1.09f Trapezium rule: numerical integration

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { x + 2 } , x \geqslant - 2\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the line \(x = 6\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { x + 2 }\)

1. Complete the table above, giving the missing value of \(y\) to 4 decimal places.
2. Use the trapezium rule, with all of the values of \(y\) in the completed table, to find an approximate value for the area of \(R\), giving your answer to 3 decimal places. Use your answer to part (b) to find approximate values of
3. 1. \(\int _ { - 2 } ^ { 6 } \frac { \sqrt { x + 2 } } { 2 } \mathrm {~d} x\)
   2. \(\int _ { - 2 } ^ { 6 } ( 2 + \sqrt { x + 2 } ) \mathrm { d } x\)

1. The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 1 } { \sqrt { ( x + 1 ) } }\), with the values
   for \(y\) rounded to 3 decimal places where necessary.

1. Complete the table by giving the value of \(y\) corresponding to \(x = 15\)
2. Use the trapezium rule with all the values of \(y\) from the completed table to find an approximate value for $$\int _ { 0 } ^ { 15 } \frac { 1 } { \sqrt { ( x + 1 ) } } \mathrm { d } x$$ giving your answer to 2 decimal places.

7. (a) Sketch the graph of \(y = \sin \left( x + \frac { \pi } { 6 } \right) , \quad 0 \leqslant x \leqslant 2 \pi\) Show the coordinates of the points where the graph crosses the \(x\)-axis. The table below gives corresponding values of \(x\) and \(y\) for \(y = \sin \left( x + \frac { \pi } { 6 } \right)\).
The values of \(y\) are rounded to 3 decimal places where necessary. (b) Use the trapezium rule with all the values of \(y\) from the table to find an approximate value for $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin \left( x + \frac { \pi } { 6 } \right) \mathrm { d } x$$ Give your answer to 2 decimal places.

1. (a) Sketch the graph of \(y = 3 ^ { x - 2 } , x \in \mathbb { R }\)

Give the exact values for the coordinates of the point where your graph crosses the \(y\)-axis. The table below gives corresponding values of \(x\) and \(y\), for \(y = 3 ^ { x - 2 }\) The values of \(y\) are rounded to 3 decimal places where necessary. (b) Use the trapezium rule with all the values of \(y\) from the table to find an approximate value for $$\int _ { 0.5 } ^ { 3 } 3 ^ { x - 2 } \mathrm {~d} x$$ Give your answer to 2 decimal places.

1. (a) Given that \(a\) is a constant, \(a > 1\), sketch the graph of

$$y = a ^ { x } , \quad x \in \mathbb { R }$$ On your diagram show the coordinates of the point where the graph crosses the \(y\)-axis.
(2) The table below shows corresponding values of \(x\) and \(y\) for \(y = 2 ^ { x }\) (b) Use the trapezium rule, with all of the values of \(y\) from the table, to find an approximate value, to 2 decimal places, for $$\int _ { - 4 } ^ { 4 } 2 ^ { x } \mathrm {~d} x$$ (c) Use the answer to part (b) to find an approximate value for

1. \(\int _ { - 4 } ^ { 4 } 2 ^ { x + 2 } \mathrm {~d} x\)
2. \(\int _ { - 4 } ^ { 4 } \left( 3 + 2 ^ { x } \right) \mathrm { d } x\)
   \includegraphics[max width=\textwidth, alt={}, center]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-23_86_47_2617_1886}

6. (a) Sketch the graph of \(y = \left( \frac { 1 } { 2 } \right) ^ { x } , x \in \mathbb { R }\), showing the coordinates of the point at which the graph crosses the \(y\)-axis. The table below gives corresponding values of \(x\) and \(y\), for \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) The values of \(y\) are rounded to 3 decimal places. (b) Use the trapezium rule with all the values of \(y\) from the table to find an approximate value for $$\int _ { - 0.9 } ^ { - 0.5 } \left( \frac { 1 } { 2 } \right) ^ { x } d x$$ II

