1.09f Trapezium rule: numerical integration

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 \(Q\).

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.

2. (a) Given that \(y = \sec x\), complete the table with the values of \(y\) corresponding to \(x = \frac { \pi } { 16 } , \frac { \pi } { 8 }\) and \(\frac { \pi } { 4 }\). (b) Use the trapezium rule, with all the values for \(y\) in the completed table, to obtain an estimate for \(\int _ { 0 } ^ { \frac { \pi } { 4 } } \sec x \mathrm {~d} x\). Show all the steps of your working, and give your answer to 4 decimal places. The exact value of \(\int _ { 0 } ^ { \frac { \pi } { 4 } } \sec x \mathrm {~d} x\) is \(\ln ( 1 + \sqrt { } 2 )\).
(c) Calculate the \% error in using the estimate you obtained in part (b).

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 \(R\).

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.

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with equation \(y = ( x - 1 ) \ln x , \quad x > 0\).

1. Complete the table with the values of \(y\) corresponding to \(x = 1.5\) and \(x = 2.5\).

   \(x\)	1	1.5	2	2.5	3
   \(y\)	0		\(\ln 2\)		\(2 \ln 3\)

   Given that \(I = \int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x\),
2. use the trapezium rule
   1. with values of \(y\) at \(x = 1,2\) and 3 to find an approximate value for \(I\) to 4 significant figures,
   2. with values of \(y\) at \(x = 1,1.5,2,2.5\) and 3 to find another approximate value for \(I\) to 4 significant figures.
3. Explain, with reference to Figure 3, why an increase in the number of values improves the accuracy of the approximation.
4. Show, by integration, that the exact value of \(\int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x\) is \(\frac { 3 } { 2 } \ln 3\).

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 \(R\).

1. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = \sqrt { } \left( 0.75 + \cos ^ { 2 } x \right)\). 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 \(x = \frac { \pi } { 3 }\).

1. Complete the table with values of \(y\) corresponding to \(x = \frac { \pi } { 6 }\) and \(x = \frac { \pi } { 4 }\).

   \(x\)	0	\(\frac { \pi } { 12 }\)	\(\frac { \pi } { 6 }\)	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 3 }\)
   \(y\)	1.3229	1.2973			1

2. Use the trapezium rule
   1. with the values of \(y\) at \(x = 0 , x = \frac { \pi } { 6 }\) and \(x = \frac { \pi } { 3 }\) to find an estimate of the area of \(R\). Give your answer to 3 decimal places.
   2. with the values of \(y\) at \(x = 0 , x = \frac { \pi } { 12 } , x = \frac { \pi } { 6 } , x = \frac { \pi } { 4 }\) and \(x = \frac { \pi } { 3 }\) to find a further estimate of the area of \(R\). Give your answer to 3 decimal places.
      (6) \section*{LU}

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 \(R\).

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

1. Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of \(R\), giving your answer to 2 decimal places.
2. Find \(\int x ^ { \frac { 1 } { 2 } } \ln 2 x \mathrm {~d} x\).
3. Hence find the exact area of \(R\), giving your answer in the form \(a \ln 2 + b\), where \(a\) and \(b\) are exact constants.

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 $$y = ( 2 - x ) \mathrm { e } ^ { 2 x } , \quad x \in \mathbb { R }$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the \(y\)-axis. The table below shows corresponding values of \(x\) and \(y\) for \(y = ( 2 - x ) \mathrm { e } ^ { 2 x }\)

1. Use the trapezium rule with all the values of \(y\) in the table, to obtain an approximation for the area of \(R\), giving your answer to 2 decimal places.
2. Explain how the trapezium rule can be used to give a more accurate approximation for the area of \(R\).
3. Use calculus, showing each step in your working, to obtain an exact value for the area of \(R\). Give your answer in its simplest form.

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.]

2. $$f ( x ) = 10 - 2 x - \frac { 1 } { 2 \sqrt { x } } - \frac { 1 } { x ^ { 3 } } \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [0.4, 0.5]
2. Determine \(\mathrm { f } ^ { \prime } ( x )\).
3. Using \(x _ { 0 } = 0.5\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has another root \(\beta\) in the interval [4.8, 4.9]
   [0pt]
4. Use linear interpolation once on the interval [4.8, 4.9] to find an approximation to \(\beta\), giving your answer to 3 decimal places.

7. $$f ( x ) = x ^ { \frac { 3 } { 2 } } + x - 3$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root, \(\alpha\), in the interval \([ 1,2 ]\) [0pt]
2. Starting with the interval [1, 2], use interval bisection twice to show that \(\alpha\) lies in the interval [1.25, 1.5]
3. 1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
   2. Using 1.375 as a first approximation for \(\alpha\), apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to determine a second approximation for \(\alpha\), giving your answer to 3 decimal places.
      [0pt]
4. Use linear interpolation once on the interval [1.25,1.5] to obtain a different approximation for \(\alpha\), giving your answer to 3 decimal places.