1.08g Integration as limit of sum: Riemann sums

3 \includegraphics[max width=\textwidth, alt={}, center]{fa2213b3-480c-44cb-8ba0-ebd2b94d3d90-04_851_805_251_616} The diagram shows the curve with equation \(\mathrm { y } = \mathrm { x } ^ { 3 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } x ^ { 3 } d x < U _ { n }\), where $$\mathrm { U } _ { \mathrm { n } } = \left( \frac { \mathrm { n } + 1 } { 2 \mathrm { n } } \right) ^ { 2 }$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(L _ { n }\) for \(\int _ { 0 } ^ { 1 } x ^ { 3 } d x\).
3. Find the least value of \(n\) such that \(\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } < 10 ^ { - 3 }\).

3 \includegraphics[max width=\textwidth, alt={}, center]{fd247a71-4680-45d8-89d2-fef17ed3a5e9-04_851_805_251_616} The diagram shows the curve with equation \(\mathrm { y } = \mathrm { x } ^ { 3 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } x ^ { 3 } d x < U _ { n }\), where $$\mathrm { U } _ { \mathrm { n } } = \left( \frac { \mathrm { n } + 1 } { 2 \mathrm { n } } \right) ^ { 2 }$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(L _ { n }\) for \(\int _ { 0 } ^ { 1 } x ^ { 3 } d x\).
3. Find the least value of \(n\) such that \(\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } < 10 ^ { - 3 }\).

3 \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-04_540_1511_276_274} The diagram shows the curve \(\mathrm { y } = \frac { \mathrm { x } } { 2 \mathrm { x } ^ { 2 } - 1 }\) for \(x \geqslant 1\), together with a set of \(N - 1\) rectangles of unit
width. width.

1. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 1 } ^ { N } \frac { r } { 2 r ^ { 2 } - 1 } < \frac { 1 } { 4 } \ln \left( 2 N ^ { 2 } - 1 \right) + 1$$
2. Use a similar method to find, in terms of \(N\), a lower bound for \(\sum _ { r = 1 } ^ { N } \frac { r } { 2 r ^ { 2 } - 1 }\).

4 The diagram shows the curve with equation \(\mathrm { y } = 2 ^ { \mathrm { x } }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac { 1 } { N }\). \includegraphics[max width=\textwidth, alt={}, center]{114ece0d-558d-4c02-8a77-034b3681cff9-06_824_1161_376_450}

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } 2 ^ { x } d x < U _ { N }\), where $$\mathrm { U } _ { \mathrm { N } } = \frac { 2 ^ { \frac { 1 } { \mathrm {~N} } } } { \mathrm {~N} \left( 2 ^ { \frac { 1 } { \mathrm {~N} } } - 1 \right) }$$
2. Use a similar method to find, in terms of \(N\), a lower bound \(\mathrm { L } _ { \mathrm { N } }\) for \(\int _ { 0 } ^ { 1 } 2 ^ { x } \mathrm {~d} x\).
3. Find the least value of \(N\) such that \(\mathrm { U } _ { \mathrm { N } } - \mathrm { L } _ { \mathrm { N } } < 10 ^ { - 4 }\).

4 The diagram shows the curve with equation \(\mathrm { y } = 2 ^ { \mathrm { x } }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac { 1 } { N }\). \includegraphics[max width=\textwidth, alt={}, center]{69c540e1-1dad-45a1-9809-7629d16260e0-06_824_1161_376_450}

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } 2 ^ { x } d x < U _ { N }\), where $$\mathrm { U } _ { \mathrm { N } } = \frac { 2 ^ { \frac { 1 } { \mathrm {~N} } } } { \mathrm {~N} \left( 2 ^ { \frac { 1 } { \mathrm {~N} } } - 1 \right) }$$
2. Use a similar method to find, in terms of \(N\), a lower bound \(\mathrm { L } _ { \mathrm { N } }\) for \(\int _ { 0 } ^ { 1 } 2 ^ { x } \mathrm {~d} x\).
3. Find the least value of \(N\) such that \(\mathrm { U } _ { \mathrm { N } } - \mathrm { L } _ { \mathrm { N } } < 10 ^ { - 4 }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{23b06b1c-997f-425d-ae3d-bd4cc1295605-10_771_1146_260_497} The diagram shows the curve with equation \(\mathrm { y } = \ln ( 1 + \mathrm { x } )\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles each of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \ln ( 1 + x ) d x < U _ { n }\), where $$U _ { n } = \frac { 1 } { n } \ln \frac { ( 2 n ) ! } { n ! } - \ln n$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(\mathrm { L } _ { \mathrm { n } }\) for \(\int _ { 0 } ^ { 1 } \ln ( 1 + x ) \mathrm { d } x\).
3. By simplifying \(\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } }\), show that \(\lim _ { \mathrm { n } \rightarrow \infty } \left( \mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

7

1. Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\). \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-12_778_1548_1007_296} The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
2. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
3. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).

