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      <text>14.1.1.3 Sum of Areas</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4><p>We continue from the end of the pervious exercise and use the limit of a sum to calculate \(\displaystyle {\int_0^1 x^2 \, \mathrm{d}x}\).</p>
<p>To find the total area <span class="math inline">\(A_n\)</span> of the <span class="math inline">\(n\)</span> rectangles we must add up all these individual rectangular areas: \[A_n=\sum_{k=1}^n {(k-1)^2\over n^3}\]</p>
<p>This sum can be simplified and then calculated as follows. You will need to make use of the formulas for the sum of the first <span class="math inline">\(n\)</span> integers, and the sum of the squares of the first <span class="math inline">\(n\)</span>    integers:</p>

<p>\[\sum_{k=1}^n1=n,\qquad \sum_{k=1}^nk=\frac{1}{2}n(n+1),\qquad \sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1)\]
</p>

<p>Then, the total area of the rectangles is given by \[ \begin{aligned} A_n&amp;=\sum_{k=1}^n {(k-1)^2\over n^3}\\&amp;= {1\over n^3}\sum_{k=1}^n (k-1)^2\\ &amp;= {1\over n^3}\sum_{k=1}^n (k^2 -2k +1) \\ &amp;= {1\over n^3}\left(\sum_{k=1}^n k^2 -2\sum_{k=1}^nk
    +\sum_{k=1}^n1 \right)\\ &amp;= {1\over n^3}\left({n\over 6}(n+1)(2n+1) - 2{n\over 2}(n+1) + n\right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &amp;= {1\over n^2}\left({(n+1)(2n+1)\over 6} - (n+1)+1\right)\\ &amp;= {1\over n^2}\left({(n+1)(2n+1)\over
    6}-n\right)\\ &amp;= {1\over 6n^2}\left(2n^2-3n+1\right)= {1\over 3} - {1\over 2n} +{1\over 6n^2} \end{aligned} \]</p>

<p>Note that this is a formula for the <span><strong>exact</strong></span> total area of the <span class="math inline">\(n\)</span> rectangles. It is an <span><strong>estimate</strong></span> of the area under the graph of <span class="math inline">\(y=x^2\)</span>.
    However, as <span class="math inline">\(n\)</span> gets larger, the terms <span class="math inline">\(\dfrac{1}{2n}\)</span> and <span class="math inline">\(\dfrac{1}{6n^2}\)</span> become small and will eventually tend to zero. If we let <span class="math inline">\(n\)</span>    tend to infinity we obtain the exact answer of <span class="math inline">\({\dfrac13}\)</span>.</p>

<p>The required area is <span class="math inline">\(\dfrac13\)</span>. It has been found as <span><strong>the limit of a sum</strong></span> and of course agrees with that calculated by integration.</p>

<p>In the calculations which follow in future work the need to evaluate complicated limits like this is avoided by performing integration techniques. Nevertheless it will sometimes be necessary to go through the process of dividing a region into small sections,
    performing a calculation on each section and then adding the results, in order to formulate the integral required. When numerical methods of integration are studied this summation method will prove fundamental.</p>

<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><b>Pulley Belt Tension</b></p>
<p>Consider that a belt is partially wound around a pulley so that there is a difference in the tension either side of the pulley (see the figure below). The pulley will be stationary as long as the friction between belt and pulley is sufficient. The frictional
    force on the pulley will depend on the extent of the contact between belt and pulley i.e. on the angle <span class="math inline">\(\theta\)</span> shown in the figure below. Given that the tensions on either side of the belt are <span class="math inline">\(T_2\)</span>    and <span class="math inline">\(T_1\)</span> and that the coefficient of friction between belt and pulley is <span class="math inline">\(\mu\)</span>, find an expression for <span class="math inline">\(T_2\)</span> in terms of <span class="math inline">\(T_1\)</span>,
    <span class="math inline">\(\mu\)</span> and <span class="math inline">\(\theta\)</span>.</p>


<h4 class="HELM_solution">Solution</h4>
<p>Consider a small element of the belt, at angle <span class="math inline">\(\theta\)</span> where the tension is <span class="math inline">\(T\)</span>. Changing the angle by a small amount <span class="math inline">\(\Delta \theta\)</span> changes the
    tension from <span class="math inline">\(T\)</span> to <span class="math inline">\(T+\Delta T\)</span>.</p>
<center>
    <p><img src="@@PLUGINFILE@@/Figure%204.svg" alt="Pulley system as described in the text above. Changes in theta are shown." width="291" height="146" class="img-responsive atto_image_button_text-bottom"><br></p>
</center>


<p>Take moments about the centre of the pulley, denoting the radius of the pulley by <span class="math inline">\(R\)</span> and assuming that the frictional force is <span class="math inline">\(\mu T\)</span> per unit length. For the pulley to remain stationary,
    \[R\Delta \theta\mu T=R(T+\Delta T)-RT \quad \text{or} \quad \Delta \theta=\dfrac{\Delta T}{\mu T}.\]</p>

<p>Using integration as the limit of a sum, \[\theta=\int_{T_1}^{T_2}\frac{\mathrm{d}T}{\mu T}=\frac{1}{\mu}\bigg[\ln T\bigg]_{T_1}^{T_2} =\frac{1}{\mu}\ln\left(\frac{T_2}{T_1}\right). \quad \; \text{So} \; T_2=T_1 e^{\mu\theta}.\]</p>]]></text>
<file name="Figure 4.svg" path="/" encoding="base64">[encoded data, 7884 characters]</file>
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