1.09f Trapezium rule: numerical integration

7 $$f ( x ) = 2 + \cos x + 3 \sin x$$

1. Express \(\mathrm { f } ( x )\) in the form $$\mathrm { f } ( x ) = a + b \cos ( x - c )$$ where \(a , b\) and \(c\) are constants, \(b > 0\) and \(0 < c < \frac { \pi } { 2 }\).
2. Solve the equation \(\mathrm { f } ( x ) = 0\) for \(x\) in the interval \(0 \leq x \leq 2 \pi\).
3. Use Simpson's rule with four strips, each of width 0.5 , to find an approximate value for $$\int _ { 0 } ^ { 2 } f ( x ) d x$$

1 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10 -second intervals. \begin{figure}[h] \end{figure} The measured speeds are shown below.

1. Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line \(t = 60\) and the axes. [This area represents the distance travelled by the car.]
2. Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area. The speed of the car may be modelled by \(v = 28 - t + 0.015 t ^ { 2 }\).
3. Show that the difference between the value given by the model when \(t = 10\) and the measured value is less than \(3 \%\) of the measured value.
4. According to this model, the distance travelled by the car is $$\int _ { 0 } ^ { 60 } \left( 28 \quad t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

2 At a place where a river is 7.5 m wide, its depth is measured every 1.5 m across the river. The table shows the results. Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.

3 Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O . \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.
2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. The curve of the roof may be modelled by \(y = - 0.013 x ^ { 3 } + 0.16 x ^ { 2 } - 0.082 x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.
3. Use integration to find the area of the cross-section according to this model.
4. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\).

4 Fig. 2 shows the coordinates at certain points on a curve. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to calculate an estimate of the area of the region bounded by this curve and the axes.

2 \begin{figure}[h] \end{figure} Fig. 4 shows a curve which passes through the points shown in the following table. Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve, the lines \(x = 1\) and \(x = 4\), and the \(x\)-axis. State, with a reason, whether the trapezium rule gives an overestimate or an underestimate of the area of this region.
[0pt] [5]

3

1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_512_819_493_700} \captionsetup{labelformat=empty} \caption{Figure 9.1}
   \end{figure} Using this model,
   (A) find the greatest height of the tunnel,
   (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
   [0pt] [5]
2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_506_942_1703_629} \captionsetup{labelformat=empty} \caption{Not to scale}
   \end{figure} Fig. 9.2 Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
   Hence estimate the volume of earth removed when the tunnel is re-shaped.

1. The diagram shows the curve with equation \(y = \ln ( 2 + \cos x ) , x \geq 0\).
The smallest value of \(x\) for which the curve meets the \(x\)-axis is \(a\) as shown.

1. Find the value of \(a\).
2. Use Simpson's rule with four strips of equal width to estimate the area of the region bounded by the curve in the interval \(0 \leq x \leq a\) and the coordinate axes.

2 \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]{4dcf71fc-2585-4247-a21d-8b14f11ce0d0-1_415_1662_1081_240} \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.

7. $$\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { 4 x + 1 } , \quad x \in \mathbb { R } , \quad x \neq - \frac { 1 } { 4 }$$

1. Find and simplify an expression for \(\mathrm { f } ^ { \prime } ( x )\).
2. Find the set of values of \(x\) for which \(\mathrm { f } ( x )\) is increasing.
3. Use Simpson's rule with six strips to find an approximate value for $$\int _ { 0 } ^ { 6 } f ( x ) d x$$

1. Use Simpson's rule with four strips to estimate the value of the integral

$$\int _ { 0 } ^ { 3 } \mathrm { e } ^ { \cos x } \mathrm {~d} x$$

8 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-4_787_742_276_719} The diagram shows part of the curve \(y = \ln \left( 5 - x ^ { 2 } \right)\) which meets the \(x\)-axis at the point \(P\) with coordinates \(( 2,0 )\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(P Q\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\).
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures.
3. Deduce an approximation to the area of the region \(B\).

