1.09f Trapezium rule: numerical integration

9 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_437_640_470_715} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
   \end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50-metre length of this ditch.
2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_574_808_1402_632} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
   (A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\).
   (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

9 \begin{figure}[h] \end{figure}

1. Jean is designing a model aeroplane. Fig. 9.1 shows her first sketch of the wing's cross-section. Calculate angle A and the area of the cross-section.
2. Jean then modifies her design for the wing. Fig. 9.2 shows the new cross-section, with 1 unit for each of \(x\) and \(y\) representing one centimetre. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5c7ac296-a911-451b-ad18-5ade3ac23e74-3_431_1682_970_194} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} Here are some of the coordinates that Jean used to draw the new cross-section.

   Upper surface		Lower surface
   \(x\)	\(y\)	\(x\)	\(y\)
   0	0	0	0
   4	1.45	4	-0.85
   8	1.56	8	-0.76
   12	1.27	12	-0.55
   16	1.04	16	-0.30
   20	0	20	0

   Use the trapezium rule with trapezia of width 4 cm to calculate an estimate of the area of this cross-section.

2

1. Use Simpson's rule with four strips to find an approximation to $$\int _ { 4 } ^ { 12 } \ln x \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Deduce an approximation to \(\int _ { 4 } ^ { 12 } \ln \left( x ^ { 10 } \right) \mathrm { d } x\).

7 The function f is defined for \(x > 0\) by \(\mathrm { f } ( x ) = \ln x\) and the function g is defined for all real values of \(x\) by \(\mathrm { g } ( x ) = x ^ { 2 } + 8\).

1. Find the exact, positive value of \(x\) which satisfies the equation \(\operatorname { fg } ( x ) = 8\).
2. State which one of f and g has an inverse and define that inverse function.
3. Find the exact value of the gradient of the curve \(y = \operatorname { gf } ( x )\) at the point with \(x\)-coordinate \(\mathrm { e } ^ { 3 }\).
4. Use Simpson's rule with four strips to find an approximate value of $$\int _ { - 4 } ^ { 4 } \mathrm { fg } ( x ) \mathrm { d } x$$ giving your answer correct to 3 significant figures.

5 \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-3_844_837_242_621} It is given that f is a one-one function defined for all real values. The diagram shows the curve with equation \(y = \mathrm { f } ( x )\). The coordinates of certain points on the curve are shown in the following table.

1. State the value of \(\mathrm { ff } ( 6 )\) and the value of \(\mathrm { f } ^ { - 1 } ( 8 )\).
2. On the copy of the diagram, sketch the curve \(y = \mathrm { f } ^ { - 1 } ( x )\), indicating how the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) are related.
3. Use Simpson's rule with 6 strips to find an approximation to \(\int _ { 2 } ^ { 14 } \mathrm { f } ( x ) \mathrm { d } x\).

6 The value of \(\int _ { 0 } ^ { 8 } \ln \left( 3 + x ^ { 2 } \right) \mathrm { d } x\) obtained by using Simpson's rule with four strips is denoted by \(A\).

1. Find the value of \(A\) correct to 3 significant figures.
2. Explain why an approximate value of \(\int _ { 0 } ^ { 8 } \ln \left( 9 + 6 x ^ { 2 } + x ^ { 4 } \right) \mathrm { d } x\) is \(2 A\).
3. Explain why an approximate value of \(\int _ { 0 } ^ { 8 } \ln \left( 3 \mathrm { e } + \mathrm { e } x ^ { 2 } \right) \mathrm { d } x\) is \(A + 8\).

3

1. Use Simpson's rule with four strips to find an approximation to $$\int _ { 0 } ^ { 2 } \mathrm { e } ^ { \sqrt { x } } \mathrm {~d} x$$ giving your answer correct to 3 significant figures.
2. Deduce an approximation to \(\int _ { 0 } ^ { 2 } \left( 1 + 10 \mathrm { e } ^ { \sqrt { x } } \right) \mathrm { d } x\).

7

1. Find the exact value of \(\int _ { 1 } ^ { 9 } ( 7 x + 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\).
2. Use Simpson's rule with two strips to show that an approximate value of \(\int _ { 1 } ^ { 9 } ( 7 x + 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\) can be expressed in the form \(m + n \sqrt [ 3 ] { 36 }\), where the values of the constants \(m\) and \(n\) are to be stated.
3. Use the results from parts (i) and (ii) to find an approximate value of \(\sqrt [ 3 ] { 36 }\), giving your answer in the form \(\frac { p } { q }\) where \(p\) and \(q\) are integers. \section*{Question 8 begins on page 4.}

1

1. Use the trapezium rule with four strips to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\), showing your working. Fig. 1 shows a sketch of \(y = \sqrt { 1 + \mathrm { e } ^ { x } }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f657e167-e6f8-4df2-901b-067c32835877-02_535_1074_571_532} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure}
2. Suppose that the trapezium rule is used with more strips than in part (i) to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\). State, with a reason but no further calculation, whether this would give a larger or smaller estimate.

