1.09f Trapezium rule: numerical integration

1 Fig. 3 shows the curve \(y = x ^ { 3 } + \sqrt { ( \sin x ) }\) for \(0 \leqslant x \leqslant \frac { \pi } { 4 }\). \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), giving your answer to 3 decimal places.
2. Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say.

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

1. The following table gives some values of \(x\) and \(y\).

   \(x\)	0	0.25	0.5	0.75	1
   \(y\)	1	1.0308		1.25	1.4142

   Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.

3 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h] \end{figure}

1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. 1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.

      \(x\)	0	0.5	1	1.5	2
      \(y\)		1.9283	2.8964	4.5919

   2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

4

1. Complete the table of values for the curve \(y = \sqrt { \cos x }\).

   \(X\)	0	\(\frac { \pi } { 8 }\)	\(\frac { \pi } { 4 }\)	\(\frac { 3 \pi } { 8 }\)	\(\frac { \pi } { 2 }\)
   \(y\)		0.9612	0.8409

   Hence use the trapezium rule with strip width \(h = \frac { \pi } { 8 }\) to estimate the value of the integral \(\int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \cos x } \mathrm {~d} x\), giving your answer to 3 decimal places. Fig. 4 shows the curve \(y = \sqrt { \cos x }\) for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-4_459_751_799_638} \captionsetup{labelformat=empty} \caption{Fig. 4}
   \end{figure}
2. State, with a reason, whether the trapezium rule with a strip width of \(\frac { \pi } { 16 }\) would give a larger or smaller estimate of the integral.

5

1. Use the trapezium rule with four strips to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\), showing your working. Fig. 1 shows a sketch of \(y = \sqrt { 1 + \mathrm { e } ^ { x } }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-5_533_1074_441_565} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure}
2. Suppose that the trapezium rule is used with more strips than in part (i) to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\). State, with a reason but no further calculation, whether this would give a larger or smaller estimate.
   [0pt] [2]

6 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.

1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.

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

1. The following table gives some values of \(x\) and \(y\).

   \(x\)	0	0.25	0.5	0.75	1
   \(y\)	1	1.0308		1.25	1.4142

   Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.

5 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h] \end{figure}

1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. 1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.

      \(x\)	0	0.5	1	1.5	2
      \(y\)		1.9283	2.8964	4.5919

   2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

3 \includegraphics[max width=\textwidth, alt={}, center]{268b605f-eb86-40df-946a-210da1355e83-2_686_967_998_589} The diagram shows the curve with equation \(y = \mathrm { e } ^ { x ^ { 2 } }\), for \(0 \leqslant x \leqslant 1\). The region under the curve between these limits is divided into four strips of equal width. The area of this region under the curve is \(A\).

1. By considering the set of rectangles indicated in the diagram, show that an upper bound for \(A\) is 1.71 .
2. By considering an appropriate set of four rectangles, find a lower bound for \(A\).

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.

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 $$\lim _ { x \rightarrow 0 } \frac { 16 ^ { x } - 1 } { x }$$

7. \begin{figure}[h] \end{figure} Diagram not drawn to scale Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 1 } { \sqrt { 2 x + 5 } } , x > - 2.5\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 2\) and \(x = 5\)

1. Use the trapezium rule with three strips of equal width to find an estimate for the area of \(R\), giving your answer to 3 decimal places.
2. Use calculus to find the exact area of \(R\).
3. Hence calculate the magnitude of the error of the estimate found in part (a), giving your answer to one significant figure.

4

1. Use the trapezium rule, with 4 strips each of width 0.5 , to find an approximate value for $$\int _ { 3 } ^ { 5 } \log _ { 10 } ( 2 + x ) d x$$ giving your answer correct to 3 significant figures.
2. Use your answer to part (i) to deduce an approximate value for \(\int _ { 3 } ^ { 5 } \log _ { 10 } \sqrt { 2 + x } \mathrm {~d} x\), showing your method clearly.

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.

