Trapezium rule with reasoning

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 \includegraphics[max width=\textwidth, alt={}, center]{9cf008d5-c15f-4491-9e4d-4bd070f896d5-06_446_832_260_653} The diagram shows part of the curve with equation \(y = \frac { 5 x } { 4 x ^ { 3 } + 1 }\). The shaded region is bounded by the curve and the lines \(x = 1 , x = 3\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of the maximum point.
2. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
3. State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.

2 \includegraphics[max width=\textwidth, alt={}, center]{af7c2b6e-1293-4744-8ea0-927fea5ab4ec-2_531_949_431_598} The diagram shows part of the curve \(y = x \mathrm { e } ^ { - x }\). The shaded region \(R\) is bounded by the curve and by the lines \(x = 2 , x = 3\) and \(y = 0\).

1. Use the trapezium rule with two intervals to estimate the area of \(R\), giving your answer correct to 2 decimal places.
2. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of \(R\).

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.

2 \includegraphics[max width=\textwidth, alt={}, center]{0f73e750-18a0-49ad-b4cb-fd6d14f0789e-2_531_700_395_719} The diagram shows the curve \(y = \sqrt { } \left( 1 + 2 \tan ^ { 2 } x \right)\) for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\).

1. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sqrt { } \left( 1 + 2 \tan ^ { 2 } x \right) \mathrm { d } x$$ giving your answer correct to 2 decimal places.
2. The estimate found in part (i) is denoted by \(E\). Explain, without further calculation, whether another estimate found using the trapezium rule with six intervals would be greater than \(E\) or less than \(E\).

7 \includegraphics[max width=\textwidth, alt={}, center]{7f6f82c3-37d3-48da-9958-e4ef366a6467-10_389_488_258_831} The diagram shows a sketch of the curve \(y = \frac { \mathrm { e } ^ { \frac { 1 } { 2 } x } } { x }\) for \(x > 0\), and its minimum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 1 } ^ { 3 } \frac { \mathrm { e } ^ { \frac { 1 } { 2 } x } } { x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
3. The estimate found in part (ii) is denoted by \(E\). Explain, without further calculation, whether another estimate found using the trapezium rule with four intervals would be greater than \(E\) or less than \(E\).

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).

1 \includegraphics[max width=\textwidth, alt={}, center]{746d2c39-7d78-4478-bc36-15ea5e3ba72a-02_460_807_258_667} The diagram shows a sketch of the curve \(y = \frac { 3 } { \sqrt { } \left( 9 - x ^ { 3 } \right) }\) for values of \(x\) from - 1.2 to 1.2 .

1. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { - 1.2 } ^ { 1.2 } \frac { 3 } { \sqrt { \left( 9 - x ^ { 3 } \right) } } \mathrm { d } x$$ 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.

8 \includegraphics[max width=\textwidth, alt={}, center]{5b5ed7d1-028e-4f9a-ae9e-26071d0df678-14_604_497_262_822} The diagram shows the graph of \(y = \sec x\) for \(0 \leqslant x < \frac { 1 } { 2 } \pi\).

1. Use the trapezium rule with 2 intervals to estimate the value of \(\int _ { 0 } ^ { 1.2 } \sec x \mathrm {~d} x\), giving your answer correct to 2 decimal places.
2. Explain, with reference to the diagram, whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).
3. \(P\) is the point on the part of the curve \(y = \sec x\) for \(0 \leqslant x < \frac { 1 } { 2 } \pi\) at which the gradient is 2 . By first differentiating \(\frac { 1 } { \cos x }\), find the \(x\)-coordinate of \(P\), giving your answer correct to 3 decimal places.

3

1. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sec x \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Using a sketch of the graph of \(y = \sec x\) for \(0 \leqslant x \leqslant \frac { 1 } { 3 } \pi\), explain whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (i).

3 \includegraphics[max width=\textwidth, alt={}, center]{322eb555-d40a-460c-8c71-5780f5772bcd-2_535_1041_573_552} The diagram shows the curve \(y = x - 2 \ln x\) and its minimum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 2 } ^ { 5 } ( x - 2 \ln x ) \mathrm { d } x$$ giving your answer correct to 2 decimal places.
3. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).