4 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h] \end{figure}

1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. 1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.

      \(x\)	0	0.5	1	1.5	2
      \(y\)		1.9283	2.8964	4.5919

   2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

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 \(A\) ?

3

1. It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = ( 2 x + 1 ) \ln ( x + y )$$ and $$y ( 0 ) = 2$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.1\), to obtain an approximation to \(y ( 0.1 )\), giving your answer to three decimal places.
2. It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) \ln ( x + y )$$ and \(y = 2\) when \(x = 0\).
   1. Use implicit differentiation to find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), giving your answer in terms of \(x\) and \(y\).
   2. Hence find the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(y ( x )\). Give your answer in an exact form.
   3. Use your answer to part (b)(ii) to obtain an approximation to \(y ( 0.1 )\), giving your answer to three decimal places.
      [0pt] [1 mark]

2

1. Use the trapezium rule, with four strips each of width 0.5 , to estimate the value of $$\int _ { 0 } ^ { 2 } \mathrm { e } ^ { x ^ { 2 } } \mathrm {~d} x$$ giving your answer correct to 3 significant figures.
2. Explain how the trapezium rule could be used to obtain a more accurate estimate.

10 \includegraphics[max width=\textwidth, alt={}, center]{febe231d-200a-4957-b41b-de5b9be98b0a-7_352_545_258_239} The diagram shows the curve \(y = \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right)\), for \(1 \leqslant x \leqslant 2\).

1. Use rectangles of width 0.25 to find upper and lower bounds for \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x\). Give your answers correct to 3 significant figures.
2. 1. Use the substitution \(t = \sqrt { x - 1 }\) to show that \(\int \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = \int 2 t \sin \left( \frac { 1 } { 2 } t \right) \mathrm { d } t\).
   2. Hence show that \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = 8 \sin \frac { 1 } { 2 } - 4 \cos \frac { 1 } { 2 }\).

1 \includegraphics[max width=\textwidth, alt={}, center]{38b515c2-4764-4b51-a1f5-9b48d46610f0-4_303_451_358_242} The diagram shows part of the curve \(y = \sqrt { x ^ { 2 } - 1 }\).

1. Use the trapezium rule with 4 intervals to find an estimate for \(\int _ { 1 } ^ { 3 } \sqrt { x ^ { 2 } - 1 } \mathrm {~d} x\). Give your answer correct to \(\mathbf { 3 }\) significant figures.
2. State whether the value from part (a) is an under-estimate or an over-estimate, giving a reason for your answer.
3. Explain how the trapezium rule could be used to obtain a more accurate estimate.

5

1. Use the trapezium rule, with two strips of equal width, to show that $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x \approx \frac { 11 } { 4 } - \sqrt { 2 }$$
2. Use the substitution \(x = u ^ { 2 }\) to find the exact value of $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x$$
3. Using your answers to parts (i) and (ii), show that $$\ln 2 \approx k + \frac { \sqrt { 2 } } { 4 }$$ where \(k\) is a rational number to be determined.

1. $$y = \sqrt { \left( 2 ^ { x } + x \right) }$$ a. Complete the table below, giving the values of \(y\) to 3 decimal places. (1)
b. Use the trapezium rule with all the values of \(y\) from your table to find an approximation for the value of $$\int _ { 0 } ^ { 1 } \sqrt { \left( 2 ^ { x } + x \right) } \mathrm { d } x$$ giving your answer to 3 significant figures. Using your answer to part (b) and making your method clear, estimate
c. \(\int _ { 0 } ^ { 0.5 } \sqrt { \left( 2 ^ { 2 x } + 2 x \right) } \mathrm { d } x\)

1. The table below shows corresponding values of \(x\) and \(y\) for \(y = \log _ { 3 } ( x )\) The values of \(y\) are given to 2 decimal places as appropriate.

a. Obtain an estimate for \(\int _ { 3 } ^ { 9 } \log _ { 3 } ( x ) \mathrm { d } x\), giving your answer to two decimal places. Use your answer to part (a) and making your method clear, estimate
b. i) \(\int _ { 3 } ^ { 9 } \log _ { 3 } \sqrt { x } \mathrm {~d} x\) ii) \(\int _ { 3 } ^ { 18 } \log _ { 3 } \left( 9 x ^ { 3 } \right) \mathrm { d } x\)

1. A continuous curve has equation \(y = \mathrm { f } ( x )\).

The table shows corresponding values of \(x\) and \(y\) for this curve, where \(a\) and \(b\) are constants. The trapezium rule is used, with all the \(y\) values in the table, to find an approximate area under the curve between \(x = 3\) and \(x = 4\) Given that this area is 17.59

1. show that \(a + 2 b = 51\) Given also that the sum of all the \(y\) values in the table is 97.2
2. find the value of \(a\) and the value of \(b\)

1. In this question you should show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

1. Given that \(1 + \cos 2 \theta + \sin 2 \theta \neq 0\) prove that $$\frac { 1 - \cos 2 \theta + \sin 2 \theta } { 1 + \cos 2 \theta + \sin 2 \theta } \equiv \tan \theta$$
2. Hence solve, for \(0 < x < 180 ^ { \circ }\) $$\frac { 1 - \cos 4 x + \sin 4 x } { 1 + \cos 4 x + \sin 4 x } = 3 \sin 2 x$$ giving your answers to one decimal place where appropriate.

14. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 5 , \quad x > 0$$ The finite region \(S\), shown shaded in Figure 4, is bounded by the curve \(C\), the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = 3\) The table below shows corresponding values of \(x\) and \(y\) with the values of \(y\) given to 4 decimal places as appropriate.

1. Use the trapezium rule, with all the values of \(y\) in the table, to obtain an estimate for the area of \(S\), giving your answer to 3 decimal places.
2. Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of \(S\).
3. Show that the exact area of \(S\) can be written in the form \(\frac { a } { b } + \ln c\), where \(a , b\) and \(c\) are integers to be found.
   (In part c, solutions based entirely on graphical or numerical methods are not acceptable.)

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \frac { x } { 1 + \sqrt { x } } , x \geqslant 0\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = 3\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { x } { 1 + \sqrt { } x }\)

1. Use the trapezium rule, with all the values of \(y\) in the table, to find an estimate for the area of \(R\), giving your answer to 3 decimal places.
2. Explain how the trapezium rule can be used to give a better approximation for the area of \(R\).
3. Giving your answer to 3 decimal places in each case, use your answer to part (a) to deduce an estimate for
   1. \(\int _ { 1 } ^ { 3 } \frac { 5 x } { 1 + \sqrt { x } } \mathrm {~d} x\)
   2. \(\int _ { 1 } ^ { 3 } \left( 6 + \frac { x } { 1 + \sqrt { x } } \right) \mathrm { d } x\)

1. The table below shows corresponding values of \(x\) and \(y\) for \(y = \log _ { 3 } 2 x\) The values of \(y\) are given to 2 decimal places as appropriate.

1. Using the trapezium rule with all the values of \(y\) in the table, find an estimate for $$\int _ { 3 } ^ { 9 } \log _ { 3 } 2 x \mathrm {~d} x$$ Using your answer to part (a) and making your method clear, estimate
2. 1. \(\int _ { 3 } ^ { 9 } \log _ { 3 } ( 2 x ) ^ { 10 } \mathrm {~d} x\)
   2. \(\int _ { 3 } ^ { 9 } \log _ { 3 } 18 x \mathrm {~d} x\)

1 The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { \frac { x } { 1 + x } }\) The values of \(y\) are given to 4 significant figures.

1. Use the trapezium rule, with all the values of \(y\) in the table, to find an estimate for $$\int _ { 0.5 } ^ { 2.5 } \sqrt { \frac { x } { 1 + x } } \mathrm {~d} x$$ giving your answer to 3 significant figures.
2. Using your answer to part (a), deduce an estimate for \(\int _ { 0.5 } ^ { 2.5 } \sqrt { \frac { 9 x } { 1 + x } } \mathrm {~d} x\) Given that $$\int _ { 0.5 } ^ { 2.5 } \sqrt { \frac { 9 x } { 1 + x } } \mathrm {~d} x = 4.535 \text { to } 4 \text { significant figures }$$
3. comment on the accuracy of your answer to part (b).

13 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt [ 3 ] { 27 - 8 x ^ { 3 } }\). Jenny uses her scientific calculator to create a table of values for \(\mathrm { f } ( x )\) and \(\mathrm { f } ^ { \prime } ( x )\).

1. Use calculus to find an expression for \(\mathrm { f } ^ { \prime } ( x )\) and hence explain why the calculator gives an error for \(\mathrm { f } ^ { \prime } ( 1.5 )\).
2. Find the first three terms of the binomial expansion of \(\mathrm { f } ( x )\).
3. Jenny integrates the first three terms of the binomial expansion of \(\mathrm { f } ( x )\) to estimate the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x\). Explain why Jenny's method is valid in this case. (You do not need to evaluate Jenny's approximation.)
4. Use the trapezium rule with 4 strips to obtain an estimate for \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x\). The calculator gives 2.92117438 for \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x\). The graph of \(y = \mathrm { f } ( x )\) is shown in Fig. 13. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{904025c9-6d68-4344-bd41-8c0fccfcf92f-08_490_906_1505_568} \captionsetup{labelformat=empty} \caption{Fig. 13}
   \end{figure}
5. Explain why the trapezium rule gives an underestimate.