Numbas demo: video

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Christian Lawson-Perfect

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Christian Lawson-Perfect

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Ready to use

8 years, 9 months ago

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Christian Lawson-Perfect 8 years, 9 months ago

Gave some feedback: Ready to use

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Christian Lawson-Perfect 9 years, 9 months ago

Created this as a copy of Integration by partial fractions. (Video).

Name	Status	Author	Last Modified
Integration by partial fractions. (Video)	Ready to use	Bill Foster	01/06/2016 09:47
Numbas demo: video	Ready to use	Christian Lawson-Perfect	25/08/2020 15:00
YJ's copy of Numbas demo: video	Ready to use	YJ Kim	01/06/2016 09:47
Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 14:06
Simon's copy of Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 14:09
Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 15:18
Simon's copy of Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 15:31
Simon's copy of Simon's copy of Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 15:33

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Question statement

It's easy to include videos in Numbas questions. In this question, if the student gets stuck they can click on "Show steps" to be given a hint, and shown a video of someone working through a similar problem.

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Find the following integral.

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

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
c	integer	7
b	integer	8
d	integer	-7
s3	integer	-1
s2	integer	1
s1	integer	1
b1	integer	8
d1	integer	-7

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Value

xxxxxxxxxx
 
s1*random(1..9)

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Generated value: `integer`

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Parts

Gap-fill
1. Gap Mathematical expression
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1. Step Information only

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$I=$ Unnamed gap

Enter the constant of integration as $C$ .

Click on Show steps for help if you need it: you'll be given a hint, and see a video which solves a similar example.

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

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

Multiplying both sides of the equation by $\displaystyle \simplify[std]{1/((x +{a})*(x+{b}))}$ , we obtain

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

Coefficients of similar powers of $x$ on each side of the equation must be equal, so we can write down two new equations identifying the coefficients on each side:

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

Coefficent of $x$ : $\simplify[std]{A+B={c}}$ 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

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

$\begin{align} I &= \simplify[std]{int(({c}*x+{d})/(x^2+{a+b}*x+{a*b}),x )} \\[0.5em] &= \simplify[std]{int(({c}*x+{d})/((x +{a})*(x+{b})),x )} \\[0.5em] &= \simplify[std]{({d-a*c}/{b-a})*(int(1/(x+{a}),x)) +({d-b*c}/{a-b})int(1/(x+{b}),x)} \\[0.5em] &= \simplify[std]{({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C} \end{align}$

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Christian Lawson-Perfect 8 years, 9 months ago

Christian Lawson-Perfect 9 years, 9 months ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Numbas demo: video

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 9 months ago

Christian Lawson-Perfect 9 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`