1.09f Trapezium rule: numerical integration

2 Use the trapezium rule with four ordinates (three strips) to find an approximate value for $$\int _ { 0 } ^ { 3 } \sqrt { 2 ^ { x } } \mathrm {~d} x$$ giving your answer to three decimal places.

6 The diagram shows a sketch of the curve with equation \(y = 3 \left( 2 ^ { x } + 1 \right)\). \includegraphics[max width=\textwidth, alt={}, center]{ad574bde-3bf1-45be-a454-9c723088b357-5_465_851_390_607} The curve \(y = 3 \left( 2 ^ { x } + 1 \right)\) intersects the \(y\)-axis at the point \(A\).

1. Find the \(y\)-coordinate of the point \(A\).
2. Use the trapezium rule with four ordinates (three strips) to find an approximate value for \(\int _ { 0 } ^ { 6 } 3 \left( 2 ^ { x } + 1 \right) d x\).
3. The line \(y = 21\) intersects the curve \(y = 3 \left( 2 ^ { x } + 1 \right)\) at the point \(P\).
   1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$2 ^ { x } = 6$$
   2. Use logarithms to find the \(x\)-coordinate of \(P\), giving your answer to three significant figures.

2 Use Simpson's rule with 5 ordinates ( 4 strips) to find an approximation to $$\int _ { 1 } ^ { 3 } \frac { 1 } { \sqrt { 1 + x ^ { 3 } } } \mathrm {~d} x$$ giving your answer to three significant figures.

2 Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to $$\int _ { 1 } ^ { 3 } \frac { 1 } { \sqrt { 1 + x ^ { 3 } } } \mathrm {~d} x$$ giving your answer to three significant figures.

1 Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to \(\int _ { 1 } ^ { 9 } \frac { 1 } { 1 + \sqrt { x } } \mathrm {~d} x\), giving your answer to three significant figures.

5

1. Use the mid-ordinate rule with four strips to find an estimate for \(\int _ { 0 } ^ { 12 } \ln \left( x ^ { 2 } + 5 \right) \mathrm { d } x\), giving your answer to three significant figures.
2. A curve has equation \(y = \ln \left( x ^ { 2 } + 5 \right)\).
   1. Show that this equation can be rewritten as \(x ^ { 2 } = \mathrm { e } ^ { y } - 5\).
   2. The region bounded by the curve, the lines \(y = 5\) and \(y = 10\) and the \(y\)-axis is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find the exact value of the volume of the solid generated.
3. The graph with equation \(y = \ln \left( x ^ { 2 } + 5 \right)\) is stretched with scale factor 4 parallel to the \(x\)-axis, and then translated through \(\left[ \begin{array} { l } 0 \\ 3 \end{array} \right]\) to give the graph with equation \(y = \mathrm { f } ( x )\). Write down an expression for \(\mathrm { f } ( x )\).

4 [Figure 1, printed on the insert, is provided for use in this question.]

1. Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to \(\int _ { 1 } ^ { 2 } 3 ^ { x } \mathrm {~d} x\), giving your answer to three significant figures.
2. The curve \(y = 3 ^ { x }\) intersects the line \(y = x + 3\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0.5 and 1.5.
   2. Show that the equation \(3 ^ { x } = x + 3\) can be rearranged into the form $$x = \frac { \ln ( x + 3 ) } { \ln 3 }$$
   3. Use the iteration \(x _ { n + 1 } = \frac { \ln \left( x _ { n } + 3 \right) } { \ln 3 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\) to two significant figures.
   4. The sketch on Figure 1 shows part of the graphs of \(y = \frac { \ln ( x + 3 ) } { \ln 3 }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.

1

1. Use the mid-ordinate rule with four strips to find an estimate for \(\int _ { 1.5 } ^ { 5.5 } \mathrm { e } ^ { 2 - x } \ln ( 3 x - 2 ) \mathrm { d } x\), giving your answer to three decimal places.
   [0pt] [4 marks]
2. Find the exact value of the gradient of the curve \(y = \mathrm { e } ^ { 2 - x } \ln ( 3 x - 2 )\) at the point on the curve where \(x = 2\).
   [0pt] [4 marks]

2

1. Use the trapezium rule, with four strips each of width 0.25 , to find an approximate value for \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \mathrm {~d} x\).
2. Explain how the trapezium rule might be used to give a better approximation to the integral given in part (a).