5. $$y = \frac { 5 } { 3 x ^ { 2 } - 2 }$$ The table below gives values of \(y\) rounded to 3 decimal places where necessary. Use the trapezium rule, with all the values of \(y\) from the table above, to find an approximate value for $$\int _ { 2 } ^ { 3 } \frac { 5 } { 3 x ^ { 2 } - 2 } d x$$ © Pearson Education Limited 2013
Sample Assessment Materials

1. The table below shows corresponding values of \(x\) and \(y\) for \(y = \log _ { 2 } ( 2 x )\)

The values of \(y\) are given to 2 decimal places as appropriate. Using the trapezium rule with all the values of \(y\) in the given table,

1. obtain an estimate for \(\int _ { 2 } ^ { 14 } \log _ { 2 } ( 2 x ) \mathrm { d } x\), giving your answer to one decimal place. Using your answer to part (a) and making your method clear, estimate
2. 1. \(\quad \int _ { 2 } ^ { 14 } \frac { \log _ { 2 } \left( 4 x ^ { 2 } \right) } { 5 } \mathrm {~d} x\)
   2. \(\int _ { 2 } ^ { 14 } \log _ { 2 } \left( \frac { 2 } { x } \right) \mathrm { d } x\)

      \(x\)	2	5	8	11	14
      \(y\)	2	3.32	4	4.46	4.81

7. Figure 1 Solar panels are installed on the roof of a building. The power, \(P\), produced on a particular day, in kW , can be modelled by the equation $$P = 0.95 + 2 ^ { t - 12 } + 2 ^ { 12 - t } - ( t - 12 ) ^ { 2 } \quad 8.5 \leqslant t \leqslant 15.2$$ where \(t\) is the time in hours after midnight. The graph of \(P\) against \(t\) is shown in Figure 1. A table of values of \(t\) and \(P\) is shown below, with the values of \(P\) given to 4 significant figures where appropriate.

1. Use the given equation to complete the table, giving the values of \(P\) to 4 significant figures where appropriate. The amount of energy, in kWh , produced between 10:00 and 12:00 can be found by calculating the area of region \(R\), shown shaded in Figure 1.
2. Use the trapezium rule, with all the values of \(P\) in the completed table, to find an estimate for the amount of energy produced between 10:00 and 12:00. Give your answer to 2 decimal places.
   7. \includegraphics[max width=\textwidth, alt={}, center]{52c90d0e-a5e4-45fa-95a4-9523287e7588-20_769_1038_116_450}

1. The table below shows corresponding values of \(x\) and \(y\) for

$$y = 2 ^ { 5 - \sqrt { x } }$$ The values of \(y\) are given to 3 decimal places. Using the trapezium rule with all the values of \(y\) in the given table,

1. obtain an estimate for $$\int _ { 5 } ^ { 7 } 2 ^ { 5 - \sqrt { x } } \mathrm {~d} x$$ giving your answer to 2 decimal places.
2. Using your answer to part (a) and making your method clear, estimate
   1. \(\quad \int _ { 5 } ^ { 7 } 2 ^ { 6 - \sqrt { x } } \mathrm {~d} x\)
   2. \(\int _ { 5 } ^ { 7 } \left( 3 + 2 ^ { 5 - \sqrt { x } } \right) \mathrm { d } x\)

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) The table below shows some corresponding values of \(x\) and \(y\) for this curve.
The values of \(y\) are given to 3 decimal places. Using the trapezium rule with all the values of \(y\) in the given table,

1. obtain an estimate for $$\int _ { - 1 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x$$ giving your answer to 2 decimal places.
2. Use your answer to part (a) to estimate
   1. \(\int _ { - 1 } ^ { 1 } ( \mathrm { f } ( x ) - 2 ) \mathrm { d } x\)
   2. \(\int _ { 1 } ^ { 3 } \mathrm { f } ( x - 2 ) \mathrm { d } x\)

1. (a) Sketch the curve with equation

$$y = a ^ { - x } + 4$$ where \(a\) is a constant and \(a > 1\) On your sketch show

• the coordinates of the point of intersection of the curve with the \(y\)-axis
• the equation of the asymptote to the curve.

