Rectangle bounds for definite integral

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 \(\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } < 10 ^ { - 3 }\).

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 \(\lim _ { \mathrm { n } \rightarrow \infty } \left( \mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

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 \(\lim _ { n \rightarrow \infty } \left( \mathrm { U } _ { n } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

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 \(\ln 2\) .

3 \includegraphics[max width=\textwidth, alt={}, center]{268b605f-eb86-40df-946a-210da1355e83-2_686_967_998_589} The diagram shows the curve with equation \(y = \mathrm { e } ^ { x ^ { 2 } }\), for \(0 \leqslant x \leqslant 1\). The region under the curve between these limits is divided into four strips of equal width. The area of this region under the curve is \(A\).

1. By considering the set of rectangles indicated in the diagram, show that an upper bound for \(A\) is 1.71 .
2. By considering an appropriate set of four rectangles, find a lower bound for \(A\).

3 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-2_643_787_1621_680} The diagram shows the curve with equation \(y = \sqrt { 1 + x ^ { 3 } }\), for \(2 \leqslant x \leqslant 3\). The region under the curve between these limits has area \(A\).

1. Explain why \(3 < A < \sqrt { 28 }\).
2. The region is divided into 5 strips, each of width 0.2 . By using suitable rectangles, find improved lower and upper bounds between which \(A\) lies. Give your answers correct to 3 significant figures.

6 \includegraphics[max width=\textwidth, alt={}, center]{debf6581-25ff-4692-bdfb-154675a3cdb0-3_608_1134_258_504} The diagram shows the curve \(y = \mathrm { f } ( x )\), defined by $$f ( x ) = \begin{cases} x ^ { x } & \text { for } 0 < x \leqslant 1 , \\ 1 & \text { for } x = 0 . \end{cases}$$

1. By first taking logarithms, show that the curve has a stationary point at \(x = \mathrm { e } ^ { - 1 }\). The area under the curve from \(x = 0.5\) to \(x = 1\) is denoted by \(A\).
2. By considering the set of three rectangles shown in the diagram, show that a lower bound for \(A\) is 0.388 .
3. By considering another set of three rectangles, find an upper bound for \(A\), giving 3 decimal places in your answer. The area under the curve from \(x = 0\) to \(x = 0.5\) is denoted by \(B\).
4. Draw a diagram to show rectangles which could be used to find lower and upper bounds for \(B\), using not more than three rectangles for each bound. (You are not required to find the bounds.)

1 \includegraphics[max width=\textwidth, alt={}, center]{cf77e51a-1d3f-423a-be59-96ec60fbeb67-2_568_959_269_593} The diagram shows the curve with equation \(y = \ln ( \cos x )\), for \(0 \leqslant x \leqslant 1.5\). The region bounded by the curve, the \(x\)-axis and the line \(x = 1.5\) has area \(A\). The region is divided into five strips, each of width 0.3 .

1. By considering the set of rectangles indicated in the diagram, find an upper bound for \(A\). Give the answer correct to 3 decimal places.
2. By considering another set of five suitable rectangles, find a lower bound for \(A\). Give the answer correct to 3 decimal places.
3. How could you reduce the difference between the upper and lower bounds for \(A\) ?

6 \includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-4_656_1017_251_525} The diagram shows part of the curve \(y = \ln ( \ln ( x ) )\). The region between the curve and the \(x\)-axis for \(3 \leqslant x \leqslant 6\) is shaded.

1. By considering \(n\) rectangles of equal width, show that a lower bound, \(L\), for the area of the shaded region is \(\frac { 3 } { n } \sum _ { r = 0 } ^ { n - 1 } \ln \left( \ln \left( 3 + \frac { 3 r } { n } \right) \right)\).
2. By considering another set of \(n\) rectangles of equal width, find a similar expression for an upper bound, \(U\), for the area of the shaded region.
3. Find the least value of \(n\) for which \(U - L < 0.001\).