6

1. Find \(\int ( \sin x - \cos x ) ^ { 2 } \mathrm {~d} x\).
2. 1. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \operatorname { cosec } x d x$$ giving your answer correct to 3 decimal places.
   2. Using a sketch of the graph of \(y = \operatorname { cosec } x\) for \(0 < x \leqslant \frac { 1 } { 2 } \pi\), explain whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (i).

5 \includegraphics[max width=\textwidth, alt={}, center]{8b051aee-4920-42a0-8b74-cbfa9f3c1ab1-3_373_623_260_760} The diagram shows the curve \(y = \sqrt { } \left( 1 + \mathrm { e } ^ { \frac { 1 } { 3 } x } \right)\) for \(0 \leqslant x \leqslant 6\). 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.
2. With reference to the diagram, explain why this estimate is greater than the exact area of \(R\).
3. The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced.

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 \(R\).

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 \(\frac { 3 } { 2 } \ln 3\).

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}

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.

4 \includegraphics[max width=\textwidth, alt={}, center]{e429080f-8634-46bc-b451-7b13b871e518-2_543_857_1155_644} The diagram shows the curve \(\mathrm { y } = \sqrt { 4 \mathrm { X } + 1 }\).

1. Use the trapezium rule, with strips of width 0.5 , to find an approximate value for the area of the region bounded by the curve \(y = \sqrt { 4 x + 1 }\), the \(x\)-axis, and the lines \(x = 1\) and \(x = 3\). Give your answer correct to 3 significant figures.
2. State with a reason whether this approximation is an under-estimate or an over-estimate.

8 \includegraphics[max width=\textwidth, alt={}, center]{73d67b39-3611-4afb-9470-2f813115abb5-4_415_714_287_678} The diagram shows the curve \(y = 1.25 ^ { x }\).

1. A point on the curve has \(y\)-coordinate 2. Calculate its \(x\)-coordinate.
2. Use the trapezium rule with 4 intervals to estimate the area of the shaded region, bounded by the curve, the axes, and the line \(x = 4\).
3. State, with a reason, whether the estimate found in part (ii) is an overestimate or an underestimate.
4. Explain briefly how the trapezium rule could be used to find a more accurate estimate of the area of the shaded region.

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.

3. \includegraphics[max width=\textwidth, alt={}, center]{faa66f88-9bff-4dc9-955f-80cdab3fdd34-1_474_863_1283_520} The diagram shows the curve with equation \(y = \frac { 4 x } { ( x + 1 ) ^ { 2 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 1\).

1. Use the trapezium rule with four intervals, each of width 0.25 , to find an estimate for the area of the shaded region.
2. State, with a reason, whether your answer to part (a) is an under-estimate or an over-estimate of the true area.

2 \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.
[0pt] [5]

3 \includegraphics[max width=\textwidth, alt={}, center]{c52fe7e9-0442-4b3e-b924-2e5e4b3e98f5-02_510_791_991_678} The diagram shows the curve \(y = \sqrt { x - 3 }\).

1. Use the trapezium rule, with 4 strips each of width 0.5 , to find an approximate value for the area of the region bounded by the curve, the \(x\)-axis and the line \(x = 5\). Give your answer correct to 3 significant figures.
2. State, with a reason, whether this approximation is an underestimate or an overestimate.

2 \includegraphics[max width=\textwidth, alt={}, center]{ad3083ae-caa6-42d8-a1f2-e984150cb104-2_536_917_1016_577} The diagram shows the curve \(y = \log _ { 10 } ( 2 x + 1 )\).

1. Use the trapezium rule with 4 strips each of width 1.5 to find an approximation to the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 4\) and \(x = 10\). Give your answer correct to 3 significant figures.
2. Explain why this approximation is an under-estimate.

6

1. Use the trapezium rule, with 2 strips each of width 4 , to show that an approximate value of \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\) is \(32 + 16 \sqrt { 5 }\).
2. Use a sketch graph to explain why the actual value of \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\) is greater than \(32 + 16 \sqrt { 5 }\).
3. Use integration to find the exact value of \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\).