The table above shows corresponding values of \(x\) and \(y\) for \(y = 3 ^ { - \frac { 1 } { 2 } x } + 4\) The values of \(y\) are given to four significant figures, as appropriate.
Using the trapezium rule with all the values of \(y\) in the table,
(b) find an approximate value for $$\int _ { - 4 } ^ { 8.5 } \left( 3 ^ { - \frac { 1 } { 2 } x } + 4 \right) d x$$ giving your answer to two significant figures.
(c) Using the answer to part (b), find an approximate value for

1. \(\int _ { - 4 } ^ { 8.5 } \left( 3 ^ { - \frac { 1 } { 2 } x } \right) \mathrm { d } x\)
2. \(\int _ { - 4 } ^ { 8.5 } \left( 3 ^ { - \frac { 1 } { 2 } x } + 4 \right) \mathrm { d } x + \int _ { - 8.5 } ^ { 4 } \left( 3 ^ { \frac { 1 } { 2 } x } + 4 \right) \mathrm { d } x\)

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the graph of the curves \(C _ { 1 }\) and \(C _ { 2 }\) The curves intersect when \(x = 2.5\) and when \(x = 4\) A table of values for some points on the curve \(C _ { 1 }\) is shown below, with \(y\) values given to 3 decimal places as appropriate. Using the trapezium rule with all the values of \(y\) in the table,

1. find, to 2 decimal places, an estimate for the area bounded by the curve \(C _ { 1 }\), the line with equation \(x = 2.5\), the \(x\)-axis and the line with equation \(x = 4\) The curve \(C _ { 2 }\) has equation $$y = x ^ { \frac { 3 } { 2 } } - 3 x + 9 \quad x > 0$$
2. Find \(\int \left( x ^ { \frac { 3 } { 2 } } - 3 x + 9 \right) \mathrm { d } x\) The region \(R\), shown shaded in Figure 2, is bounded by the curves \(C _ { 1 }\) and \(C _ { 2 }\)
3. Use the answers to part (a) and part (b) to find, to one decimal place, an estimate for the area of the region \(R\).
   (3)

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

A table of values of \(x\) and \(y\) for \(y = \mathrm { f } ( x )\) is shown below. Use the trapezium rule with all the values of \(y\) in the table to find an approximation for $$\int _ { 4 } ^ { 5 } f ( x ) d x$$ giving your answer to 3 decimal places.

1. (a) Sketch the curve with equation

$$y = a ^ { x } + 4$$ where \(a\) is a positive constant greater than 1
On your sketch, show

• the coordinates of the point of intersection of the curve with the \(y\)-axis
• the equation of the asymptote of the curve

The table shows corresponding values of \(x\) and \(y\) for $$y = 2 ^ { x } - 2 x$$ with the values of \(y\) given to 4 decimal places as appropriate.
Using the trapezium rule with all the values of \(y\) in the given table,
(b) obtain an estimate for \(\int _ { 2 } ^ { 3.5 } \left( 2 ^ { x } - 2 x \right) \mathrm { d } x\), giving your answer to 2 decimal places.
(c) Using your answer to part (b) and making your method clear, estimate

1. \(\int _ { 2 } ^ { 3.5 } \left( 2 ^ { x } + 2 x \right) \mathrm { d } x\)
2. \(\int _ { 2 } ^ { 3.5 } \left( 2 ^ { x + 1 } - 4 x \right) \mathrm { d } x\)