7

1. Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\). \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-12_778_1548_1007_296} The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
2. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
3. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).

6 \includegraphics[max width=\textwidth, alt={}, center]{d421652f-576d-4843-abbf-54404e225fec-10_1015_988_260_577} The diagram shows the curve with equation \(\mathrm { y } = ( 1 - \mathrm { x } ) ^ { 2 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } ( 1 - x ) ^ { 2 } d x < U _ { n }\), where $$U _ { n } = \frac { 2 n ^ { 2 } + 3 n + 1 } { 6 n ^ { 2 } }$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(L _ { n }\) for \(\int _ { 0 } ^ { 1 } ( 1 - x ) ^ { 2 } d x\).
3. Show that \(\lim _ { n \rightarrow \infty } \left( U _ { n } - L _ { n } \right) = 0\).

5 \includegraphics[max width=\textwidth, alt={}, center]{bca7281b-a6a9-4b4c-94e5-3da2a561ad86-08_663_1152_260_452} The diagram shows the curve with equation \(\mathrm { y } = 2 \mathrm { x } - \mathrm { x } ^ { 2 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \left( 2 x - x ^ { 2 } \right) d x < U _ { n }\), where $$U _ { n } = \left( 1 + \frac { 1 } { n } \right) \left( \frac { 2 } { 3 } - \frac { 1 } { 6 n } \right) .$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(L _ { n }\) for \(\int _ { 0 } ^ { 1 } \left( 2 x - x ^ { 2 } \right) d x\).
3. Show that \(\lim _ { n \rightarrow \infty } \left( \mathrm { U } _ { n } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

5 \includegraphics[max width=\textwidth, alt={}, center]{114be67d-a57f-4c36-8f1c-974a2719c1f1-08_663_1152_260_452} The diagram shows the curve with equation \(\mathrm { y } = 2 \mathrm { x } - \mathrm { x } ^ { 2 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \left( 2 x - x ^ { 2 } \right) d x < U _ { n }\), where $$U _ { n } = \left( 1 + \frac { 1 } { n } \right) \left( \frac { 2 } { 3 } - \frac { 1 } { 6 n } \right) .$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(L _ { n }\) for \(\int _ { 0 } ^ { 1 } \left( 2 x - x ^ { 2 } \right) d x\).
3. Show that \(\lim _ { n \rightarrow \infty } \left( \mathrm { U } _ { n } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

4 \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-08_408_1433_296_315} The diagram shows the curve with equation \(y = x ^ { - 2 }\) for \(2 \leqslant x \leqslant N\) together with a set of ( \(N - 2\) ) rectangles of unit width.

1. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 1 } ^ { N } \frac { 1 } { r ^ { 2 } } > \frac { 3 } { 2 } - \frac { 1 } { N } + \frac { 1 } { N ^ { 2 } }$$ \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-08_2718_35_141_2012}
2. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { r = 1 } ^ { N } \frac { 1 } { r ^ { 2 } }\).
3. Deduce lower and upper bounds for \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } }\).

3 The curve \(C\) has equation $$x y ^ { 3 } - 4 x ^ { 3 } y = 3$$

1. Show that, at the point \(( - 1,1 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = 11\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( - 1,1 )\). \includegraphics[max width=\textwidth, alt={}, center]{59982339-c496-4bd7-8dcd-9b257f3afc02-06_535_1584_276_276} The diagram shows the curve with equation \(\mathrm { y } = \frac { \ln \mathrm { x } } { \mathrm { x } ^ { 2 } }\) for \(x \geqslant 2\), together with a set of \(( N - 2 )\) rectangles
   of unit width.

6 \includegraphics[max width=\textwidth, alt={}, center]{323ac7a5-4690-441d-87fc-325a393098fa-10_585_1349_258_358} The diagram shows the curve \(\mathrm { y } = \frac { 1 } { \sqrt { \mathrm { x } ^ { 2 } + 2 \mathrm { x } } }\) for \(x > 0\), together with a set of \(( n - 1 )\) rectangles of unit
width. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { \sqrt { r ^ { 2 } + 2 r } } < \ln \left( n + 1 + \sqrt { n ^ { 2 } + 2 n } \right) + \frac { 1 } { 3 } \sqrt { 3 } - \ln ( 2 + \sqrt { 3 } )$$

