Luis's copy of Integration by parts

Error

There was an error loading the page.

Contributors

History

○

Luis Hernandez 6 years, 4 months ago

Created this as a copy of Integration by parts.

Name	Status	Author	Last Modified
Integration by parts	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Luis's copy of Integration by parts	draft	Luis Hernandez	30/11/2018 18:35

There are 17 other versions that do you not have access to.

Question statement

Find the following indefinite integrals.

Input all numbers as fractions or integers and not decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
c	integer	-2
b	integer	2
s3	integer	-1
s2	integer	-1
s1	integer	1
a1	integer	4
a2	integer	-2

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..5)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
1. Step Information only
Part b) Gap-fill
1. Gap Mathematical expression
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$I=\displaystyle \int \simplify[std]{({a}x+{b})*e^({c}x)} dx$
You are given that the answer is of the form

$I=g(x)e^{\var{c}x}+C$ for a polynomial

$g(x)$ . You have to find

$g(x)$ .

$g(x)=\;$ Unnamed gap

Input all numbers as fractions or integers and not decimals.

You can get help by clicking on Show steps. You will lose 1 mark if you do.

Advice

The formula for integrating by parts is

$\int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.$

We choose $u = \simplify[std]{{a}x+{b}}$ and $\displaystyle\frac{dv}{dx} = \simplify[std]{e^({c}x)}$ .

So $\displaystyle \frac{du}{dx} = \var{a}$ and $\displaystyle v = \simplify[std]{(1/{c})*e^({c}*x)}$ .

Hence,

$\begin{eqnarray} \int \simplify[std]{({a}*x+{b})*e^({c}*x)} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{({1}/{c})*({a}x+{b})*e^({c}x) - (1/{c})*Int(({a})*e^({c}x),x)} \\ &=& \simplify[std]{(1/{c})*({a}x+{b})*e^({c}x) -({a}/{c^2})*e^({c}x) + C}\\ &=& \simplify[std]{(({a}x+{b})/{c}-{a}/{c^2})*e^({c}*x) + C}\\ &=& \simplify[std]{(({a}/{c})x+{b*c-a}/{c^2})*e^({c}*x) + C} \end{eqnarray}$

Hence $\displaystyle \simplify[std]{g(x)=({a}/{c})*x+{c*b-a}/{c^2}}$

For this part we choose $u = \simplify[std]{({a}x+{b})^2}$ and $\displaystyle \frac{dv}{dx} = \simplify[std]{e^({c}x)}$ .

So $\displaystyle \frac{du}{dx}$ = $\simplify[std]{{2*a}*({a}*(x)+{b})}$ and $\displaystyle v = \simplify[std]{(1/{c})*e^({c}*x)}$ .

Hence,

$\begin{eqnarray*}I= \int \simplify[std]{({a}*x+{b})^2*e^({c}*x)} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{({1}/{c})*({a}x+{b})^2*e^({c}x) - (1/{c})*Int({2*a}*({a}x+{b})*e^({c}x),x)} \\ &=& \simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*Int(({a}x+{b})*e^({c}x),x)}\dots (*) \end{eqnarray*}$

But in Part a we have aready worked out $\displaystyle \simplify[std]{Int(({a}x+{b})*e^({c}*x),x)=(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}$

So on substituting this in equation (*) we find:

$\begin{eqnarray*}I&=& \simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}\\ &=& \simplify[std]{({a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3})*e^({c}x) +C} \end{eqnarray*}$

Hence $\displaystyle \simplify[std]{h(x)={a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3}}$

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet
Information only
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet

This question is not used in any exams.

Loading...

Error

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Luis Hernandez 6 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

Loading...

Error

Luis's copy of Integration by parts

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Luis Hernandez 6 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`