5. (a) Given \(0 < a < 1\), sketch the curve with equation $$y = a ^ { x }$$ showing the coordinates of the point at which the curve crosses the \(y\)-axis. The table above shows corresponding values of \(x\) and \(y\) for \(y = x ^ { 2 } + \left( \frac { 1 } { 2 } \right) ^ { x }\) The values of \(y\) are given to 4 significant figures as appropriate.
Using the trapezium rule with all the values of \(y\) in the given table,
(b) obtain an estimate for \(\int _ { 2 } ^ { 4 } \left( x ^ { 2 } + \left( \frac { 1 } { 2 } \right) ^ { x } \right) \mathrm { d } x\) Using your answer to part (b) and making your method clear, estimate
(c) \(\quad \int _ { 2 } ^ { 4 } \left( x ( x - 3 ) + \left( \frac { 1 } { 2 } \right) ^ { x } \right) \mathrm { d } x\)

2. $$y = \frac { 2 ^ { x } } { \sqrt { \left( 5 x ^ { 2 } + 3 \right) } }$$

1. Complete the table below,giving the values of \(y\) to 3 decimal places.

   \(x\)	- 0.25	0	0.25	0.5	0.75
   \(y\)	0.462		0.653		0.698

2. Use the trapezium rule,with all the values of \(y\) from the completed table,to find an approximate value for
   . $$\int _ { - 0.25 } ^ { 0.75 } \frac { 2 ^ { x } } { \sqrt { \left( 5 x ^ { 2 } + 3 \right) } } \mathrm { d } x$$

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \log _ { 10 } x\) The region \(R\), shown shaded in Figure 1, is bounded by the curve, the line with equation \(x = 2\), the \(x\)-axis and the line with equation \(x = 14\) Using the trapezium rule with four strips of equal width,

1. show that the area of \(R\) is approximately 10.10
2. Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of \(R\).
3. Using the answer to part (a) and making your method clear, estimate the value of
   1. \(\quad \int _ { 2 } ^ { 14 } \log _ { 10 } \sqrt { x } \mathrm {~d} x\)
   2. \(\int _ { 2 } ^ { 14 } \log _ { 10 } 100 x ^ { 3 } \mathrm {~d} x\)

1. The curve \(C _ { 1 }\) has equation \(y = \mathrm { f } ( x )\).

A table of values of \(x\) and \(y\) for \(y = \mathrm { f } ( x )\) is shown below, with the \(y\) values rounded to 4 decimal places where appropriate.

1. Use the trapezium rule with all the values of \(y\) in the table to find an approximation for $$\int _ { 0 } ^ { 2 } f ( x ) d x$$ giving your answer to 3 decimal places. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6f926d53-c6de-4eb7-9d18-596f61ec26e1-16_629_592_1105_402} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6f926d53-c6de-4eb7-9d18-596f61ec26e1-16_540_456_1194_1192} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The region \(R\), shown shaded in Figure 1, is bounded by
   • the curve \(C _ { 1 }\)
   • the curve \(C _ { 2 }\) with equation \(y = 2 - \frac { 1 } { 4 } x ^ { 2 }\)
   • the line with equation \(x = 2\)
   • the \(y\)-axis
   The region \(R\) forms part of the design for a logo shown in Figure 2.
   The design consists of the shaded region \(R\) inside a rectangle of width 2 and height 3 Using calculus and the answer to part (a),
2. calculate an estimate for the percentage of the logo which is shaded.

6. \begin{figure}[h] \end{figure} A river is being studied.
At one particular place, the river is 15 m wide.
The depth, \(y\) metres, of the river is measured at a point \(x\) metres from one side of the river. Figure 1 shows a plot of the cross-section of the river and the coordinate values \(( x , y )\)

1. Use the trapezium rule with all the \(y\) values given in Figure 1 to estimate the cross-sectional area of the river. The water in the river is modelled as flowing at a constant speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) across the whole of the cross-section.
2. Use the model and the answer to part (a) to estimate the volume of water flowing through this section of the river each minute, giving your answer in \(\mathrm { m } ^ { 3 }\) to 2 significant figures. Assuming the model,
3. state, giving a reason for your answer, whether your answer for part (b) is an overestimate or an underestimate of the true volume of water flowing through this section of the river each minute.

