Numerical integration

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

\(x\)	0	0.5	1	1.5	2
\(y\)	1	1.216	\(p\)	1.413	\(q\)

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 $$I = \int_0^2 \sqrt{1 + \sin x} \, dx.$$ [4]

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.

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 \(R\).

1 Use the trapezium rule, with 3 strips each of width 2 , to estimate the value of $$\int _ { 5 } ^ { 11 } \frac { 8 } { x } \mathrm {~d} x .$$

6. (a) Sketch the graph of \(y = 1 + \cos x , \quad 0 \leqslant x \leqslant 2 \pi\) Show on your sketch the coordinates of the points where your graph meets the coordinate axes.
(b) Use the trapezium rule, with 6 strips of equal width, to find an approximate value for $$\int _ { 0 } ^ { 2 \pi } ( 1 + \cos x ) d x$$

A particle of mass 4 kg moves horizontally in a straight line. At time \(t\) seconds the velocity of the particle is \(v\) m s\(^{-1}\) The following horizontal forces act on the particle: • a constant driving force of magnitude 1.8 newtons • another driving force of magnitude \(30\sqrt{t}\) newtons • a resistive force of magnitude \(0.08v^2\) newtons When \(t = 70\), \(v = 54\) Use Euler's method with a step length of 0.5 to estimate the velocity of the particle after 71 seconds. Give your answer to four significant figures. [6 marks]

1

1. Solve the following simultaneous equations. $$\begin{aligned} & x + \quad y = 1 \\ & x + 0.99 y = 2 \end{aligned}$$
2. The coefficient 0.99 is correct to two decimal places. All other coefficients in the equations are exact. With the aid of suitable calculations, explain why your answer to part (i) is unreliable.

2 Fig. 7 shows a curve and the coordinates of some points on it. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive \(x\) - and \(y\)-axes.

7 Fig. 7.1 shows two values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h] \end{table}

1. Use the forward difference method to calculate an estimate of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), giving your answer correct to 4 decimal places. Fig. 7.2 shows some spreadsheet output with additional values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h]

   \(x\)	3	3.00001	3.0001	3.001	3.01	3.1
   \(\mathrm { f } ( x )\)	6.082763	6.08274	6.082541	6.08054	6.060454	5.848846

   \captionsetup{labelformat=empty} \caption{Fig. 7.2}
   \end{table} These values have been used to produce a sequence of estimates of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), together with some further analysis. This is shown in the spreadsheet output in Fig. 7.3. \begin{table}[h]

   \(h\)	0.1	0.01	0.001	0.0001	0.00001
   estimate	-2.339165	-2.230883	-2.220532	-2.219501	-2.219398
   difference	0.1082815	0.010352	0.0010307	0.000103
   ratio	0.095602	0.099567	0.0999568

   \captionsetup{labelformat=empty} \caption{Fig. 7.3}
   \end{table} Tommy states that the differences between successive estimates is decreasing so rapidly that the order of convergence of this sequence of estimates is much faster than first order.
2. Explain whether or not Tommy is correct.
3. Use extrapolation to determine the value of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\) as accurately as possible, justifying the precision quoted.
4. Calculate an estimate of the absolute error when \(\mathrm { f } ( 3 )\) is used as an approximation to \(\mathrm { f } ( 3.02 )\).

1. Use Simpson's rule with 4 intervals to estimate

$$\int _ { 0.4 } ^ { 2 } e ^ { x ^ { 2 } } d x$$

4 The integral I is defined by $$I = \int _ { 0 } ^ { 13 } ( 2 x + 1 ) ^ { \frac { 1 } { 3 } } d x$$

1. Use integration to find the exact value of I .
2. Use Simpson's rule with two strips to find an approximate value for I. Give your answer correct to 3 significant figures.

9 The trapezium rule is used to calculate 3 approximations to \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) with 1,2 and 4 strips respectively. The results are shown in Table 9.1. \begin{table}[h] \end{table}

1. Use these results to determine two approximations to \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) using Simpson's rule.
2. Use your answers to part (a) to state the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) as accurately as you can, justifying the precision quoted. Table 9.2 shows some further approximations found using the trapezium rule, together with some analysis of these approximations. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Table 9.2}

   \(n\)	\(\mathrm { T } _ { n }\)	difference	ratio
   1	0.5276437
   2	0.6661765	0.138533
   4	0.7253427	0.059166	0.42709
   8	0.7498821	0.024539	0.41475
   16	0.7598858	0.010004	0.40766
   32	0.7639221	0.004036	0.40348
   64	0.7655404	0.001618	0.40095

   \end{table}
3. Explain what can be deduced about the order of the method in this case.
4. Use extrapolation to obtain the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) as accurately as you can, justifying the precision quoted.

3 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-2_643_787_1621_680} The diagram shows the curve with equation \(y = \sqrt { 1 + x ^ { 3 } }\), for \(2 \leqslant x \leqslant 3\). The region under the curve between these limits has area \(A\).

