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    <name>
      <text>29.1.1.3 line integral with constant scalar field</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p>Find
<ol class="HELM_parts_inline">
<li>\(\displaystyle \int_C F(x,y)\ \mathrm{d}x,\)</li>
<li>\(\displaystyle \int_C F(x,y)\ \mathrm{d}y,\)</li>
<li>\(\displaystyle \int_C F(x,y)\ \mathrm{d}s\)</li>
</ol>
where \(F (x, y) = 1\) and \(C\) is the curve \(y = \frac{1}{2} x^2 − \frac{1}{4} \ln x\) from \(({@aa@}, \frac{1}{2} )\) to \(({@aa+1@}, 2 − \frac{1}{4} \ln ({@bb@}))\).</p>

<p>\(\displaystyle \int_C F(x,y)\ \mathrm{d}x =\)[[input:ans1]] [[validation:ans1]][[feedback:prt1]]</p>
<p>\(\displaystyle \int_C F(x,y)\ \mathrm{d}y =\)[[input:ans2]] [[validation:ans2]][[feedback:prt2]]</p>
<p>\(\displaystyle \int_C F(x,y)\ \mathrm{d}s =\)[[input:ans3]] [[validation:ans3]][[feedback:prt3]]</p>]]></text>
    </questiontext>
    <generalfeedback format="moodle_auto_format">
      <text><![CDATA[<p>We know that \(F (x, y) = 1\) and \(C\) is the curve \(y = \frac{1}{2} x^2 − \frac{1}{4} \ln x\) from \(({@aa@}, \frac{1}{2} )\) to \(({@aa+1@}, 2 − \frac{1}{4} \ln {@bb@})\), and aim to find: (a)\(\displaystyle \int_C F(x,y)\ \mathrm{d}x,\) (b)\(\displaystyle \int_C F(x,y)\ \mathrm{d}y,\) and (c)\(\displaystyle \int_C F(x,y)\ \mathrm{d}s\).</p>

<p><strong>(a)</strong>
\[\begin{aligned}
\int_C F(x,y)\ \mathrm{d}x &= \int^{@aa+1@}_{@aa@} 1\ \mathrm{d}x\\
&= \left[x\right]^{@aa+1@}_{@aa@}\\
&= {@aa+1@}-{@aa@}\\
&=1\end{aligned}\]</p>

<p><strong>(b)</strong>
\[\begin{aligned}
\int_C F(x,y)\ \mathrm{d}y &= \int^{2 - \frac{1}{4} \ln {@bb@}}_\frac{1}{2} 1 \mathrm{d}y \\
&= (2-\frac{1}{4}\ln{@bb@})-(\frac{1}{2}) \\
& = {@3/2- bb/4@}\end{aligned}\]</p>

<p><strong>(c)</strong> we first aim to express everything in terms of x
\[\begin{aligned}
\mathrm{d}s &= \sqrt{1+(\frac{\mathrm{d}y}{\mathrm{d}x})^2}\mathrm{d}x\\
&= \sqrt{1+(x-\frac{1}{4x})^2}\mathrm{d}x
\end{aligned}\]</p>

Hence:
\[\begin{aligned}
\int_C F(x,y)\ \mathrm{d}s &= \int^{@aa+1@}_{@aa@}\sqrt{1+(x-\frac{1}{4x})^2}\mathrm{d}x \\
&= \int^{@aa+1@}_{@aa@}\sqrt{x^2+\frac{1}{2}+\frac{1}{16x^2}}\mathrm{d}x\\
&=\int^{@aa+1@}_{@aa@}\left(x+\frac{1}{4x}\right)\ \mathrm{d}x\\
&=\left({@aa+1@}+\frac{1}{{@4*(aa+1)@}}\right) - \left({@aa@}+\frac{1}{@4*aa@}\right)\\
&={@1+1/(4*(aa+1)) - 1/(4*aa)@}\end{aligned}\]</p>]]></text>
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      <text>aa:rand(4)+1;
bb:rand_with_prohib(-3,2,[0])

ta1:1;
ta2:3/2- bb/4;
ta3:1+1/(4*(aa+1)) - 1/(4*aa);</text>
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    <questionnote format="html">
      <text>\(C\) from \(({@aa@}, \frac{1}{2} )\) to \(({@aa+1@}, 2 − \frac{1}{4} \ln {@bb@}), \quad \displaystyle \int_C F(x,y)\ \mathrm{d}y = {@ta2@}, \quad \displaystyle \int_C F(x,y)\ \mathrm{d}s = {@ta3@}\)</text>
    </questionnote>
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      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:orange;"><i class="fa fa-adjust"></i></span> Your answer is partially correct.]]></text>
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    <prtincorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:red;"><i class="fa fa-times"></i></span> Incorrect answer.]]></text>
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