11 The region \(R\) enclosed by the lines \(x = 1 , x = 6 , y = 0\) and the curve $$y = \ln ( 8 - x )$$ is shown shaded in Figure 3 below. \begin{figure}[h] \end{figure} All distances are measured in centimetres.
11

1. Use a single trapezium to find an approximate value of the area of the shaded region, giving your answer in \(\mathrm { cm } ^ { 2 }\) to two decimal places.
   [0pt] [2 marks]
   \section*{Question 11 continues on the next page} 11
2. Shape \(B\) is made from four copies of region \(R\) as shown in Figure 4 below. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 4} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-18_707_711_438_667}
   \end{figure} Shape \(B\) is cut from metal of thickness 2 mm
   The metal has a density of \(10.5 \mathrm {~g} / \mathrm { cm } ^ { 3 }\) Use the trapezium rule with six ordinates to calculate an approximate value of the mass of Shape B. Give your answer to the nearest gram.
   11
3. Without further calculation, give one reason why the mass found in part (b) may be:
   11 (c) (i) an underestimate.
   11 (c) (ii) an overestimate.

14 The region bounded by the curve $$y = ( 2 x - 8 ) \ln x$$ and the \(x\)-axis is shaded in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-26_867_908_543_566} 14

1. Use the trapezium rule with 5 ordinates to find an estimate for the area of the shaded region. Give your answer correct to three significant figures.
   14
2. Show that the exact area is given by $$32 \ln 2 - \frac { 33 } { 2 }$$ Fully justify your answer.

5

1. Use the trapezium rule with 6 ordinates ( 5 strips) to find an approximate value for the shaded area. Give your answer to four decimal places.
   5
2. Using your answer to part (a) deduce an estimate for \(\int _ { 1 } ^ { 4 } \frac { 20 } { \mathrm { e } ^ { x } - 1 } \mathrm {~d} x\)

6 The diagram below shows part of the graph of \(y = \mathrm { f } ( x )\) The line \(T P Q\) is a tangent to the graph of \(y = \mathrm { f } ( x )\) at the point \(P \left( \frac { a + b } { 2 } , \mathrm { f } \left( \frac { a + b } { 2 } \right) \right)\) The points \(S ( a , 0 )\) and \(T\) lie on the line \(x = a\) The points \(Q\) and \(R ( b , 0 )\) lie on the line \(x = b\) \includegraphics[max width=\textwidth, alt={}, center]{74b8239a-1f46-45e7-ad20-2dce7bf4baf6-05_748_696_669_671} Sharon uses the mid-ordinate rule with one strip to estimate the value of the integral \(\int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x\) By considering the area of the trapezium QRST, state, giving reasons, whether you would expect Sharon's estimate to be an under-estimate or an over-estimate.

1. (a) Use Simpson's rule with 4 intervals to find an estimate for

$$\int _ { 0 } ^ { 2 } \mathrm { e } ^ { \sin ^ { 2 } x } \mathrm {~d} x$$ Give your answer to 3 significant figures. Given that \(\int _ { 0 } ^ { 2 } \mathrm { e } ^ { \mathrm { sin } ^ { 2 } x } \mathrm {~d} x = 3.855\) to 4 significant figures,
(b) comment on the accuracy of your answer to part (a).

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 $$\int _ { 0.5 } ^ { 5.5 } ( \mathrm { f } ( x ) + 4 x ) \mathrm { d } x$$

\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]

\includegraphics{figure_4} The diagram shows the curve with equation \(y = \sqrt{1 + e^{0.5x}}\). The shaded region is bounded by the curve and the straight lines \(x = 0\), \(x = 6\) and \(y = 0\).

1. Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures. [3]
2. The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced. [4]

\includegraphics{figure_5} The diagram shows the curve \(y = \sqrt{1 + e^{4x}}\) for \(0 \leq x \leq 6\). The region bounded by the curve and the lines \(x = 0\), \(x = 6\) and \(y = 0\) is denoted by \(R\).

