial fractions.

Name	Status	Author	Last Modified
Integration by partial fractions	Ready to use	Bill Foster	01/06/2016 09:47
CF Maths In class test three question 8 Integration by partial fractions with limits	Ready to use	Jinhua Mathias	28/08/2024 12:15
Andrew's copy of CF Maths In class test three question 8 Integration by partial fractions with limits	Ready to use	Andrew Dunbar	01/06/2016 13:58
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 14:06
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 14:07
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 15:10

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Question statement

Find the following integral.

$I = \simplify[std]{defint(({c}*x+{d})/((x +{a})*(x+{b})),x,10,12 )}$

Input all numbers as fractions or integers and not decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	7
c	integer	3
b	integer	2
d	integer	7
s3	integer	1
s2	integer	1
s1	integer	1
b1	integer	2
d1	integer	6

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s1*random(1..9)

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Parts

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1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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3. Gap Gap 2. Number entry
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Ask the student a question, and give any hints about how they should answer this part.

First, input the sum of partial fraction

Gap 0.

Second, input the integral without the limit

$I=\;$ Gap 1.

Input all numbers as fractions or integers and not decimals.

Do not need to input $C$ , the constant of integration.

The last, substitute the limits to the integral and calculate it.

Gap 2., accurate to 2 decimal places.

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

Advice

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:

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

Identifying coefficients:

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

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

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

Which gives:

$\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\; =\simplify[std]{ ({d-a*c}/{b-a})*(1/(x+{a}) )+({d-b*c}/{a-b})*(1/(x+{b}))}$

$\begin{eqnarray*} I &=& \simplify[std]{Int(({c}*x+{d})/((x +{a})*(x+{b})),x )}\\ &=& \simplify[std]{({d-a*c}/{b-a})*(Int(1/(x+{a}),x)) +({d-b*c}/{a-b})Int(1/(x+{b}),x)}\\ &=& \simplify[std]{({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})} \end{eqnarray*}$

For the last step substitute the upper limt and lower limit separately to the integral, and the value from upper limt take away from the lower one to give the final answer.

$\simplify[std]{({d-a*c}/{b-a})*ln(12+{a})+({d-b*c}/{a-b})*ln(12+{b})}$ take away

$\simplify[std]{({d-a*c}/{b-a})*ln(10+{a})+({d-b*c}/{a-b})*ln(10+{b})}$

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CF Maths In class test three question 8 Integration by partial fractions with limits

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Christian Lawson-Perfect 8 years, 9 months ago

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CF Maths In class test three question 8 Integration by partial fractions with limits

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Taxonomy: mathcentre

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Christian Lawson-Perfect 8 years, 9 months ago

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Generated value: integer

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Generated value: `integer`