Integration by Partial Fractions 1 - double root

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joshua boddy

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Jinhua Mathias 4 years, 2 months ago

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joshua boddy 9 years, 7 months ago

Created this as a copy of Partial Fractions 3 - double root.

Name	Status	Author	Last Modified
Partial Fractions	draft	Bill Foster	06/08/2020 08:42
Partial Fractions 1	Ready to use	joshua boddy	01/06/2016 09:47
Partial Fractions 2	Ready to use	joshua boddy	01/06/2016 09:47
Partial Fractions 3 - double root	Ready to use	joshua boddy	01/06/2016 09:47
Integration with Partial Fractions 2	Ready to use	joshua boddy	01/06/2016 09:47
Integration by Partial Fractions 1 - double root	Has some problems	joshua boddy	23/02/2021 13:55
Partial Fractions: Simple Linear Factors	Ready to use	Michael Proudman	07/12/2021 10:59
Partial Fractions: Repeated Factor	Ready to use	Michael Proudman	07/12/2021 10:59
CF Maths In class test three mick paper question 7	Ready to use	Jinhua Mathias	28/08/2024 12:15
CF Maths January test mock paper Partial Fractions 3 - double root	Ready to use	Jinhua Mathias	28/08/2024 12:15
CF Maths Januaray test mock paper Partial Fractions 1	Ready to use	Jinhua Mathias	28/08/2024 12:15
Partial Fractions: Simple Linear Factors [L5]	draft	Abbi Mullins	25/07/2018 14:02
cormac's copy of CF Maths January test mock paper Partial Fractions 3 - double root	draft	cormac breen	30/10/2018 15:53
Blathnaid's copy of Partial Fractions: Simple Linear Factors	draft	Blathnaid Sheridan	13/11/2018 10:42
Blathnaid's copy of Michael's copy of Partial Fractions 3 - double root	draft	Blathnaid Sheridan	13/11/2018 10:43
Laura's copy of Partial Fractions: Repeated Factor	draft	Laura Horton	12/02/2019 16:22

There are 43 other versions that do you not have access to.

Question statement

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			Name	Type	Generated Value

a	integer	3
c	integer	-4
b	integer	-1
nb	string	taking away
s1	integer	-1
a_	integer	2
c_	integer	-4
b_	integer	5
nb_	string	taking away
a1_	integer	1
a2_	integer	1
s1_	integer	-1
new	integer	1
a1	integer	1
a2	integer	4
a3	integer	-1
d	integer	1
d_	integer	1
b1	integer	1
b2	integer	5
b3	integer	-1
p	integer	9
q	integer	4
c1	integer	-2
c2	integer	-3
c3	integer	3
p1	integer	5
q1	integer	-3
i1	integer	3
i0	integer	1

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Value

xxxxxxxxxx
 
random(1..9)

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Generated value: `integer`

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Parts

Part a) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Gap-fill

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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.

Input the partial fractions here: Gap 0..

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

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

Advice

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

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

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

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

Identifying coefficients:

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

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

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

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

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

Integration:

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 $\int \frac{1}{(x+a)^2} \mathrm{d}x = \frac{-1}{x+a}$

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

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

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This question is used in the following exam:

Simple integration and integration using partial fraction by Jinhua Mathias in Maths 2.

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Integration by Partial Fractions 1 - double root

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Jinhua Mathias 4 years, 2 months ago

joshua boddy 9 years, 7 months ago

joshua boddy 9 years, 7 months ago

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Integration by Partial Fractions 1 - double root

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Jinhua Mathias 4 years, 2 months ago

joshua boddy 9 years, 7 months ago

joshua boddy 9 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`