8

1. State the sum of the series \(1 + z + z ^ { 2 } + \ldots + z ^ { n - 1 }\), for \(z \neq 1\).
2. By letting \(z = \cos \theta + i \sin \theta\), where \(\cos \theta \neq 1\), show that $$1 + \cos \theta + \cos 2 \theta + \ldots + \cos ( n - 1 ) \theta = \frac { 1 } { 2 } \left( 1 - \cos n \theta + \frac { \sin n \theta \sin \theta } { 1 - \cos \theta } \right)$$ \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-15_833_785_214_680} The diagram shows the curve with equation \(\mathrm { y } = \cos \mathrm { x }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).
3. By considering the sum of the areas of these rectangles, show that $$\int _ { 0 } ^ { 1 } \cos x d x < \frac { 1 } { 2 n } \left( 1 - \cos 1 + \frac { \sin 1 \sin \frac { 1 } { n } } { 1 - \cos \frac { 1 } { n } } \right)$$
4. Use a similar method to find, in terms of \(n\), a lower bound for \(\int _ { 0 } ^ { 1 } \cos x d x\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

6 \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-12_533_1532_278_264} The diagram shows the curve with equation \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac { 1 } { N }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x > L _ { N }\), where $$L _ { N } = \frac { 1 } { 2 N \left( 2 ^ { \frac { 1 } { N } } - 1 \right) }$$ \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-12_2717_38_109_2009}
2. Use a similar method to find, in terms of \(N\), an upper bound \(U _ { N }\) for \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x\).
3. Find the least value of \(N\) such that \(U _ { N } - L _ { N } \leqslant 10 ^ { - 3 }\).
4. Given that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x = \frac { 1 } { 2 \ln 2 }\) ,use the value of \(N\) found in part(c)to find upper and lower bounds for \(\ln 2\) .

6 \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-12_533_1532_278_264} The diagram shows the curve with equation \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac { 1 } { N }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x > L _ { N }\), where $$L _ { N } = \frac { 1 } { 2 N \left( 2 ^ { \frac { 1 } { N } } - 1 \right) }$$ \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-12_2717_38_109_2009}
2. Use a similar method to find, in terms of \(N\), an upper bound \(U _ { N }\) for \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x\).
3. Find the least value of \(N\) such that \(U _ { N } - L _ { N } \leqslant 10 ^ { - 3 }\).
4. Given that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x = \frac { 1 } { 2 \ln 2 }\) ,use the value of \(N\) found in part(c)to find upper and lower bounds for \(\ln 2\) .

4 \includegraphics[max width=\textwidth, alt={}, center]{b5503355-3952-47dc-91f4-80a674349b4a-06_538_949_269_557} The diagram shows the curve with equation \(y = \frac { 1 } { x ^ { 2 } }\) for \(x > 0\), together with a set of \(( n - 1 )\) rectangles of unit width.

1. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } } < \frac { 2 n - 1 } { n } .$$
2. Use a similar method to find, in terms of \(n\), a lower bound for \(\sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } }\).

1. The curve \(C\) has equation

$$y = \frac { 1 } { 2 } \ln ( \operatorname { coth } x ) , \quad x > 0$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosech } 2 x$$ The points \(A\) and \(B\) lie on \(C\). The \(x\) coordinates of \(A\) and \(B\) are \(\ln 2\) and \(\ln 3\) respectively.
2. Find the length of the arc \(A B\), giving your answer in the form \(p \ln q\), where \(p\) and \(q\) are rational numbers.
   (6)

6 \includegraphics[max width=\textwidth, alt={}, center]{52b43f20-e0e6-4ddd-9518-bea9782982bf-3_623_1354_262_392} The diagram shows the curve with equation \(y = 3 ^ { x }\) for \(0 \leqslant x \leqslant 1\). The area \(A\) under the curve between these limits is divided into \(n\) strips, each of width \(h\) where \(n h = 1\).

1. By using the set of rectangles indicated on the diagram, show that \(A > \frac { 2 h } { 3 ^ { h } - 1 }\).
2. By considering another set of rectangles, show that \(A < \frac { ( 2 h ) 3 ^ { h } } { 3 ^ { h } - 1 }\).
3. Given that \(h = 0.001\), use these inequalities to find values between which \(A\) lies.

6 \includegraphics[max width=\textwidth, alt={}, center]{dd0e327e-6125-4970-8cfa-cefcedfec06f-3_822_1373_264_386} The diagram shows the curve with equation \(y = \frac { 1 } { x ^ { 2 } }\) for \(x > 0\), together with a set of \(n\) rectangles of unit width, starting at \(x = 1\).

