Complete table then apply trapezium rule

5. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the curve with equation $$y = x \mathrm { e } ^ { 2 x } , \quad x \geqslant 0$$ The finite region \(R\) bounded by the lines \(x = 1\), the \(x\)-axis and the curve is shown shaded in Figure 1.

1. Use integration to find the exact value for the area of \(R\).
2. Complete the table with the values of \(y\) corresponding to \(x = 0.4\) and 0.8 .

   \(x\)	0	0.2	0.4	0.6	0.8	1
   \(y = x \mathrm { e } ^ { 2 x }\)	0	0.29836		1.99207		7.38906

3. Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures.

8. $$I = \int _ { 0 } ^ { 5 } \mathrm { e } ^ { \sqrt { } ( 3 x + 1 ) } \mathrm { d } x$$

1. Given that \(y = \mathrm { e } ^ { \sqrt { } ( 3 x + 1 ) }\), complete the table with the values of \(y\) corresponding to \(x = 2\), 3 and 4.

   \(x\)	0	1	2	3	4	5
   \(y\)	\(\mathrm { e } ^ { 1 }\)	\(\mathrm { e } ^ { 2 }\)				\(\mathrm { e } ^ { 4 }\)

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the original integral \(I\), giving your answer to 4 significant figures.
3. Use the substitution \(t = \sqrt { } ( 3 x + 1 )\) to show that \(I\) may be expressed as \(\int _ { a } ^ { b } k t e ^ { t } \mathrm {~d} t\), giving the values of \(a , b\) and \(k\).
4. Use integration by parts to evaluate this integral, and hence find the value of \(I\) correct to 4 significant figures, showing all the steps in your working.

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

1. Complete the table below with the values of \(y\) corresponding to \(x = \frac { \pi } { 4 }\) and \(\frac { \pi } { 2 }\), 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 in the completed table, to obtain an estimate for the area of the region \(R\). Give your answer to 4 decimal places.

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

1. Complete the table with the values of \(y\) corresponding to \(x = 2\) and \(x = 2.5\), giving your answers to 3 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. 1. Use integration by parts to find \(\int x \ln x \mathrm {~d} x\).
   2. Hence find the exact area of \(R\), giving your answer in the form \(\frac { 1 } { 4 } ( a \ln 2 + b )\), where \(a\) and \(b\) are integers.

7. $$I = \int _ { 2 } ^ { 5 } \frac { 1 } { 4 + \sqrt { } ( x - 1 ) } \mathrm { d } x$$

1. Given that \(y = \frac { 1 } { 4 + \sqrt { } ( x - 1 ) }\), complete the table below with values of \(y\) corresponding to \(x = 3\) and \(x = 5\). Give your values to 4 decimal places.

   \(x\)	2	3	4	5
   \(y\)	0.2		0.1745

2. Use the trapezium rule, with all of the values of \(y\) in the completed table, to obtain an estimate of \(I\), giving your answer to 3 decimal places.
3. Using the substitution \(x = ( u - 4 ) ^ { 2 } + 1\), or otherwise, and integrating, find the exact value of \(I\).

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with equation \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) } , 0 \leqslant x \leqslant \frac { \pi } { 2 }\).
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve and the \(x\)-axis. The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) }\).

1. Complete the table above 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 obtain an estimate for the area of \(R\), giving your answer to 4 decimal places.
3. Using the substitution \(u = 1 + \cos x\), or otherwise, show that $$\int \frac { 2 \sin 2 x } { ( 1 + \cos x ) } d x = 4 \ln ( 1 + \cos x ) - 4 \cos x + k$$ where \(k\) is a constant.
4. Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.

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

1. Complete the table with the value of \(y\) corresponding to \(x = 3\), giving your answer to 4 decimal places.
   (1)

   \(x\)	1	2	3	4
   \(y\)	0.5	0.8284		1.3333

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate of the area of the region \(R\), giving your answer to 3 decimal places.
3. Use the substitution \(u = 1 + \sqrt { } x\), to find, by integrating, the exact area of \(R\).
   \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a98d4a7f-1e6d-4294-9b5c-c945e8fbe83e-07_743_1568_219_182} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = 1 - \frac { 1 } { 2 } t , \quad y = 2 ^ { t } - 1$$ The curve crosses the \(y\)-axis at the point \(A\) and crosses the \(x\)-axis at the point \(B\).

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

1. Complete the table above by giving the missing 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 2 decimal places.
3. 1. Using the substitution \(u = 1 + 3 \mathrm { e } ^ { - x }\), or otherwise, find $$\int \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) } \mathrm { d } x$$
   2. Hence find the value of the area of \(R\).

7. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = \sqrt { } ( \tan x )\). The finite region \(R\), which is bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), is shown shaded in Figure 1.

1. Given that \(y = \sqrt { } ( \tan x )\), complete the table with the values of \(y\) corresponding to \(x = \frac { \pi } { 16 } , \frac { \pi } { 8 }\) and \(\frac { 3 \pi } { 16 }\), giving your answers to 5 decimal places.

   \(x\)	0	\(\frac { \pi } { 16 }\)	\(\frac { \pi } { 8 }\)	\(\frac { 3 \pi } { 16 }\)	\(\frac { \pi } { 4 }\)
   \(y\)	0				1

2. Use the trapezium rule with all the values of \(y\) in the completed table to obtain an estimate for the area of the shaded region \(R\), giving your answer to 4 decimal places. The region \(R\) is rotated through \(2 \pi\) radians around the \(x\)-axis to generate a solid of revolution.
3. Use integration to find an exact value for the volume of the solid generated. \section*{LO}

1. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = \mathrm { e } ^ { 0.5 x ^ { 2 } }\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\).

1. Complete the table with the values of \(y\) corresponding to \(x = 0.8\) and \(x = 1.6\).

   \(x\)	0	0.4	0.8	1.2	1.6	2
   \(y\)	\(\mathrm { e } ^ { 0 }\)	\(\mathrm { e } ^ { 0.08 }\)		\(\mathrm { e } ^ { 0.72 }\)		\(\mathrm { e } ^ { 2 }\)

2. Use the trapezium rule with all the values in the table to find an approximate value for the area of \(R\), giving your answer to 4 significant figures.

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

1. Complete the table above giving the missing value of \(y\) to 5 decimal places.
2. Using the trapezium rule, with all the values of \(y\) from the completed table, find an approximation for the area of \(R\), giving your answer to 3 decimal places.
3. Use integration to find the exact area of \(R\).

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 \(R\).

5. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(x = 4 t \mathrm { e } ^ { - \frac { 1 } { 3 } t } + 3\). The finite region \(R\) shown shaded in Figure 1 is bounded by the curve, the \(x\)-axis, the \(t\)-axis and the line \(t = 8\).

1. Complete the table with the value of \(x\) corresponding to \(t = 6\), giving your answer to 3 decimal places.

   \(t\)	0	2	4	6	8
   \(x\)	3	7.107	7.218		5.223

2. Use the trapezium rule with all the values of \(x\) in the completed table to obtain an estimate for the area of the region \(R\), giving your answer to 2 decimal places.
3. Use calculus to find the exact value for the area of \(R\).
4. Find the difference between the values obtained in part (b) and part (c), giving your answer to 2 decimal places.

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.

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

1. Complete the table above, giving the missing 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 to find the exact value for the area of \(R\).

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.

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.

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.

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)

3. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = \ln ( 2 + \cos x ) , 0 \leq x \leq \pi\).

1. Complete the table below for points on the curve, giving the \(y\) values to 4 decimal places.
2. Giving your answers to 3 decimal places, find estimates for the area of the region bounded by the curve and the coordinate axes using the trapezium rule with
   1. 1 strip,
   2. 2 strips,
   3. 4 strips.
3. Making your reasoning clear, suggest a value to 2 decimal places for the actual area of the region bounded by the curve and the coordinate axes.

   \(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
   \(y\)	1.0986				0

   1. continued
   \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{3056ad22-f87b-46c3-86cf-d46939927465-06_563_983_146_379} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows the curve with parametric equations $$x = \tan \theta , \quad y = \cos ^ { 2 } \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$ The shaded region bounded by the curve, the \(x\)-axis and the lines \(x = - 1\) and \(x = 1\) is rotated through \(2 \pi\) radians about the \(x\)-axis.

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation \(y = \frac{10}{2x + 5\sqrt{x}}\), \(x > 0\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, and the lines with equations \(x = 1\) and \(x = 4\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac{10}{2x + 5\sqrt{x}}\)

1. [(a)] Complete the table above by giving the missing value of \(y\) to 5 decimal places. \hfill [1]
2. [(b)] 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. \hfill [3]
3. [(c)] By reference to the curve in Figure 1, state, giving a reason, whether your estimate in part (b) is an overestimate or an underestimate for the area of \(R\). \hfill [1]
4. [(d)] Use the substitution \(u = \sqrt{x}\), or otherwise, to find the exact value of $$\int_1^4 \frac{10}{2x + 5\sqrt{x}} dx$$ \hfill [6]