3. Change of Type (2 to 1)

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Question statement

Consider the integral

$\int_{\frac{1}{\var{c}}\simplify{e^({a-d}/{b})}}^{\var{q}} \int_\var{a}^{\simplify{{b}ln({c}x)+{d}}} f(x,y) \, \mathrm{d}y \, \mathrm{d}x \,.$

The domain of integration is shown below.

{plot()}

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			Name	Type	Generated Value

gcd0	number	1
gcd1	number	1
k	rational	-57
gcd2	number	1
gcd3a	number	1
gcd3b	number	3
gcd4	number	1

			Name	Type	Generated Value

a	integer	0
b	integer	3
c	integer	5
d	integer	-3
p	number	0.5436563657
q	integer	1
s	integer	1
t	integer	2

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Name

Data type

Value

xxxxxxxxxx
 
gcd((s*q+2*t)*d*q,2)

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Generated value: `number`

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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Mathematical expression
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Part c) Mathematical expression

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Of which type is this integral?

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This integral is type 2, because the order of integration is $y$, then $x$.

Sketch the graph, looks like this:

From looking at the graph we can see that the curve $y=\simplify{{b}ln({c}x)+{d}}$, which was the upper limit for $y$ in type 2, is now the lower limit for $x$ in type 1. So we need to make $x$ the subject of the equation:

\begin{align}
\simplify{{b}ln({c}x)+{d}}=\,& y \\
\var{b}\ln(\var{c}x) =\,& \simplify{y-{d}} \\
\ln(\var{c}x) =\,& \simplify{(y-{d})/{b}} \\
\var{c}x =\,& e^{\simplify{(y-{d})/{b}}} \\
x =\,& \frac{1}{\var{c}}e^{\simplify{(y-{d})/{b}}}.
\end{align}

The upper limit is just the straight line: $x=\var{q}$.

Now for $y$ the bottom limit is simply the line $y=\var{a}$, and the top limit is the point of intersection between $y=\simplify{{b}ln({c}x)+{d}}$ and $x=\var{q}$, so $y=\simplify{{b}ln{c*q}+{d}}$.

Finally, make sure you also swap $\mathrm{d}x \, \mathrm{d}y$ since the integral is now in type 1.

Answer:

\[ \int_\var{a}^{\simplify{{b}ln{c*q}+{d}}} \int_{\frac{1}{\var{c}}e^{\simplify[fractionNumbers]{(y-{d})/{b}}}}^{\var{q}} f(x,y) \, \mathrm{d}x \, \mathrm{d}y \]

\begin{align}
\int_\var{a}^{\simplify{{b}ln{c*q}+{d}}} \int_{\frac{1}{\var{c}}e^{\simplify{(y-{d})/{b}}}}^{\var{q}} \simplify{{s}x}+\var{t} \, \mathrm{d}x \, \mathrm{d}y

&=\int_\var{a}^{\simplify{{b}ln{c*q}+{d}}} \simplify{{s}/{2}x^2}+\simplify{{t}x} \, \bigg|_{x=\frac{1}{\var{c}}e^{\simplify{(y-{d})/{b}}}}^{x=\var{q}} \, \mathrm{d}y
\\
&=\int_\var{a}^{\simplify{{b}ln{c*q}+{d}}} \simplify[fractionNumbers]{{s*q^2/2}}+\var{t*q} - \simplify[fractionNumbers]{{s/2}}\cdot\frac{1}{\var{c}^2}e^{2\simplify{(y-{d})/{b}}}-\frac{\var{t}}{\var{c}}e^{\simplify{(y-{d})/{b}}} \, \mathrm{d}y
\\
&=\int_\var{a}^{\simplify{{b}ln{c*q}+{d}}} \simplify[fractionNumbers]{{s*q^2/2+t*q}} - \simplify[fractionNumbers]{{s/(2*c^2)}}e^{\simplify{2*(y-{d})/{b}}}-\simplify[fractionNumbers]{{t/c}}e^\simplify{(y-{d})/{b}} \, \mathrm{d}y
\\
\end{align}

\begin{align}
\int_\var{a}^{\simplify{{b}ln{c*q}+{d}}} \simplify[fractionNumbers]{{s*q^2/2+t*q}} - \simplify[fractionNumbers]{{s/(2*c^2)}}e^{\simplify{2*(y-{d})/{b}}}-\simplify[fractionNumbers]{{t/c}}e^\simplify{(y-{d})/{b}} \, \mathrm{d}y

&=\simplify[fractionNumbers]{{s*q^2/2+t*q}}y - \simplify[fractionNumbers]{{b*s/(4c^2)}}e^{\simplify{2*(y-{d})/{b}}}-\simplify[fractionNumbers]{{b*t/c}}e^\simplify{(y-{d})/{b}} \bigg|_{y=\var{a}}^{y=\simplify{{b}ln{c*q}+{d}}}
\\
&=\simplify[fractionNumbers]{{(s*q/2+t)*b*q}ln{c*q}} + \simplify{({(s*q+2*t)*d*q/gcd0}/{2/gcd0})} - \simplify[fractionNumbers]{{b*s/(4c^2)}}(\var{c*q})^2-\simplify[fractionNumbers]{{b*t/c}}\cdot\var{c*q} - \simplify[fractionNumbers]{{(s*q/2+t)*a*q}} + \simplify[fractionNumbers]{{b*s/(4c^2)}}\simplify{e^({2*(a-d)}/{b})} + \simplify[fractionNumbers]{{b*t/c}}\simplify{e^({a-d}/{b})}
\\
&=\simplify[fractionNumbers]{{(s*q/2+t)*b*q}ln{c*q}} + \simplify{({k/gcd2}/{4/gcd2})} + \simplify[fractionNumbers]{{b*s/(4c^2)}}\simplify{e^({2*(a-d)}/{b})} + \simplify[fractionNumbers]{{b*t/c}}\simplify{e^({a-d}/{b})}
\end{align}

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This question is used in the following exam:

Test exam by Tyson Ritter in Tyson's workspace.

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3. Change of Type (2 to 1)

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Nicholas Barker 7 years, 1 month ago

Nicholas Barker 7 years, 1 month ago

Nicholas Barker 7 years, 5 months ago

Error in variable testing condition

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Generated value: `number`