1. By considering the areas of these rectangles, explain why $$\frac { 1 } { 1 ^ { 2 } } + \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \ldots + \frac { 1 } { n ^ { 2 } } > \int _ { 1 } ^ { n + 1 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$$
2. By considering the areas of another set of rectangles, explain why $$\frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 4 ^ { 2 } } + \ldots + \frac { 1 } { n ^ { 2 } } < \int _ { 1 } ^ { n } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$$
3. Hence show that $$1 - \frac { 1 } { n + 1 } < \sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } } < 2 - \frac { 1 } { n }$$
4. Hence give bounds between which \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } }\) lies.

9

1. Prove that \(\int _ { 0 } ^ { N } \ln ( 1 + x ) \mathrm { d } x = ( N + 1 ) \ln ( N + 1 ) - N\), where \(N\) is a positive constant.
2. \includegraphics[max width=\textwidth, alt={}, center]{63a316f6-1c18-4224-930f-0b58112c9f71-4_616_1261_406_482} The diagram shows the curve \(y = \ln ( 1 + x )\), for \(0 \leqslant x \leqslant 70\), together with a set of rectangles of unit width.
   1. By considering the areas of these rectangles, explain why $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 < \int _ { 0 } ^ { 70 } \ln ( 1 + x ) d x$$
   2. By considering the areas of another set of rectangles, show that $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 > \int _ { 0 } ^ { 69 } \ln ( 1 + x ) d x$$
   3. Hence find bounds between which \(\ln ( 70 ! )\) lies. Give the answers correct to 1 decimal place.

7

1. By using a set of rectangles of unit width to approximate an area under the curve \(y = \frac { 1 } { x }\), show that \(\sum _ { x = 1 } ^ { \infty } \frac { 1 } { x }\) is infinite.
2. By using a set of rectangles of unit width to approximate an area under the curve \(y = \frac { 1 } { x ^ { 2 } }\), find an upper limit for the series \(\sum _ { x = 1 } ^ { \infty } \frac { 1 } { x ^ { 2 } }\).

5 \includegraphics[max width=\textwidth, alt={}, center]{e4e1c424-8dd5-4d18-9950-e902de0301b0-3_444_999_1258_539} The diagram shows the curve \(y = \frac { 1 } { x + 1 }\) together with four rectangles of unit width.

1. Explain how the diagram shows that $$\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 } < \int _ { 0 } ^ { 4 } \frac { 1 } { x + 1 } \mathrm {~d} x$$ The curve \(y = \frac { 1 } { x + 2 }\) passes through the top left-hand corner of each of the four rectangles shown.
2. By considering the rectangles in relation to this curve, write down a second inequality involving \(\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 }\) and a definite integral.
3. By considering a suitable range of integration and corresponding rectangles, show that $$\ln ( 500.5 ) < \sum _ { r = 2 } ^ { 1000 } \frac { 1 } { r } < \ln ( 1000 ) .$$

6. \begin{figure}[h] \end{figure} Figure 2 shows the first few iterations in the construction of a curve, \(L\).
Starting with a straight line \(L _ { 0 }\) of length 4 , the middle half of this line is replaced by three sides of a trapezium above \(L _ { 0 }\) as shown, such that the length of each of these sides is \(\frac { 1 } { 4 }\) of the length of \(L _ { 0 }\) After the first iteration each line segment has length one.
In subsequent iterations, each line segment parallel to \(L _ { 0 }\) similarly has its middle half replaced by three sides of a trapezium above that line segment, with each side \(\frac { 1 } { 4 }\) the length of that line segment. Line segments in \(L _ { n }\) are either parallel to \(L _ { 0 }\) or are sloped.

1. Show that the length of \(L _ { 2 }\) is \(\frac { 23 } { 4 }\)
2. Write down the number of
   1. line segments in \(L _ { n }\) that are parallel to \(L _ { 0 }\)
   2. sloped line segments in \(L _ { 2 }\) that are not in \(L _ { 1 }\)
   3. new sloped line segments that are created by the ( \(n + 1\) )th iteration.
3. Hence find the length of \(L _ { n }\) as \(n \rightarrow \infty\) The area enclosed between \(L _ { 0 }\) and \(L _ { n }\) is \(A _ { n }\)
4. Find the value of \(A _ { 1 }\)
5. Find, in terms of \(n\), an expression for \(A _ { n + 1 } - A _ { n }\)
6. Hence find the value of \(A _ { n }\) as \(n \rightarrow \infty\) The same construction as described above is applied externally to the three sides of an equilateral triangle of side length \(a\).
   Given that the limit of the area of the resulting shape is \(26 \sqrt { 3 }\)
7. find the value of \(a\).