1.09f Trapezium rule: numerical integration

3. $$\mathrm { f } ( x ) = x ^ { 3 } - 5 \sqrt { x } - 4 x + 7 \quad x \geqslant 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 0.25,1 ]\)

1. Use linear interpolation once on the interval [ \(0.25,1\) ] to determine an approximation to \(\alpha\), giving your answer to 3 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval [1.5, 2.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\)
3. Hence, using \(x _ { 0 } = 1.75\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation to \(\beta\), giving your answer to 3 decimal places.

2. $$f ( x ) = 7 \sqrt { x } - \frac { 1 } { 2 } x ^ { 3 } - \frac { 5 } { 3 x } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval [2.8, 2.9]
2. 1. Find \(\mathrm { f } ^ { \prime } ( x )\).
   2. Hence, using \(x _ { 0 } = 2.8\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to calculate a second approximation to \(\alpha\), giving your answer to 3 decimal places.
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3. Use linear interpolation once on the interval [2.8, 2.9] to find another approximation to \(\alpha\). Give your answer to 3 decimal places.

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2. $$f ( x ) = 5 x ^ { 2 } - 4 x ^ { \frac { 3 } { 2 } } - 6 , \quad x \geqslant 0$$ The root \(\alpha\) of the equation \(\mathrm { f } ( x ) = 0\) lies in the interval \([ 1.6,1.8 ]\)

1. Use linear interpolation once on the interval \([ 1.6,1.8 ]\) to find an approximation to \(\alpha\). Give your answer to 3 decimal places.
2. Taking 1.7 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.

\begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the graph of \(y = \frac { 12 } { \sqrt { \left( x ^ { 2 } - 2 \right) } } , x \geqslant 2\) The table below gives values of \(y\) rounded to 3 decimal places.

1. Use the trapezium rule with all the values of \(y\) from the table to find an approximate value, to 2 decimal places, for $$\int _ { 2 } ^ { 11 } \frac { 12 } { \sqrt { \left( x ^ { 2 } - 2 \right) } } \mathrm { d } x$$
2. Use your answer to part (a) to estimate a value for $$\int _ { 2 } ^ { 11 } \left( 1 + \frac { 6 } { \sqrt { \left( x ^ { 2 } - 2 \right) } } \right) d x$$

The trapezium rule, with the table below, was used to estimate the area between the curve \(y = \sqrt { x ^ { 3 } + 1 }\), the lines \(x = 1 , x = 3\) and the \(x\)-axis.

1. Calculate, to 3 decimal places, the values of \(y\) for \(x = 2.5\) and \(x = 3\).
2. Use the values from the table and your answers to part (a) to find an estimate, to 2 decimal places, for this area.

2 Use the trapezium rule, with 3 strips each of width 2, to estimate the value of $$\int _ { 1 } ^ { 7 } \sqrt { x ^ { 2 } + 3 } \mathrm {~d} x$$

9

1. 1. Write down the exact values of \(\cos \frac { 1 } { 6 } \pi\) and \(\tan \frac { 1 } { 3 } \pi\) (where the angles are in radians). Hence verify that \(x = \frac { 1 } { 6 } \pi\) is a solution of the equation $$2 \cos x = \tan 2 x$$
   2. Sketch, on a single diagram, the graphs of \(y = 2 \cos x\) and \(y = \tan 2 x\), for \(x\) (radians) such that \(0 \leqslant x \leqslant \pi\). Hence state, in terms of \(\pi\), the other values of \(x\) between 0 and \(\pi\) satisfying the equation $$2 \cos x = \tan 2 x$$
2. 1. Use the trapezium rule, with 3 strips, to find an approximate value for the area of the region bounded by the curve \(y = \tan x\), the \(x\)-axis, and the lines \(x = 0.1\) and \(x = 0.4\). (Values of \(x\) are in radians.)
   2. State with a reason whether this approximation is an underestimate or an overestimate.

9

1. Sketch the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\), and state the coordinates of any point where the curve crosses an axis.
2. Use the trapezium rule, with 4 strips of width 0.5 , to estimate the area of the region bounded by the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\), the axes, and the line \(x = 2\).
3. The point \(P\) on the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) has \(y\)-coordinate equal to \(\frac { 1 } { 6 }\). Prove that the \(x\)-coordinate of \(P\) may be written as $$1 + \frac { \log _ { 10 } 3 } { \log _ { 10 } 2 }$$

4 \includegraphics[max width=\textwidth, alt={}, center]{e429080f-8634-46bc-b451-7b13b871e518-2_543_857_1155_644} The diagram shows the curve \(\mathrm { y } = \sqrt { 4 \mathrm { X } + 1 }\).

1. Use the trapezium rule, with strips of width 0.5 , to find an approximate value for the area of the region bounded by the curve \(y = \sqrt { 4 x + 1 }\), the \(x\)-axis, and the lines \(x = 1\) and \(x = 3\). Give your answer correct to 3 significant figures.
2. State with a reason whether this approximation is an under-estimate or an over-estimate.

9

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]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_506_812_676_653} \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.
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]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_513_1256_1894_575} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} 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.

9

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]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_506_812_676_653} \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.
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]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_513_1256_1894_575} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} 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.

4 \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.

