Trapezium rule estimation

4 \includegraphics[max width=\textwidth, alt={}, center]{c473f577-1e96-4d11-a0d5-cdfa4873c295-06_460_1445_260_349} The diagram shows the curve with equation \(y = \frac { x - 2 } { x ^ { 2 } + 8 }\). The shaded region is bounded by the curve and the lines \(x = 14\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence determine the exact \(x\)-coordinates of the stationary points.
2. Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give the answer correct to 2 significant figures.

6

1. Show that \(\int _ { 6 } ^ { 16 } \frac { 6 } { 2 x - 7 } \mathrm {~d} x = \ln 125\).
2. Use the trapezium rule with four intervals to find an approximation to $$\int _ { 1 } ^ { 17 } \log _ { 10 } x d x$$ giving your answer correct to 3 significant figures.

3

1. Find \(\int 4 \cos \left( \frac { 1 } { 3 } x + 2 \right) \mathrm { d } x\).
2. Use the trapezium rule with three intervals to find an approximation to $$\int _ { 0 } ^ { 12 } \sqrt { } \left( 4 + x ^ { 2 } \right) \mathrm { d } x$$ giving your answer correct to 3 significant figures.

1 Use the trapezium rule with three intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \ln ( 1 + \sin x ) \mathrm { d } x$$ giving your answer correct to 2 decimal places.

6 It is given that \(I = \int _ { 0 } ^ { 0.3 } \left( 1 + 3 x ^ { 2 } \right) ^ { - 2 } \mathrm {~d} x\).

1. Use the trapezium rule with 3 intervals to find an approximation to \(I\), giving the answer correct to 3 decimal places.
2. For small values of \(x , \left( 1 + 3 x ^ { 2 } \right) ^ { - 2 } \approx 1 + a x ^ { 2 } + b x ^ { 4 }\). Find the values of the constants \(a\) and \(b\). Hence, by evaluating \(\int _ { 0 } ^ { 0.3 } \left( 1 + a x ^ { 2 } + b x ^ { 4 } \right) \mathrm { d } x\), find a second approximation to \(I\), giving the answer correct to 3 decimal places.

8 \includegraphics[max width=\textwidth, alt={}, center]{8f815127-61b2-4a7f-8687-747950ea6597-3_693_1061_262_541} The diagram shows the curve \(y = x ^ { 2 } \mathrm { e } ^ { - x }\) and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Show that the tangent to the curve at the point where \(x = 1\) passes through the origin.
3. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { 1 } ^ { 3 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.

6

1. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { 1 } \frac { 1 } { 6 + 2 \mathrm { e } ^ { x } } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Find \(\int \frac { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } { \mathrm { e } ^ { 2 x } } \mathrm {~d} x\).

6. (a) Sketch the graph of \(y = 1 + \cos x , \quad 0 \leqslant x \leqslant 2 \pi\) Show on your sketch the coordinates of the points where your graph meets the coordinate axes.
(b) Use the trapezium rule, with 6 strips of equal width, to find an approximate value for $$\int _ { 0 } ^ { 2 \pi } ( 1 + \cos x ) d x$$

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.

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 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. 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.

13. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 1 } { 2 x } \ln 2 x , \quad x > \frac { 1 } { 2 }$$ The finite region \(R\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = \mathrm { e }\) and \(x = 5 \mathrm { e }\). The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 1 } { 2 x } \ln 2 x\). The values for \(y\) are given to 4 significant figures.

1. Use the trapezium rule with all the \(y\) values in the table to find an approximate value for the area of \(R\), giving your answer to 3 significant figures.
2. Using the substitution \(u = \ln 2 x\), or otherwise, find \(\int \frac { 1 } { 2 x } \ln 2 x \mathrm {~d} x\)
3. Use your answer to part (b) to find the true area of \(R\), giving your answer to 3 significant figures.
4. Using calculus, find an equation for the tangent to the curve at the point where \(x = \frac { \mathrm { e } ^ { 2 } } { 2 }\), giving your answer in the form \(y = m x + c\) where \(m\) and \(c\) are exact multiples of powers of e.

8. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { \frac { x } { x ^ { 2 } + 1 } } , \quad x \geqslant 0\) The finite 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 \(x = 7\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { \frac { x } { x ^ { 2 } + 1 } }\)

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. The region \(R\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis to form a solid of revolution.
2. Use calculus to find the exact volume of the solid of revolution formed. Write your answer in its simplest form. \includegraphics[max width=\textwidth, alt={}, center]{29b56d51-120a-4275-a761-8b8aed7bca54-24_2255_47_314_1979}

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. 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 $$1 + \frac { \log _ { 10 } 3 } { \log _ { 10 } 2 }$$

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 \(x = 4\).

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.

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 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]

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?

1 Fig. 3 shows the curve \(y = x ^ { 3 } + \sqrt { ( \sin x ) }\) for \(0 \leqslant x \leqslant \frac { \pi } { 4 }\). \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), giving your answer to 3 decimal places.
2. Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say.