Integration by parts 1 with limits

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Christian Lawson-Perfect

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Christian Lawson-Perfect 8 years, 9 months ago

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joshua boddy 9 years, 6 months ago

Created this as a copy of Integration by parts 1.

Name	Status	Author	Last Modified
Integration by parts	Ready to use	Bill Foster	01/06/2016 09:47
Katie's copy of Integration by parts	Ready to use	Katie Dexter	01/06/2016 09:47
Integration by parts 1	Ready to use	joshua boddy	01/06/2016 09:47
Integration by parts 1 with limits	Ready to use	joshua boddy	01/06/2016 09:47
Integration by parts - Ch 1	Ready to use	Katie Dexter	01/06/2016 09:47
Ch1: Integration by parts	Ready to use	Graham Wynn	01/06/2016 09:47

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Question statement

Using integration by parts, find the following indefinite integral $I=\displaystyle \int^\var{i1}_\var{i0} \simplify[std]{x*({b}x+{c})^{m}} dx$

You need to know that for $n \neq -1$ :

$\int (ax+b)^n dx = \frac{1}{a(n+1)}(ax+b)^{n+1}+C$

Give any introductory information the student needs.

			Name	Type	Generated Value

s3	integer	-1
c	integer	-7
b	integer	3
m	integer	6
i1	integer	3
i0	integer	1

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Generated value: `integer`

-1

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1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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The formula for integrating by parts is

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

What is the most suitable choice for $u$ and $\frac{dv}{dx}$ ?

$u=\;$ Gap 0.

$\frac{dv}{dx}=\;$ Gap 1. $+C$

Advice

The formula for integrating by parts is

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

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

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

Hence,

$\begin{eqnarray*} \displaystyle \int^\var{i1}_\var{i0} \simplify[std]{x*({b}x+{c})^{m}} dx &=& [uv]^\var{i1}_\var{i0} - \int^\var{i1}_\var{i0} v \frac{du}{dx} dx \\ &=& \simplify[std]{[(x/{(m+1)*b})*({b}*x+{c})^{m+1}]}^\var{i1}_\var{i0} -\simplify{ (1/{(m+1)*b})}*\int^\var{i1}_\var{i0}(\var{b}*x+\var{c})^{\var{m}+1} \mathrm{d}x \\ &=& \simplify[std]{[(x/{(m+1)*b})*({b}*x+{c})^{m+1}]}^\var{i1}_\var{i0} - \simplify{(1/{(m+1)*(m+2)*b^2})*({b}*x+{c})^{m+2}}^\var{i1}_\var{i0} \\ &=&[\simplify[std]{({b}*x+{c})^{m+1}/{(m+1)*(m+2)*b^2}*({b*(m+2)}x - ({b}x+{c}))}]^\var{i1}_\var{i0}\\ &=&[\simplify[std]{({b}*x+{c})^{m+1}/{(m+1)*(m+2)*b^2}*({b*(m+1)}x - {c})}]^\var{i1}_\var{i0} \end{eqnarray*}$

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Christian Lawson-Perfect 8 years, 9 months ago

joshua boddy 9 years, 6 months ago

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Integration by parts 1 with limits

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 9 months ago

joshua boddy 9 years, 6 months ago

Error in variable testing condition

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Generated value: integer

Parts

Gap-fill

Generated value: `integer`