9

1. Sketch the graph of \(y = 4 k ^ { x }\), where \(k\) is a constant such that \(k > 1\). State the coordinates of any points of intersection with the axes.
2. The point \(P\) on the curve \(y = 4 k ^ { x }\) has its \(y\)-coordinate equal to \(20 k ^ { 2 }\). Show that the \(x\)-coordinate of \(P\) may be written as \(2 + \log _ { k } 5\).
3. (a) Use the trapezium rule, with two strips each of width \(\frac { 1 } { 2 }\), to find an expression for the approximate value of $$\int _ { 0 } ^ { 1 } 4 k ^ { x } \mathrm {~d} x$$ (b) Given that this approximate value is equal to 16 , find the value of \(k\).

2

1. Use the trapezium rule, with 3 strips each of width 3 , to estimate the area of the region bounded by the curve \(y = \sqrt [ 3 ] { 7 + x }\), the \(x\)-axis, and the lines \(x = 1\) and \(x = 10\). Give your answer correct to 3 significant figures.
2. Explain how the trapezium rule could be used to obtain a more accurate estimate of the area.

8 \includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-4_417_931_255_607} The diagram shows the curve \(y = 2 ^ { x } - 3\).

1. Describe the geometrical transformation that transforms the curve \(y = 2 ^ { x }\) to the curve \(y = 2 ^ { x } - 3\).
2. State the \(y\)-coordinate of the point where the curve \(y = 2 ^ { x } - 3\) crosses the \(y\)-axis.
3. Find the \(x\)-coordinate of the point where the curve \(y = 2 ^ { x } - 3\) crosses the \(x\)-axis, giving your answer in the form \(\log _ { a } b\).
4. The curve \(y = 2 ^ { x } - 3\) passes through the point ( \(p , 62\) ). Use logarithms to find the value of \(p\), correct to 3 significant figures.
5. Use the trapezium rule, with 2 strips each of width 0.5 , to find an estimate for \(\int _ { 3 } ^ { 4 } \left( 2 ^ { x } - 3 \right) \mathrm { d } x\). Give your answer correct to 3 significant figures.

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 \(\int _ { 1 } ^ { 9 } 4 \sqrt { x } \mathrm {~d} x\).

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

9 \includegraphics[max width=\textwidth, alt={}, center]{9e95415c-00f5-4b52-a443-0b946602b3b4-4_387_624_287_717} The diagram shows part of the curve \(y = - 3 + 2 \sqrt { x + 4 }\). The point \(P ( 5,3 )\) lies on the curve. Region \(A\) is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 5\). Region \(B\) is bounded by the curve, the \(y\)-axis and the line \(y = 3\).

1. Use the trapezium rule, with 2 strips each of width 2.5 , to find an approximate value for the area of region \(A\), giving your answer correct to 3 significant figures.
2. Use your answer to part (i) to deduce an approximate value for the area of region \(B\).
3. By first writing the equation of the curve in the form \(x = \mathrm { f } ( y )\), use integration to show that the exact area of region \(B\) is \(\frac { 14 } { 3 }\). \section*{END OF QUESTION PAPER} \section*{OCR \(^ { \text {N } }\)}

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.

8

1. The curve \(y = 3 ^ { x }\) can be transformed to the curve \(y = 3 ^ { x - 2 }\) by a translation. Give details of the translation.
2. Alternatively, the curve \(y = 3 ^ { x }\) can be transformed to the curve \(y = 3 ^ { x - 2 }\) by a stretch. Give details of the stretch.
3. Sketch the curve \(y = 3 ^ { x - 2 }\), stating the coordinates of any points of intersection with the axes.
4. The point \(P\) on the curve \(y = 3 ^ { x - 2 }\) has \(y\)-coordinate equal to 180 . Use logarithms to find the \(x\)-coordinate of \(P\), correct to 3 significant figures.
5. Use the trapezium rule, with 2 strips each of width 1.5, to find an estimate for \(\int _ { 1 } ^ { 4 } 3 ^ { x - 2 } \mathrm {~d} x\). Give your answer correct to 3 significant figures.

2 Fig. 2 shows the coordinates at certain points on a curve. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to calculate an estimate of the area of the region bounded by this curve and the axes.

3 At a place where a river is 7.5 m wide, its depth is measured every 1.5 m across the river. The table shows the results. Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.