8 \includegraphics[max width=\textwidth, alt={}, center]{1216a06e-7e14-48d7-a7ca-7acd8d71af5f-4_538_1443_262_351} The diagram shows the curve with equation \(y = x ^ { 8 } \mathrm { e } ^ { - x ^ { 2 } }\). The curve has maximum points at \(P\) and \(Q\). The shaded region \(A\) is bounded by the curve, the line \(y = 0\) and the line through \(Q\) parallel to the \(y\)-axis. The shaded region \(B\) is bounded by the curve and the line \(P Q\).

1. Show by differentiation that the \(x\)-coordinate of \(Q\) is 2 .
2. Use Simpson's rule with 4 strips to find an approximation to the area of region \(A\). Give your answer correct to 3 decimal places.
3. Deduce an approximation to the area of region \(B\).

8 The definite integral \(I\) is defined by $$I = \int _ { 0 } ^ { 6 } 2 ^ { x } \mathrm {~d} x$$

1. Use Simpson's rule with 6 strips to find an approximate value of \(I\).
2. By first writing \(2 ^ { x }\) in the form \(\mathrm { e } ^ { k x }\), where the constant \(k\) is to be determined, find the exact value of \(I\).
3. Use the answers to parts (i) and (ii) to deduce that \(\ln 2 \approx \frac { 9 } { 13 }\).

4

1. \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-2_579_785_1279_721} The diagram shows the curve \(y = \frac { 2 } { \sqrt { } x }\). The region \(R\), shaded in the diagram, is bounded by the curve and by the lines \(x = 1 , x = 5\) and \(y = 0\). The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed.
2. Use Simpson's rule, with 4 strips, to find an approximate value for $$\int _ { 1 } ^ { 5 } \sqrt { } \left( x ^ { 2 } + 1 \right) \mathrm { d } x ,$$ giving your answer correct to 3 decimal places.

4 The integral I is defined by $$I = \int _ { 0 } ^ { 13 } ( 2 x + 1 ) ^ { \frac { 1 } { 3 } } d x$$

1. Use integration to find the exact value of I .
2. Use Simpson's rule with two strips to find an approximate value for I. Give your answer correct to 3 significant figures.

9 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_424_707_1260_724} The diagram shows the curve \(y = \tan ^ { - 1 } x\) and its asymptotes \(y = \pm a\).

1. State the exact value of \(a\).
2. Find the value of \(x\) for which \(\tan ^ { - 1 } x = \frac { 1 } { 2 } a\). The equation of another curve is \(y = 2 \tan ^ { - 1 } ( x - 1 )\).
3. Sketch this curve on a copy of the diagram, and state the equations of its asymptotes in terms of \(a\).
4. Verify by calculation that the value of \(x\) at the point of intersection of the two curves is 1.54 , correct to 2 decimal places. Another curve (which you are not asked to sketch) has equation \(y = \left( \tan ^ { - 1 } x \right) ^ { 2 }\).
5. Use Simpson's rule, with 4 strips, to find an approximate value for \(\int _ { 0 } ^ { 1 } \left( \tan ^ { - 1 } x \right) ^ { 2 } \mathrm {~d} x\).

A vehicle of mass 900 kg moves along a straight horizontal road. At time \(t\) seconds the resultant force acting on the vehicle has magnitude \(\frac { 63000 } { k t ^ { 2 } } \mathrm {~N}\), where \(k\) is a positive constant. The force acts in the direction of motion of the vehicle. At time \(t\) seconds, \(t \geqslant 1\), the speed of the vehicle is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the vehicle is a distance \(x\) metres from a fixed point \(O\) on the road. When \(t = 1\) the vehicle is at rest at \(O\) and when \(t = 4\) the speed of the vehicle is \(10.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(v = 14 - \frac { 14 } { t }\)
2. Hence deduce that the speed of the vehicle never reaches \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. Use the trapezium rule, with 4 equal intervals, to estimate the value of \(x\) when \(v = 7\)

6. \includegraphics[max width=\textwidth, alt={}, center]{687756c0-2038-4077-8c5c-fe0ca0f6ce65-2_444_825_1571_516} The diagram shows the curve with equation \(y = \sqrt { \frac { x } { x + 1 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 3\).