1. Explain why \(3 < A < \sqrt { 28 }\).
2. The region is divided into 5 strips, each of width 0.2 . By using suitable rectangles, find improved lower and upper bounds between which \(A\) lies. Give your answers correct to 3 significant figures.

6 Alex is investigating the area, \(A\), under the graph of \(y = x ^ { 2 }\) between \(x = 1\) and \(x = 1.5\). They draw the graph, together with rectangles of width \(\delta x = 0.1\), and varying heights \(y\). \includegraphics[max width=\textwidth, alt={}, center]{7298e7b9-ad52-480c-bc2b-8289aeab9ebb-06_531_714_356_251}

1. Use the rectangles in the diagram to show that lower and upper bounds for the area \(A\) are 0.73 and 0.855 respectively.
2. Alex finds lower and upper bounds for the area \(A\), using widths \(\delta x\) of decreasing size. The results are shown in the table. Where relevant, values are given correct to 3 significant figures.

   Width \(\delta x\)	0.1	0.05	0.025	0.0125
   Lower bound for area \(A\)	0.73	0.761	0.776	0.784
   Upper bound for area \(A\)	0.855	0.823	0.807	0.799

   Use Alex's results to estimate the value of \(A\) correct to \(\mathbf { 2 }\) significant figures. Give a brief justification for your estimate.
3. Write down an expression, in terms of \(y\) and \(\delta x\), for the exact value of the area \(A\).

6 \includegraphics[max width=\textwidth, alt={}, center]{23b06b1c-997f-425d-ae3d-bd4cc1295605-10_771_1146_260_497} The diagram shows the curve with equation \(\mathrm { y } = \ln ( 1 + \mathrm { x } )\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles each of width \(\frac { 1 } { n }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \ln ( 1 + x ) d x < U _ { n }\), where $$U _ { n } = \frac { 1 } { n } \ln \frac { ( 2 n ) ! } { n ! } - \ln n$$
2. Use a similar method to find, in terms of \(n\), a lower bound \(\mathrm { L } _ { \mathrm { n } }\) for \(\int _ { 0 } ^ { 1 } \ln ( 1 + x ) \mathrm { d } x\).
3. By simplifying \(\mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } }\), show that \(\lim _ { \mathrm { n } \rightarrow \infty } \left( \mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

2

1. Use the trapezium rule with 3 intervals to estimate the value of $$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 2 } { 3 } \pi } \operatorname { cosec } x d x$$ giving your answer correct to 2 decimal places.
2. Using a sketch of the graph of \(y = \operatorname { cosec } x\), explain whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).

4 \begin{figure}[h] \end{figure} Fig. 4 shows a curve which passes through the points shown in the following table. Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve, the lines \(x = 1\) and \(x = 4\), and the \(x\)-axis. State, with a reason, whether the trapezium rule gives an overestimate or an underestimate of the area of this region.

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.

Fig. 7 shows a curve and the coordinates of some points on it. \includegraphics{figure_7} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive \(x\)- and \(y\)-axes. [4]

6. (a) Sketch the graph of \(y = \left( \frac { 1 } { 2 } \right) ^ { x } , x \in \mathbb { R }\), showing the coordinates of the point at which the graph crosses the \(y\)-axis. The table below gives corresponding values of \(x\) and \(y\), for \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) The values of \(y\) are rounded to 3 decimal places. (b) Use the trapezium rule with all the values of \(y\) from the table to find an approximate value for $$\int _ { - 0.9 } ^ { - 0.5 } \left( \frac { 1 } { 2 } \right) ^ { x } d x$$ II

13. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve with equation \(y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right) , x > 0\) The finite region \(R\), shown shaded in Figure 4, is bounded by the curve, the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = 2\) The table below shows corresponding values of \(x\) and \(y\) for \(y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right)\), with the values of \(y\) given to 3 significant figures.

1. Use the trapezium rule, with all the values of \(y\), to obtain an estimate for the area of \(R\), giving your answer to 2 significant figures.
2. Use the substitution \(u = x ^ { 2 }\) to show that the area of \(R\) is given by $$\int _ { 1 } ^ { 4 } 6 u ^ { \frac { 1 } { 2 } } \ln ( 2 u ) \mathrm { d } u$$
3. Hence, using calculus, find the exact area of \(R\), writing your answer in the form \(a + b \ln 2\), where \(a\) and \(b\) are constants to be found.

6 The graph shows part of the curve \(y = \frac { 1 } { 1 + x ^ { 2 } }\). \includegraphics[max width=\textwidth, alt={}, center]{62dbc58e-f498-483f-a9aa-05cb5aa44881-3_474_961_406_479} Use the trapezium rule to estimate the area between the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) using

1. 2 strips,
2. 4 strips. What can you conclude about the true value of the area?