3. $$y = \sqrt { \left( 3 ^ { x } + x \right) }$$

1. Complete the table below, giving the values of \(y\) to 3 decimal places.

   \(x\)	0	0.25	0.5	0.75	1
   \(y\)	1	1.251			2

2. Use the trapezium rule with all the values of \(y\) from your table to find an approximation for the value of $$\int _ { 0 } ^ { 1 } \sqrt { \left( 3 ^ { x } + x \right) } \mathrm { d } x$$ You must show clearly how you obtained your answer.
3. Explain how the trapezium rule could be used to obtain a more accurate estimate for the value of $$\int _ { 0 } ^ { 1 } \sqrt { \left( 3 ^ { x } + x \right) } d x$$
   \includegraphics[max width=\textwidth, alt={}]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-10_2673_1948_107_118}

1. The speed, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of a train at time \(t\) seconds is given by

$$v = \sqrt { } \left( 1.2 ^ { t } - 1 \right) , \quad 0 \leqslant t \leqslant 30$$ The following table shows the speed of the train at 5 second intervals.

1. Complete the table, giving the values of \(v\) to 2 decimal places. The distance, \(s\) metres, travelled by the train in 30 seconds is given by $$s = \int _ { 0 } ^ { 30 } \sqrt { } \left( 1.2 ^ { t } - 1 \right) \mathrm { d } t$$
2. Use the trapezium rule, with all the values from your table, to estimate the value of \(s\).
   (3)

3. \(y = \sqrt { } \left( 10 x - x ^ { 2 } \right)\).

1. Complete the table below, giving the values of \(y\) to 2 decimal places.

   \(x\)	1	1.4	1.8	2.2	2.6	3
   \(y\)	3	3.47			4.39

2. Use the trapezium rule, with all the values of \(y\) from your table, to find an approximation for the value of \(\int _ { 1 } ^ { 3 } \sqrt { } \left( 10 x - x ^ { 2 } \right) \mathrm { d } x\).

6. $$y = \frac { 5 } { 3 x ^ { 2 } - 2 }$$

1. Complete the table below, giving the values of \(y\) to 2 decimal places.

   \(x\)	2	2.25	2.5	2.75	3
   \(y\)	0.5	0.38			0.2

2. Use the trapezium rule, with all the values of \(y\) from your table, to find an approximate value for \(\int _ { 2 } ^ { 3 } \frac { 5 } { 3 x ^ { 2 } - 2 } \mathrm {~d} x\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{be8f9187-055a-476f-974d-22e8e16e9996-08_537_743_941_603} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \frac { 5 } { 3 x ^ { 2 } - 2 } , x > 1\).
   At the points \(A\) and \(B\) on the curve, \(x = 2\) and \(x = 3\) respectively.
   The region \(S\) is bounded by the curve, the straight line through \(B\) and ( 2,0 ), and the line through \(A\) parallel to the \(y\)-axis. The region \(S\) is shown shaded in Figure 2.
3. Use your answer to part (b) to find an approximate value for the area of \(S\).

6. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the curve with equation $$y = \frac { 16 } { x ^ { 2 } } - \frac { x } { 2 } + 1 , \quad x > 0$$ The finite region \(R\), bounded by the lines \(x = 1\), the \(x\)-axis and the curve, is shown shaded in Figure 1. The curve crosses the \(x\)-axis at the point \(( 4,0 )\).

1. Complete the table with the values of \(y\) corresponding to \(x = 2\) and 2.5

   \(x\)	1	1.5	2	2.5	3	3.5	4
   \(y\)	16.5	7.361			1.278	0.556	0

2. Use the trapezium rule with all the values in the completed table to find an approximate value for the area of \(R\), giving your answer to 2 decimal places.
3. Use integration to find the exact value for the area of \(R\).