10 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 $$\int _ { 0 } ^ { 60 } \left( 28 - t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

9 Fig. 9 shows \(P \quad\) The line \(y = x\) \(Q\) The curve \(y = \sqrt { \frac { 1 } { 2 } \left( x + x ^ { 2 } \right) }\) \(R \quad\) The curve \(\quad y = \sqrt { x }\). \begin{figure}[h] \end{figure}

1. Write down the area of the triangle formed by the line \(y = x\), the line \(x = 1\) and the \(x\)-axis.
2. Show that the area of the region formed by the curve \(y = \sqrt { x }\), the line \(x = 1\) and the \(x\)-axis is \(\frac { 2 } { 3 }\). An estimate is required of the Area, \(A\), of the region formed by the curve \(y = \sqrt { \frac { 1 } { 2 } \left( x + x ^ { 2 } \right) }\), the line \(x = 1\) and the \(x\)-axis.
3. Use results to parts (i) and (ii) to complete the statement $$\ldots \ldots \ldots \ldots . . < A < \ldots \ldots \ldots \ldots \ldots . .$$
4. Use the Trapezium Rule with 4 strips to find an estimate for \(A\).
5. Draw a sketch of Fig. 9. Use it to illustrate the area found as the trapezium rule estimate for \(A\).
   Explain how your diagram shows that the trapezium rule estimate must be:
   consistent with the answer to part (iv);
   an under-estimate for A .

10 Fig. 10 shows the curve with equation \(y = x ^ { 2 } + \frac { 16 } { x }\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence calculate the coordinates of the stationary point on the curve.
3. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and explain why this confirms that he stationary point is a minimum.
4. Using the trapezium rule with 4 intervals, estimate the area between the curve and the \(x\) axis between \(x = 2\) and \(x = 4\).
5. State, giving a reason, whether this estimate of the area under-estimates or over-estimates the true area beneath the curve.

6. (i) Write down the exact value of \(\cos \frac { \pi } { 6 }\). The finite region \(R\) is bounded by the curve \(y = \cos ^ { 2 } x\), where \(x\) is measured in radians, the positive coordinate axes and the line \(x = \frac { \pi } { 3 }\).
(ii) Use the trapezium rule with two intervals of equal width to estimate the area of \(R\), giving your answer to 3 significant figures. The finite region \(S\) is bounded by the curve \(y = \sin ^ { 2 } x\), where \(x\) is measured in radians, the positive coordinate axes and the line \(x = \frac { \pi } { 3 }\).
(iii) Using your answer to part (b), find an estimate for the area of \(S\).

3. \includegraphics[max width=\textwidth, alt={}, center]{faa66f88-9bff-4dc9-955f-80cdab3fdd34-1_474_863_1283_520} The diagram shows the curve with equation \(y = \frac { 4 x } { ( x + 1 ) ^ { 2 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 1\).

1. Use the trapezium rule with four intervals, each of width 0.25 , to find an estimate for the area of the shaded region.
2. State, with a reason, whether your answer to part (a) is an under-estimate or an over-estimate of the true area.

2. \includegraphics[max width=\textwidth, alt={}, center]{27703044-8bb3-4809-9454-ae6774fec060-1_485_808_973_520} The diagram shows the curve with equation \(y = \sqrt { 4 x - 1 }\).

1. Use the trapezium rule with four intervals of equal width to estimate the area of the shaded region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).
2. Explain briefly how you could use the trapezium rule to obtain a more accurate estimate of the area of the shaded region.

2. \includegraphics[max width=\textwidth, alt={}, center]{e4afa57d-5be3-42a6-ab35-39b0fdcc1681-1_554_848_685_461} The diagram shows the curve with equation \(y = 4 x + \frac { 1 } { x } , x > 0\).
Use the trapezium rule with three intervals, each of width 1 , to estimate the area of the shaded region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\).

8. The finite region \(R\) is bounded by the curve \(y = 1 + 3 \sqrt { x }\), the \(x\)-axis and the lines \(x = 2\) and \(x = 8\).

1. Use the trapezium rule with three intervals, each of width 2 , to estimate to 3 significant figures the area of \(R\).
2. Use integration to find the exact area of \(R\) in the form \(a + b \sqrt { 2 }\).
3. Find the percentage error in the estimate made in part (a).

4. \includegraphics[max width=\textwidth, alt={}, center]{e5d62032-84ad-4e0b-9b72-ccfd8f4dbac8-2_465_844_246_516} The diagram shows the curve with equation \(y = \frac { 1 } { x ^ { 2 } + 1 }\).
The shaded region \(R\) is bounded by the curve, the coordinate axes and the line \(x = 2\).

1. Use the trapezium rule with four strips of equal width to estimate the area of \(R\). The cross-section of a support for a bookshelf is modelled by \(R\) with 1 unit on each axis representing 8 cm . Given that the support is 2 cm thick,
2. find an estimate for the volume of the support.

4. (i) Use Simpson's rule with four intervals, each of width 0.25 , to estimate the value of the integral $$\int _ { 0 } ^ { 1 } x \mathrm { e } ^ { 2 x } \mathrm {~d} x$$ (ii) Find the exact value of the integral $$\int _ { \frac { 1 } { 2 } } ^ { 1 } e ^ { 1 - 2 x } d x$$

1 Oskar is designing a building. Fig. 12 shows his design for the end wall and the curve of the roof. The units for \(x\) and \(y\) are metres. \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 5 strips to estimate the area of the end wall of the building.
2. Oskar now uses the equation \(y = - 0.001 x ^ { 3 } - 0.025 x ^ { 2 } + 0.6 x + 9\), for \(0 \leqslant x \leqslant 15\), to model the curve of the roof.
   (A) Calculate the difference between the height of the roof when \(x = 12\) given by this model and the data shown in Fig. 12.
   (B) Use integration to find the area of the end wall given by this model.

2 Fig. 7 shows a curve and the coordinates of some points on it. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive \(x\) - and \(y\)-axes.

3 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]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745} \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]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656} \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.