2 Fig. 2 shows part of the curve \(y = \sqrt { 1 + x ^ { 3 } }\). \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 4 strips to estimate \(\int _ { 0 } ^ { 2 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), giving your answer correct to 3 significant figures.
2. Chris and Dave each estimate the value of this integral using the trapezium rule with 8 strips. Chris gets a result of 3.25, and Dave gets 3.30. One of these results is correct. Without performing the calculation, state with a reason which is correct.
   [0pt] [2]

7 Fig. 7 shows the graph of \(y = \mathrm { f } ( x )\) for values of \(x\) from 0 to 1 . \begin{figure}[h] \end{figure} The following spreadsheet printout shows estimates of \(\int _ { 0 } ^ { 1 } f ( x ) d x\) found using the midpoint and trapezium rules for different values of \(h\), the strip width.

1. Without doing any further calculation, write down the smallest possible interval which contains the value of the integral. Justify your answer.
2. (A) - Calculate the ratio of differences, \(r\), for the sequence of estimates calculated using the trapezium rule.
   Using a similar approach with the sequence of estimates calculated using the midpoint rule, the approximation to the integral from extrapolation was found to be 1.94427 correct to 5 decimal places.
3. Andrea uses the extrapolated midpoint rule value and the value found in part (ii) ( \(B\) ) to write down an interval which contains the value of the integral. Comment on the validity of Andrea's method.
4. Use the values from the spreadsheet output to calculate an approximation to the integral using Simpson's rule with \(h = 0.125\). Give your answer to 5 decimal places. Approximations to the integral using Simpson's rule are given in the spreadsheet output below. The number of applications of Simpson's rule is given in column N.

   N	O	P	Q
   \(n\)	Simpson	differences	ratio
   1	1.91610623	0.01810005	0.3584931
   2	1.93420628	0.00648874	0.3556525
   4	1.94069502	0.00230774	0.3544828
   8	1.94300275	0.00081805	0.3539885
   16	1.94382081	0.00028958	0.3537638
   32	1.94411039	0.00010244	0.3536568
   64	1.94421283	\(3.623 \mathrm { E } - 05\)
   128	1.94424906

5. Use the spreadsheet output to find the value of the integral as accurately as possible. Justify the precision quoted. \section*{END OF QUESTION PAPER} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
   OCR is part of the }\section*{}

8. The finite region \(R\) is bounded by the curve \(y = 1 + 3 \sqrt { x }\), the \(x\)-axis and the lines \(x = 2\) and \(x = 8\).

1. Use the trapezium rule with three intervals, each of width 2 , to estimate to 3 significant figures the area of \(R\).
2. Use integration to find the exact area of \(R\) in the form \(a + b \sqrt { 2 }\).
3. Find the percentage error in the estimate made in part (a).

2

1. Use the trapezium rule, with 4 strips each of width 1.5, to estimate the value of $$\int _ { 4 } ^ { 10 } \sqrt { 2 x - 1 } \mathrm {~d} x ,$$ giving your answer correct to 3 significant figures.
2. Explain how the trapezium rule could be used to obtain a more accurate estimate.

\includegraphics{figure_4} The diagram shows the curve with equation \(y = \ln x\) for \(x \geqslant 1\), together with a set of \((N-1)\) rectangles of unit width.

1. By considering the sum of the areas of these rectangles, show that $$\ln N! > N \ln N - N + 1.$$ [5]
2. Use a similar method to find, in terms of \(N\), an upper bound for \(\ln N!\). [3]

4

1. Use the trapezium rule with three intervals to show that the value of \(\int _ { 1 } ^ { 4 } \ln x \mathrm {~d} x\) is approximately \(\ln 12\).
2. Use a graph of \(y = \ln x\) to show that \(\ln 12\) is an under-estimate of the true value of \(\int _ { 1 } ^ { 4 } \ln x \mathrm {~d} x\).

4.(a)Use the trapezium rule with 4 strips to find an approximate value for $$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$ (b)Use the trapezium rule with \(n\) strips to write down an expression that would give an approximate value for $$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$ (c)Hence show that $$\int _ { 0 } ^ { 1 } 16 ^ { x } \mathrm {~d} x = \lim _ { n \rightarrow \infty } \left( \frac { 1 } { n } \left( 1 + 16 ^ { \frac { 1 } { n } } + \ldots + 16 ^ { \frac { n - 1 } { n } } \right) \right)$$ (d)Use integration to determine the exact value of $$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$ Given that the limit exists,
(e)use part(c)and the answer to part(d)to determine the exact value of $$\lim _ { x \rightarrow 0 } \frac { 16 ^ { x } - 1 } { x }$$

2 \includegraphics[max width=\textwidth, alt={}, center]{208eab3e-a78c-43b4-918f-a9efc9b4f47a-2_508_586_450_776} The diagram shows a sketch of the curve \(y = \frac { 1 } { 1 + x ^ { 3 } }\) for values of \(x\) from - 0.6 to 0.6 .

1. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { - 0.6 } ^ { 0.6 } \frac { 1 } { 1 + x ^ { 3 } } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Explain, with reference to the diagram, why the trapezium rule may be expected to give a good approximation to the true value of the integral in this case.