1. Use the trapezium rule with 2 strips to find an estimate of the area of \(R\), giving your answer correct to 2 decimal places. [3]
2. With reference to the diagram, explain why this estimate is greater than the exact area of \(R\). [1]
3. The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced. [4]

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \sqrt{1 + 3\cos^2(\frac{1}{2}x)}\) for \(0 \leqslant x \leqslant \pi\). The region \(R\) is bounded by the curve, the axes and the line \(x = \pi\).

1. Use the trapezium rule with two intervals to find an approximation to the area of \(R\), giving your answer correct to 3 significant figures. [3]
2. The region \(R\) is rotated completely about the \(x\)-axis. Without using a calculator, find the exact volume of the solid produced. [5]

\includegraphics{figure_1} Figure 1 shows the graph of $$y = 1 - \log_{10}(\sin x) \quad 0 < x < \pi$$ where \(x\) is in radians. The table below shows some values of \(x\) and \(y\) for this graph, with values of \(y\) given to 3 decimal places.

1. Complete the table above, giving values of \(y\) to 3 decimal places. [2]
2. Use the trapezium rule with all the \(y\) values in the completed table to find, to 2 decimal places, an estimate for $$\int_{0.5}^{3} \left(1 - \log_{10}(\sin x)\right) dx$$ [3]
3. Use your answer to part (b) to find an estimate for $$\int_{0.5}^{3} \left(3 + \log_{10}(\sin x)\right) dx$$ [3]

A river, running between parallel banks, is 20 m wide. The depth, \(y\) metres, of the river measured at a point \(x\) metres from one bank is given by the formula $$y = \frac{1}{10}x(20 - x), \quad 0 \leq x \leq 20.$$

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

   \(x\)	0	4	8	12	16	20
   \(y\)	0		2.771			0

   [2]
2. Use the trapezium rule with all the values in the table to estimate the cross-sectional area of the river. [4]

Given that the cross-sectional area is constant and that the river is flowing uniformly at 2 m s⁻¹,

1. estimate, in m³, the volume of water flowing per minute, giving your answer to 3 significant figures. [2]

The speed, \(v\) m s⁻¹, of a train at time \(t\) seconds is given by \(v = \sqrt{(1.2^t - 1)}, \quad 0 \leq t \leq 30.\) 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. [3]

The distance, \(s\) metres, travelled by the train in 30 seconds is given by $$s = \int_0^{30} \sqrt{(1.2^t - 1)} \, dt.$$

1. Use the trapezium rule, with all the values from your table, to estimate the value of \(s\). [3]

The curve \(C\) has equation \(y = x\sqrt{x^2 + 1}, \quad 0 \leq x \leq 2\).

1. Copy and 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]
2. Use the trapezium rule, with all the \(y\) values from your table, to find an approximation for the value of \(\int_0^2 x\sqrt{x^2 + 1} \, dx\), giving your answer to 3 significant figures. [4]

\includegraphics{figure_2} Figure 2 shows the curve \(C\) with equation \(y = x\sqrt{x^2 + 1}\), \(0 \leq x \leq 2\), and the straight line segment \(l\), which joins the origin and the point \((2, 6)\). The finite region \(R\) is bounded by \(C\) and \(l\).

1. Use your answer to part (b) to find an approximation for the area of \(R\), giving your answer to 3 significant figures. [3]

\includegraphics{figure_1} Figure 1 shows the graph of the curve with equation $$y = xe^x, \quad x \geq 0.$$ 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 of the area for \(R\). [5]
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
   \(y = xe^x\)	0	0.29836		1.99207

   [1]
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. [4]

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation \(y = xe^{-\frac{1}{2}x}\), \(x > 0\). 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 \(y = xe^{-\frac{1}{2}x}\).

1. Complete the table with the value of \(y\) corresponding to \(x = 2\) [1]
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. [4]
3. 1. Find \(\int xe^{-\frac{1}{2}x} \, dx\).
   2. Hence find the exact area of \(R\), giving your answer in the form \(a + be^{-2}\), where \(a\) and \(b\) are integers. [6]