Complete table then apply trapezium rule

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. 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}

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

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 \(R\).

9. \(y\) \begin{figure}[h] \end{figure} The finite region \(R\), as shown in Figure 2, is bounded by the \(x\)-axis and the curve with equation $$y = 27 - 2 x - 9 \sqrt { } x - \frac { 16 } { x ^ { 2 } } , \quad x > 0$$ The curve crosses the \(x\)-axis at the points \(( 1,0 )\) and \(( 4,0 )\).

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

   \(x\)	1	1.5	2	2.5	3	3.5	4
   \(y\)	0	5.866		5.210		1.856	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 \(R\).

5. The curve \(C\) has equation $$y = x \sqrt { } \left( x ^ { 3 } + 1 \right) , \quad 0 \leqslant x \leqslant 2$$

1. Complete the table below, giving the values of \(y\) to 3 decimal places at \(x = 1\) and \(x = 1.5\).

   \(x\)	0	0.5	1	1.5	2
   \(y\)	0	0.530			6

2. Use the trapezium rule, with all the \(y\) values from your table, to find an approximation for the value of \(\int _ { 0 } ^ { 2 } x \sqrt { } \left( x ^ { 3 } + 1 \right) \mathrm { d } x\), giving your answer to 3 significant figures. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{22ebc302-765c-4734-b312-b286ccb20be9-06_1110_644_1119_648} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows the curve \(C\) with equation \(y = x \sqrt { } \left( x ^ { 3 } + 1 \right) , 0 \leqslant x \leqslant 2\), and the straight line segment \(l\), which joins the origin and the point \(( 2,6 )\). The finite region \(R\) is bounded by \(C\) and \(l\).
3. Use your answer to part (b) to find an approximation for the area of \(R\), giving your answer to 3 significant figures.
   (3) \section*{LU}

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

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

   \(x\)	0	0.5	1	1.5	2
   \(y\)			2.646	3.630

2. Use the trapezium rule, with all the values of \(y\) from your table, to find an approximation for the value of \(\int _ { 0 } ^ { 2 } \sqrt { } \left( 5 ^ { x } + 2 \right) \mathrm { d } x\).

4. (a) Complete the table below, giving values of \(\sqrt { } \left( 2 ^ { x } + 1 \right)\) to 3 decimal places. \begin{figure}[h] \end{figure} Figure 1 shows the region \(R\) which is bounded by the curve with equation \(y = \sqrt { } \left( 2 ^ { x } + 1 \right)\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\) (b) Use the trapezium rule, with all the values from your table, to find an approximation for the area of \(R\).
(c) By reference to the curve in Figure 1 state, giving a reason, whether your approximation in part (b) is an overestimate or an underestimate for the area of \(R\).

1. $$y = 3 ^ { x } + 2 x$$

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

   \(x\)	0	0.2	0.4	0.6	0.8	1
   \(y\)	1	1.65				5

2. Use the trapezium rule, with all the values of \(y\) from your table, to find an approximate value for \(\int _ { 0 } ^ { 1 } \left( 3 ^ { x } + 2 x \right) d x\).

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

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.

2. $$y = \frac { x } { \sqrt { ( 1 + x ) } }$$

1. Complete the table below with the value of \(y\) corresponding to \(x = 1.3\), giving your answer to 4 decimal places.

   \(x\)	1	1.1	1.2	1.3	1.4	1.5
   \(y\)	0.7071	0.7591	0.8090		0.9037	0.9487

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an approximate value for $$\int _ { 1 } ^ { 1.5 } \frac { x } { \sqrt { } ( 1 + x ) } \mathrm { d } x$$ giving your answer to 3 decimal places.
   You must show clearly each stage of your working.

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { } ( 2 x - 1 ) , x \geqslant 0.5\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 2\) and \(x = 10\). The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { } ( 2 x - 1 )\).

1. Complete the table with the values of \(y\) corresponding to \(x = 4\) and \(x = 8\).
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to find an approximate value for the area of \(R\), giving your answer to 2 decimal places.
3. State whether your approximate value in part (b) is an overestimate or an underestimate for the area of \(R\).

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { } \left( x ^ { 2 } + 1 \right) , x \geqslant 0\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) The table below shows corresponding values for \(x\) and \(y\) for \(y = \sqrt { } \left( x ^ { 2 } + 1 \right)\).

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

2. The curve \(C\) has equation $$y = 8 - 2 ^ { x - 1 } , \quad 0 \leqslant x \leqslant 4$$

1. Complete the table below with the value of \(y\) corresponding to \(x = 1\)

   \(x\)	0	1	2	3	4
   \(y\)	7.5		6	4	0

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to find an approximate value for \(\int _ { 0 } ^ { 4 } \left( 8 - 2 ^ { x - 1 } \right) \mathrm { d } x\) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{582cda45-80fc-43a8-90e6-1cae08cb1534-03_650_606_1016_671} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation \(y = 8 - 2 ^ { x - 1 } , \quad 0 \leqslant x \leqslant 4\) The curve \(C\) meets the \(x\)-axis at the point \(A\) and meets the \(y\)-axis at the point \(B\).
   The region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\) and the straight line through \(A\) and \(B\).
3. Use your answer to part (b) to find an approximate value for the area of \(R\).

