Complete table then estimate

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

1. Complete the table above giving the missing values of \(y\) to 4 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places.
3. Use the substitution \(u = x ^ { 2 } + 2\) to show that the area of \(R\) is $$\frac { 1 } { 2 } \int _ { 2 } ^ { 4 } ( u - 2 ) \ln u \mathrm {~d} u$$
4. Hence, or otherwise, find the exact area of \(R\).

3. \begin{figure}[h] \end{figure} Figure 1 shows the finite region \(R\) bounded by the \(x\)-axis, the \(y\)-axis, the line \(x = \frac { \pi } { 2 }\) and the curve with equation $$y = \sec \left( \frac { 1 } { 2 } x \right) , \quad 0 \leqslant x \leqslant \frac { \pi } { 2 }$$ The table shows corresponding values of \(x\) and \(y\) for \(y = \sec \left( \frac { 1 } { 2 } x \right)\).

1. Complete the table above giving the missing value of \(y\) to 6 decimal places.
2. Using the trapezium rule, with all of the values of \(y\) from the completed table, find an approximation for the area of \(R\), giving your answer to 4 decimal places. Region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
3. Use calculus to find the exact volume of the solid formed.

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) } , x \in \mathbb { R }\) 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 \(x = 1\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) }\)

1. Complete the table above by giving the missing value of \(y\) to 5 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to find an estimate for the area of \(R\), giving your answer to 4 decimal places.
3. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that the area of \(R\) can be given by $$\int _ { a } ^ { b } \frac { 6 } { u ( u + 2 ) } \mathrm { d } u$$ where \(a\) and \(b\) are constants to be determined.
4. Hence use calculus to find the exact area of \(R\). [Solutions based entirely on graphical or numerical methods are not acceptable.]

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.

4. \includegraphics[max width=\textwidth, alt={}, center]{fe47eac1-645a-46c6-a2b9-c4ad0bcaa538-1_433_844_1416_575} The diagram shows the curve with equation \(y = \left( x - \log _ { 10 } x \right) ^ { 2 } , x > 0\).

1. Copy and complete the table below for points on the curve, giving the \(y\) values to 2 decimal places.

   \(x\)	2	3	4	5	6
   \(y\)	2.89	6.36

   The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 6\).
2. Use the trapezium rule with all the values in your table to estimate the area of the shaded region.
3. State, with a reason, whether your answer to part (b) is an under-estimate or an over-estimate of the true area.

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.

2 Fig. 2 shows the curve \(y = \overline { 1 + x ^ { 2 } }\). \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.

4

1. Complete the table of values for the curve \(y = \sqrt { \cos x }\).

   \(X\)	0	\(\frac { \pi } { 8 }\)	\(\frac { \pi } { 4 }\)	\(\frac { 3 \pi } { 8 }\)	\(\frac { \pi } { 2 }\)
   \(y\)		0.9612	0.8409

   Hence use the trapezium rule with strip width \(h = \frac { \pi } { 8 }\) to estimate the value of the integral \(\int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \cos x } \mathrm {~d} x\), giving your answer to 3 decimal places. Fig. 4 shows the curve \(y = \sqrt { \cos x }\) for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-4_459_751_799_638} \captionsetup{labelformat=empty} \caption{Fig. 4}
   \end{figure}
2. State, with a reason, whether the trapezium rule with a strip width of \(\frac { \pi } { 16 }\) would give a larger or smaller estimate of the integral.

4 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.

4

1. Complete the table of values for the curve \(y = \sqrt { \cos x }\).

   \(x\)	0	\(\frac { \pi } { 8 }\)	\(\frac { \pi } { 4 }\)	\(\frac { 3 \pi } { 8 }\)	\(\frac { \pi } { 2 }\)
   \(y\)		0.9612	0.8409

   Hence use the trapezium rule with strip width \(h = \frac { \pi } { 8 }\) to estimate the value of the integral \(\int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \cos x } \mathrm {~d} x\), giving your answer to 3 decimal places. Fig. 4 shows the curve \(y = \sqrt { \cos x }\) for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-02_457_750_1446_653} \captionsetup{labelformat=empty} \caption{Fig. 4}
   \end{figure}
2. State, with a reason, whether the trapezium rule with a strip width of \(\frac { \pi } { 16 }\) would give a larger or smaller estimate of the integral.

6. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = \left( x - \log _ { 10 } x \right) ^ { 2 } , x > 0\).

1. Copy and complete the table below for points on the curve, giving the \(y\) values to 2 decimal places.

   \(x\)	2	3	4	5	6
   \(y\)	2.89	6.36

   The shaded area is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 6\).
2. Use the trapezium rule with all the values in your table to estimate the area of the shaded region.
3. State, with a reason, whether your answer to part (b) is an under-estimate or an over-estimate of the true area.

6. \begin{figure}[h] \end{figure} Figure 2 shows the cross-section of a road tunnel and its concrete surround. The curved section of the tunnel is modelled by the curve with equation \(y = 8 \sqrt { \left( \sin \frac { \pi x } { 10 } \right) }\), in the interval \(0 \leq x \leq\) 10. The concrete surround is represented by the shaded area bounded by the curve, the \(x\)-axis and the lines \(x = - 2 , x = 12\) and \(y = 10\). The units on both axes are metres.

1. Using this model, copy and complete the table below, giving the values of \(y\) to 2 decimal places.

   \(x\)	0	2	4	6	8	10
   \(y\)	0	6.13				0

   The area of the cross-section of the tunnel is given by \(\int _ { 0 } ^ { 10 } y \mathrm {~d} x\).
2. Estimate this area, using the trapezium rule with all the values from your table.
3. Deduce an estimate of the cross-sectional area of the concrete surround.
4. State, with a reason, whether your answer in part (c) over-estimates or under-estimates the true value.
   (2)

1. The following is a table of values for \(y = \sqrt { } ( 1 + \sin x )\), where \(x\) is in radians.

1. Find the value of \(p\) and the value of \(q\).
   (2)
2. Use the trapezium rule and all the values of \(y\) in the completed table to obtain an estimate of \(I\), where $$I = \int _ { 0 } ^ { 2 } \sqrt { } ( 1 + \sin x ) \mathrm { d } x$$ (4)

9. The following is a table of values for \(y = \sqrt { } ( 1 + \sin x )\), where \(x\) is in radians.

1. Find the value of \(p\) and the value of \(q\).
2. Use the trapezium rule and all the values of \(y\) in the completed table to obtain an estimate of \(I\), where $$I = \int _ { 0 } ^ { 2 } \sqrt { } ( 1 + \sin x ) \mathrm { d } x$$