<p>The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x&gt;0$.</p>
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<ol class="i-list"><li><div class="list-item-paragraph">Evaluate $\displaystyle\int\nolimits_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}$ for $k&gt;0$, leaving your answer in terms of $k$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$ for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that</div><div class="list-item-paragraph" style="text-align: center">$\frac{1}{{{\left( k+1 \right)}^{2}}+1}&lt;{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k&lt;\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that</div><div class="list-item-paragraph" style="text-align: center">${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x&gt;y&gt;0$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">By considering parts <strong>(ii)</strong> and <strong>(iii)</strong>, prove by the method of differences that</div><div class="list-item-paragraph" style="text-align: center">$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}&lt;{{\tan }^{-1}}\left( \frac{n}{n+2} \right)&lt;\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$</div><p style="text-align: right;">[4]</p></li></ol>

The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x&gt;0$.

Evaluate $\displaystyle\displaystyle \displaystyle\int\nolimits_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\mathrm{d}x}$ for $k&gt;0$, leaving your answer in terms of $k$.

By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$ for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that

$\frac{1}{{{\left( k+1 \right)}^{2}}+1}&lt;{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k&lt;\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that

${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x&gt;y&gt;0$.

By considering parts <strong>(ii)</strong> and <strong>(iii)</strong>, prove by the method of differences that

$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}&lt;{{\tan }^{-1}}\left( \frac{n}{n+2} \right)&lt;\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

Default solution

\[\int_k^{k+1} \frac{1}{x^2 + 1} \, dx = \left[\tan^{-1} x\right]_k^{k+1} = \tan^{-1}(k+1) - \tan^{-1}(k)\]

Taking the left bound for the height of rectangle, area is an overestimation.

Let overestimated area be $A_1 = \frac{1}{1 + k^2} \times 1 = \frac{1}{1 + k^2}$

Taking the right bound for the height of rectangle, area is an underestimation.

Let underestimated area be $A_2 = \frac{1}{1 + (k+1)^2} \times 1 = \frac{1}{1 + (k+1)^2}$

$A_2 < \text{Exact Area} < A_1$

\[\frac{1}{(k+1)^2 + 1} < \tan^{-1}(k+1) - \tan^{-1}(k) < \frac{1}{k^2 + 1}\]

Need to show: $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)$

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

$A - B = \tan^{-1}\left(\frac{\tan A - \tan B}{1 + \tan A \tan B}\right)$ --- (3)

Let $x = \tan A$ and $y = \tan B$ --- (1)
$\Rightarrow A = \tan^{-1}(x)$, $B = \tan^{-1}(y)$ --- (2)

Substituting (1) and (2) into (3):
\[\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)\]

From (ii): $\frac{1}{(k+1)^2 + 1} < \tan^{-1}(k+1) - \tan^{-1}(k) < \frac{1}{k^2 + 1}$ for $k \in \mathbb{Z}^+$

Summing from $k = 1$ to $n$:
\[\sum_{k=1}^{n} \frac{1}{(k+1)^2 + 1} < \sum_{k=1}^{n} \left[\tan^{-1}(k+1) - \tan^{-1}(k)\right] < \sum_{k=1}^{n} \frac{1}{k^2 + 1}\]

The middle sum telescopes:
\[\sum_{k=1}^{n} \left[\tan^{-1}(k+1) - \tan^{-1}(k)\right] = \tan^{-1}(n+1) - \tan^{-1}(1)\]

Combining:
\[\sum_{k=1}^{n} \frac{1}{(k+1)^2 + 1} < \tan^{-1}(n+1) - \tan^{-1}(1) < \sum_{k=1}^{n} \frac{1}{k^2 + 1}\]

From (iii), $\tan^{-1}(n+1) - \tan^{-1}(1) = \tan^{-1}\left(\frac{n+1-1}{1+(n+1)(1)}\right) = \tan^{-1}\left(\frac{n}{n+2}\right)$

\[\sum_{k=1}^{n} \frac{1}{(k+1)^2 + 1} < \tan^{-1}\left(\frac{n}{n+2}\right) < \sum_{k=1}^{n} \frac{1}{k^2 + 1}\]

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Default solution