6 Alex is investigating the area, \(A\), under the graph of \(y = x ^ { 2 }\) between \(x = 1\) and \(x = 1.5\). They draw the graph, together with rectangles of width \(\delta x = 0.1\), and varying heights \(y\). \includegraphics[max width=\textwidth, alt={}, center]{7298e7b9-ad52-480c-bc2b-8289aeab9ebb-06_531_714_356_251}

1. Use the rectangles in the diagram to show that lower and upper bounds for the area \(A\) are 0.73 and 0.855 respectively.
2. Alex finds lower and upper bounds for the area \(A\), using widths \(\delta x\) of decreasing size. The results are shown in the table. Where relevant, values are given correct to 3 significant figures.

   Width \(\delta x\)	0.1	0.05	0.025	0.0125
   Lower bound for area \(A\)	0.73	0.761	0.776	0.784
   Upper bound for area \(A\)	0.855	0.823	0.807	0.799

   Use Alex's results to estimate the value of \(A\) correct to \(\mathbf { 2 }\) significant figures. Give a brief justification for your estimate.
3. Write down an expression, in terms of \(y\) and \(\delta x\), for the exact value of the area \(A\).

8 Fig. 8.1 shows the cross-section of a straight driveway 4 m wide made from tarmac. \begin{figure}[h] \end{figure} The height \(h \mathrm {~m}\) of the cross-section at a displacement \(x \mathrm {~m}\) from the middle is modelled by \(\mathrm { h } = \frac { 0.2 } { 1 + \mathrm { x } ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\). A lower bound of \(0.3615 \mathrm {~m} ^ { 2 }\) is found for the area of the cross-section using rectangles as shown in Fig. 8.2. \begin{figure}[h] \end{figure}

1. Use a similar method to find an upper bound for the area of the cross-section.
2. Use the trapezium rule with 4 strips to estimate \(\int _ { 0 } ^ { 2 } \frac { 0.2 } { 1 + x ^ { 2 } } d x\).
3. The driveway is 10 m long. Use your answer in part (b) to find an estimate of the volume of tarmac needed to make the driveway.

\includegraphics{figure_6} The diagram shows the curve with equation \(y = e^{1-x}\) 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 \(\int_0^1 e^{1-x} \, dx < U_n\), where $$U_n = \frac{e-1}{n(1-e^{-1})}.$$ [4]
2. Use a similar method to find, in terms of \(n\), a lower bound \(L_n\) for \(\int_0^1 e^{1-x} \, dx\). [4]
3. Show that \(\lim_{n \to \infty}(U_n - L_n) = 0\). [2]
4. Use the Maclaurin's series for \(e^x\) given in the list of formulae (MF19) to find the first three terms of the series expansion of \(z(1-e^{-z})\), in ascending powers of \(z\), and deduce the value of \(\lim_{n \to \infty}(U_n)\). [3]

Fig. 14.1 shows the curve with equation \(y = \frac{1}{1 + x^2}\), together with 5 rectangles of equal width. \includegraphics{figure_14_1} Fig. 14.2 shows the coordinates of the points A, B, C, D, E and F. \includegraphics{figure_14_2}

1. Use the 5 rectangles shown in Fig. 14.1 and the information in Fig. 14.2 to show that a lower bound for \(\int_0^1 \frac{1}{1 + x^2}\,dx\) is 0.7337, correct to 4 decimal places. [2]
2. Use the 5 rectangles shown in Fig. 14.1 and the information in Fig. 14.2 to calculate an upper bound for \(\int_0^1 \frac{1}{1 + x^2}\,dx\) correct to 4 decimal places. [2]
3. Hence find the length of the interval in which your answers to parts (a) and (b) indicate the value of \(\int_0^1 \frac{1}{1 + x^2}\,dx\) lies. [1]

Amit uses \(n\) rectangles, each of width \(\frac{1}{n}\), to calculate upper and lower bounds for \(\int_0^1 \frac{1}{1 + x^2}\,dx\), using different values of \(n\). His results are shown in Fig. 14.3. \includegraphics{figure_14_3}

1. Find the length of the smallest interval in which Amit now knows \(\int_0^1 \frac{1}{1 + x^2}\,dx\) lies. [2]
2. Without doing any calculation, explain how Amit could find a smaller interval which contains the value of \(\int_0^1 \frac{1}{1 + x^2}\,dx\). [1]