Apply trapezium rule to given table

4. (a) Sketch the graph of \(y = \frac { 1 } { x } , x > 0\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 1 } { x }\), with the values for \(y\) 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, to 2 decimal places, for \(\int _ { 1 } ^ { 3 } \frac { 1 } { x } \mathrm {~d} x\)

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}

13. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve with equation \(y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right) , x > 0\) The finite region \(R\), shown shaded in Figure 4, is bounded by the curve, the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = 2\) The table below shows corresponding values of \(x\) and \(y\) for \(y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right)\), with the values of \(y\) given to 3 significant figures.

1. Use the trapezium rule, with all the values of \(y\), to obtain an estimate for the area of \(R\), giving your answer to 2 significant figures.
2. Use the substitution \(u = x ^ { 2 }\) to show that the area of \(R\) is given by $$\int _ { 1 } ^ { 4 } 6 u ^ { \frac { 1 } { 2 } } \ln ( 2 u ) \mathrm { d } u$$
3. Hence, using calculus, find the exact area of \(R\), writing your answer in the form \(a + b \ln 2\), where \(a\) and \(b\) are constants to be found.

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

A table of values of \(x\) and \(y\) for \(y = \mathrm { f } ( x )\) is shown below. Using the trapezium rule with all the values of \(y\) in the given table,

1. find an estimate for $$\int _ { 0.5 } ^ { 5.5 } \mathrm { f } ( x ) \mathrm { d } x$$ giving your answer to one decimal place.
2. Using your answer to part (a) and making your method clear, estimate $$\int _ { 0.5 } ^ { 5.5 } ( \mathrm { f } ( x ) + 4 x ) \mathrm { d } x$$

A measure of the effective voltage, \(M\) volts, in an electrical circuit is given by $$M^2 = \int_0^1 V^2 \, dt$$ where \(V\) volts is the voltage at time \(t\) seconds. Pairs of values of \(V\) and \(t\) are given in the following table.

\(t\)	0	0.25	0.5	0.75	1
\(V\)	-48	207	37	-161	-29
\(V^2\)

Use the trapezium rule with five values of \(V^2\) to estimate the value of \(M\). [6]

Fig. 7 shows a curve and the coordinates of some points on it. \includegraphics{figure_7} 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]

\includegraphics{figure_1} Figure 1 shows part of the curve with equation \(y = e^{\frac{1}{5}x^2}\) for \(x \geq 0\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(y\)-axis, the \(x\)-axis, and the line with equation \(x = 2\) The table below shows corresponding values of \(x\) and \(y\) for \(y = e^{\frac{1}{5}x^2}\)

\(x\)	0	0.5	1	1.5	2
\(y\)	1	\(e^{0.05}\)	\(e^{0.2}\)	\(e^{0.45}\)	\(e^{0.8}\)

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 2 decimal places. [3]
2. Use your answer to part (a) to deduce an estimate for
   1. \(\int_0^2 \left( 4 + e^{\frac{1}{5}x^2} \right) dx\)
   2. \(\int_1^3 e^{\frac{1}{5}(x-1)^2} dx\) giving your answers to 2 decimal places. [2]