Integration by parts

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Simon Vaughan

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4 years, 9 months ago

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Simon Vaughan 4 years, 9 months ago

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Simon Vaughan 4 years, 9 months ago

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Simon Vaughan 4 years, 9 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
Integration by parts	Ready to use	Simon Vaughan	20/08/2020 13:01

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Question statement

Find the following indefinite integrals.

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ where needed.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	1
c	integer	2
b	integer	5
s3	integer	1
s2	integer	1
s1	integer	1
a1	integer	8
a2	integer	2

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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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Part c) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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$\displaystyle \int \simplify[std]{{a}x*sin({b}x+{c})} dx = \phantom{{}}$ Unnamed gap

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

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Advice

The formula for integrating by parts is

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

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

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

Hence,

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

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

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

Hence,

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

Using the results from Parts a and b, we have

$\begin{eqnarray*}I &=& \int \simplify[std]{{a1}x*sin({b}x+{c})} dx + \int \simplify[std]{{a2}x*cos({b}x+{c})} dx\\ &=& \simplify[std]{{a1}*((-{a}/{b})*x*cos({b}x+{c}) + ({a}/{b^2})*sin({b}x+{c}))+{a2}*(({a}/{b})*x*sin({b}x+{c}) +({a}/{b^2})*cos({b}x+{c}))+C}\\ &=&\simplify[std]{(-{a1}/{b})*x*cos({b}x+{c}) + ({a1}/{b^2})*sin({b}x+{c})+({a2}/{b})*x*sin({b}x+{c}) +({a2}/{b^2})*cos({b}x+{c}) + C}\\ &=&\simplify[std]{({a2}/{b}*x+{a1}/{b^2})*sin({b}x+{c})+({-a1}/{b}*x+{a2}/{b^2})*cos({b}x+{c})+C} \end{eqnarray*}$
Hence

$\displaystyle \simplify[std]{f(x) = {a2}/{b}*x+{a1}/{b^2}}$

$\displaystyle \simplify[std]{g(x) = {-a1}/{b}*x+{a2}/{b^2}}$

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Simon Vaughan 4 years, 9 months ago

Simon Vaughan 4 years, 9 months ago

Simon Vaughan 4 years, 9 months ago

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Generated value: `integer`

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Integration by parts

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Simon Vaughan 4 years, 9 months ago

Simon Vaughan 4 years, 9 months ago

Simon Vaughan 4 years, 9 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`