1. Use Simpson's rule with six strips to estimate the area of the shaded region. The shaded region is rotated through four right angles about the \(x\)-axis.
2. Show that the volume of the solid formed is \(\pi ( 3 - \ln 4 )\).

2 Fig. 2 shows part of the curve \(y = \sqrt { 1 + x ^ { 3 } }\). \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 4 strips to estimate \(\int _ { 0 } ^ { 2 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), giving your answer correct to 3 significant figures.
2. Chris and Dave each estimate the value of this integral using the trapezium rule with 8 strips. Chris gets a result of 3.25, and Dave gets 3.30. One of these results is correct. Without performing the calculation, state with a reason which is correct.
   [0pt] [2]

$$+ \mu \left( \begin{array} { r } 1 \\ 0 \\ - 2 \end{array} \right)$$ Find the acute angle between the lines. 6 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.

1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } - \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.

2 Fig. 2 shows the curve \(y = \sqrt { 1 + x ^ { 2 } }\). \begin{figure}[h] \end{figure}

1. The following table gives some values of \(x\) and \(y\).

   \(x\)	0	0.25	0.5	0.75	1
   \(y\)	1	1.0308		1.25	1.4142

   Find the missing value of \(y\), giving your answer correct to 4 decimal places.
   Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.

6 The graph shows part of the curve \(y = \frac { 1 } { 1 + x ^ { 2 } }\). \includegraphics[max width=\textwidth, alt={}, center]{62dbc58e-f498-483f-a9aa-05cb5aa44881-3_474_961_406_479} Use the trapezium rule to estimate the area between the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) using

1. 2 strips,
2. 4 strips. What can you conclude about the true value of the area?

8

1. Evaluate \(A _ { 0 } = \int _ { 0 } ^ { 2 } \left( 2 + 2 x - x ^ { 2 } \right) \mathrm { d } x\). Fig 8.1 illustrates the cross-section of a proposed tunnel. Lengths are in metres. The equation of the curved section is \(y = 2 + \sqrt { 2 x - x ^ { 2 } }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{23771896-942c-4a1d-ab95-6b6d3cc5643c-3_419_515_1155_836} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
   \end{figure} The designers need to know the area of the cross-section, \(A \mathrm {~m} ^ { 2 }\), so that they can work out the volume of the soil that will need to be removed when the tunnel is built.
2. An initial estimate, \(A _ { 1 }\), is given by the area of the 8 rectangles shown in Fig 8.2. Calculate \(A _ { 1 }\), and state whether it is an overestimate or underestimate for \(A\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{23771896-942c-4a1d-ab95-6b6d3cc5643c-3_520_645_2053_644} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure}
3. On graph paper, draw the graphs of $$y = 2 + 2 x - x ^ { 2 } \text { and } y = 2 + \sqrt { 2 x - x ^ { 2 } } \text { for } 0 \leq x \leq 2 .$$ Make it clear which equation applies to which curve.
4. State whether \(A _ { 0 }\), your answer to part (i), is an underestimate for \(A\) or an overestimate. Give a reason for your answer.
5. The designers use the trapezium rule to estimate \(A\). What values does this give when they take
   (A) 2 strips,
   (B) 4 strips,
   (C) 8 strips? What can you conclude about the value of \(A\) ?
6. The best estimate from the trapezium rule is denoted by \(A _ { 2 }\). State, with a reason, whether the true value of \(A\) is nearer \(A _ { 1 }\) or \(A _ { 2 }\).

10 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.

1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.