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = \frac { ( x + 2 ) ^ { \frac { 3 } { 2 } } } { 4 } , \quad x \geqslant - 2$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the line with equation \(x = 10\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { ( x + 2 ) ^ { \frac { 3 } { 2 } } } { 4 }\)

1. Complete the table, giving values of \(y\) corresponding to \(x = 2\) and \(x = 6\)

   \(x\)	- 2	2	6	10
   \(y\)	0			\(6 \sqrt { } 3\)

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

12. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$ The finite region \(S\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = 1\) and \(x = 3\)

1. Complete the table below with the value of \(y\) corresponding to \(x = 2\). Give your answer to 4 decimal places.

   \(x\)	1	1.5	2	2.5	3
   \(y\)	2	1.3041		0.9089	1.2958

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(S\), giving your answer to 3 decimal places.
3. Use calculus to find the exact area of \(S\). Give your answer in the form \(\frac { a } { b } + \ln c\), where \(a , b\) and \(c\) are integers.
4. Hence calculate the percentage error in using your answer to part (b) to estimate the area of \(S\). Give your answer to one decimal place.
5. Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of \(S\). \section*{Question 12 continued} \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) 13. (a) Express \(10 \cos \theta - 3 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\) Give the exact value of \(R\) and give the value of \(\alpha\) to 2 decimal places. Alana models the height above the ground of a passenger on a Ferris wheel by the equation $$H = 12 - 10 \cos ( 30 t ) ^ { \circ } + 3 \sin ( 30 t ) ^ { \circ }$$ where the height of the passenger above the ground is \(H\) metres at time \(t\) minutes after the wheel starts turning. \includegraphics[max width=\textwidth, alt={}, center]{03548211-79cb-4629-b6ca-aa9dfcc77a33-23_419_567_516_1160}
   (b) Calculate
   1. the maximum value of \(H\) predicted by this model,
   2. the value of \(t\) when this maximum first occurs. Give each answer to 2 decimal places.
      (c) Calculate the value of \(t\) when the passenger is 18 m above the ground for the first time. Give your answer to 2 decimal places.
      (d) Determine the time taken for the Ferris wheel to complete two revolutions. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \section*{Question 13 continued}

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = x \cos x , \quad x \in \mathbb { R }$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\) and the \(x\)-axis for \(\frac { 3 \pi } { 2 } \leqslant x \leqslant \frac { 5 \pi } { 2 }\)

1. Complete the table below with the exact value of \(y\) corresponding to \(x = \frac { 7 \pi } { 4 }\) and with the exact value of \(y\) corresponding to \(x = \frac { 9 \pi } { 4 }\)

   \(x\)	\(\frac { 3 \pi } { 2 }\)	\(\frac { 7 \pi } { 4 }\)	\(2 \pi\)	\(\frac { 9 \pi } { 4 }\)	\(\frac { 5 \pi } { 2 }\)
   \(y\)	0		\(2 \pi\)		0

2. Use the trapezium rule, with all five \(y\) values in the completed table, to find an approximate value for the area of \(R\), giving your answer to 4 significant figures.
3. Find $$\int x \cos x d x$$
4. Using your answer from part (c), find the exact area of the region \(R\).

13. \begin{figure}[h] \end{figure} Figure 5 shows a sketch of part of the curve with equation \(y = 2 - \ln x , x > 0\) The finite region \(R\), shown shaded in Figure 5, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \mathrm { e }\). The table below shows corresponding values of \(x\) and \(y\) for \(y = 2 - \ln x\)

1. Complete the table giving the value 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 3 decimal places.
3. Use integration by parts to show that \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x = x ( \ln x ) ^ { 2 } - 2 x \ln x + 2 x + c\) The area \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
4. Use calculus to find the exact volume of the solid generated. Write your answer in the form \(\pi \mathrm { e } ( p \mathrm { e } + q )\), where \(p\) and \(q\) are integers to be found.

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation \(y = 2 \mathrm { e } ^ { - x } \sqrt { \sin x } , 0 \leqslant x \leqslant \pi\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve and the \(x\)-axis.

1. Complete the table below with the value of \(y\) corresponding to \(x = \frac { \pi } { 2 }\), giving your answer to 5 decimal places.

   \(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
   \(y\)	0	0.76679		0.15940	0

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of the region \(R\). Give your answer to 4 decimal places.
3. Given \(y = 2 \mathrm { e } ^ { - x } \sqrt { \sin x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) for \(0 < x < \pi\). The curve \(C\) has a maximum turning point when \(x = a\).
4. Use your answer to part (c) to find the value of \(a\), giving your answer to 3 decimal places.

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { e } ^ { x } \sqrt { \sin x } , 0 \leqslant x \leqslant \pi\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve and the \(x\)-axis.

1. Complete the table below with the values of \(y\) corresponding to \(x = \frac { \pi } { 4 }\) and \(x = \frac { \pi } { 2 }\), giving your answers to 5 decimal places.

   \(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
   \(y\)	0			8.87207	0

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of the region \(R\). Give your answer to 4 decimal places. The curve \(y = \mathrm { e } ^ { x } \sqrt { \sin x } , 0 \leqslant x \leqslant \pi\), has a maximum turning point at \(Q\), shown in Figure 1.
3. Find the \(x\) coordinate of \(Q\).