Trapezium rule with reasoning

1 \includegraphics[max width=\textwidth, alt={}, center]{38b515c2-4764-4b51-a1f5-9b48d46610f0-4_303_451_358_242} The diagram shows part of the curve \(y = \sqrt { x ^ { 2 } - 1 }\).

1. Use the trapezium rule with 4 intervals to find an estimate for \(\int _ { 1 } ^ { 3 } \sqrt { x ^ { 2 } - 1 } \mathrm {~d} x\). Give your answer correct to \(\mathbf { 3 }\) significant figures.
2. State whether the value from part (a) is an under-estimate or an over-estimate, giving a reason for your answer.
3. Explain how the trapezium rule could be used to obtain a more accurate estimate.

14. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 5 , \quad x > 0$$ The finite region \(S\), shown shaded in Figure 4, is bounded by the curve \(C\), the line with equation \(x = 1\), the \(x\)-axis and the line with equation \(x = 3\) The table below shows corresponding values of \(x\) and \(y\) with the values of \(y\) given to 4 decimal places as appropriate.

1. Use the trapezium rule, with all the values of \(y\) in the table, to obtain an estimate for the area of \(S\), giving your answer to 3 decimal places.
2. Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of \(S\).
3. Show that the exact area of \(S\) can be written in the form \(\frac { a } { b } + \ln c\), where \(a , b\) and \(c\) are integers to be found.
   (In part c, solutions based entirely on graphical or numerical methods are not acceptable.)

4 Fig. 4 shows the graph of \(y = \sqrt { 1 + x ^ { 3 } }\). \begin{figure}[h] \end{figure}

1. Use the trapezium rule with \(h = 0.5\) to find an estimate of \(\int _ { - 1 } ^ { 0 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), giving your answer correct to 6 decimal places.
2. State whether your answer to part (a) is an under-estimate or an over-estimate, justifying your answer.

3. \begin{figure}[h] \end{figure} Figure 1 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 of equal width 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.

3. \includegraphics[max width=\textwidth, alt={}, center]{6e307391-198f-4ea9-99ed-6ef184fca0f7-3_826_873_246_539} Figure 2 shows part of the curve with equation $$y = \mathrm { e } ^ { x } \cos x , 0 \leq x \leq \frac { \pi } { 2 }$$ The finite region \(R\) is bounded by the curve and the coordinate axes.

1. Calculate, to 2 decimal places, the \(y\)-coordinates of the points on the curve where \(x = 0 , \frac { \pi } { 6 } , \frac { \pi } { 3 }\) and \(\frac { \pi } { 2 }\).
   (3)
2. Using the trapezium rule and all the values calculated in part (a), find an approximation for the area of \(R\).
   (4)
3. State, with a reason, whether your approximation underestimates or overestimates the area of \(R\).
   (2)

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.

2 The trapezium rule is used to estimate the area of the shaded region in each of the graphs below. Identify the graph for which the trapezium rule produces an overestimate. Tick ( \(\checkmark\) ) one box. \includegraphics[max width=\textwidth, alt={}, center]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-03_524_424_539_497} \includegraphics[max width=\textwidth, alt={}, center]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-03_506_424_1096_497} \includegraphics[max width=\textwidth, alt={}, center]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-03_512_426_1640_497}
□
□ \includegraphics[max width=\textwidth, alt={}]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-03_147_124_1256_1121} \(\square\) \includegraphics[max width=\textwidth, alt={}, center]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-03_515_431_2188_495}

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.