Simon's copy of Integration by parts

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Simon Thomas 6 years ago

Created this as a copy of Integration by parts.

Name	Status	Author	Last Modified
Integration by parts	Ready to use	Bill Foster	01/06/2016 09:47
Integration by parts, Exponential 31/10	Ready to use	Katie Dexter	01/06/2016 09:47
Phil's copy of Integration by parts, Exponential 20/10	Ready to use	Phil Barker	01/06/2016 09:47
Simon's copy of Integration by parts	draft	Simon Thomas	21/03/2019 10:32
Integration by parts, Exponential 20/10	draft	Xiaodan Leng	11/07/2019 06:43

There are 15 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	5
c	integer	-2
b	integer	7
s3	integer	-1
s2	integer	-1
s1	integer	1
a1	integer	6
a2	integer	-2

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Description

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Generated value: `integer`

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Parts

Part a) Gap-fill
1. Gap Mathematical expression
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1. Step Information only
Part b) Gap-fill
1. Gap Mathematical expression
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$I=\displaystyle \int \simplify[std]{({a}x+{b})*e^({c}x)} dx$
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

a)

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

b)

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

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This question is used in the following exams:

Methods of integration by Simon Thomas in Maths support.
Integration: Mixed questions by Simon Thomas in Calculus.

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Simon Thomas 6 years ago

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Generated value: `integer`

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Simon's copy of Integration by parts

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

a)

b)

Generated value: `integer`