Area Under & Between Curves

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 \(y = \left( \frac { 1 } { 2 } \right) ^ { x } , x \in \mathbb { R }\), 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

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

3. Find the area of the finite region enclosed by the curve \(y = 5 x - x ^ { 2 }\) and the \(x\)-axis.

1 The graph of \(y = x ^ { 2 } - 9\) is shown below. \includegraphics[max width=\textwidth, alt={}, center]{be9ad375-d3cd-488c-9e3b-a4cd035d0d1d-02_335_593_767_744} Find the area of the shaded region.
Circle your answer.
[0pt] [1 mark] \(- 18 - 6618\)

11 Four finite regions \(A , B , C\) and \(D\) are enclosed by the curve with equation $$y = x ^ { 3 } - 7 x ^ { 2 } + 11 x + 6$$ and the lines \(y = k , x = 1\) and \(x = 4\), as shown in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{1d017497-11b1-4096-b83a-63314188307e-12_865_1056_520_493} The areas of \(B\) and \(C\) are equal.
Find the value of \(k\).

1. $$y = 3 ^ { x } + 2 x$$

1. Complete the table below, giving the values of \(y\) to 2 decimal places.

   \(x\)	0	0.2	0.4	0.6	0.8	1
   \(y\)	1	1.65				5

2. Use the trapezium rule, with all the values of \(y\) from your table, to find an approximate value for \(\int _ { 0 } ^ { 1 } \left( 3 ^ { x } + 2 x \right) d x\).

4. (a) Sketch the graph of \(y = \frac { 1 } { x } , x > 0\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 1 } { x }\), with the values for \(y\) rounded to 3 decimal places where necessary. (b) Use the trapezium rule with all the values of \(y\) from the table to find an approximate value, to 2 decimal places, for \(\int _ { 1 } ^ { 3 } \frac { 1 } { x } \mathrm {~d} x\)

13. \begin{figure}[h] \end{figure} Figure 5 shows a sketch of part of the curve with equation \(y = 2 - \ln x , x > 0\) The finite region \(R\), shown shaded in Figure 5, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \mathrm { e }\). The table below shows corresponding values of \(x\) and \(y\) for \(y = 2 - \ln x\)

1. Complete the table giving the value of \(y\) to 4 decimal places.
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 3 decimal places.
3. Use integration by parts to show that \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x = x ( \ln x ) ^ { 2 } - 2 x \ln x + 2 x + c\) The area \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
4. Use calculus to find the exact volume of the solid generated. Write your answer in the form \(\pi \mathrm { e } ( p \mathrm { e } + q )\), where \(p\) and \(q\) are integers to be found.

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

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.

2

1. Use Simpson's rule with four strips to find an approximation to $$\int _ { 4 } ^ { 12 } \ln x \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
2. Deduce an approximation to \(\int _ { 4 } ^ { 12 } \ln \left( x ^ { 10 } \right) \mathrm { d } x\).

1. (a) Given that \(a\) is a constant, \(a > 1\), sketch the graph of

$$y = a ^ { x } , \quad x \in \mathbb { R }$$ On your diagram show the coordinates of the point where the graph crosses the \(y\)-axis.
(2) The table below shows corresponding values of \(x\) and \(y\) for \(y = 2 ^ { x }\) (b) Use the trapezium rule, with all of the values of \(y\) from the table, to find an approximate value, to 2 decimal places, for $$\int _ { - 4 } ^ { 4 } 2 ^ { x } \mathrm {~d} x$$ (c) Use the answer to part (b) to find an approximate value for

1. \(\int _ { - 4 } ^ { 4 } 2 ^ { x + 2 } \mathrm {~d} x\)
2. \(\int _ { - 4 } ^ { 4 } \left( 3 + 2 ^ { x } \right) \mathrm { d } x\)
   \includegraphics[max width=\textwidth, alt={}, center]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-23_86_47_2617_1886}

6. (i) Write down the exact value of \(\cos \frac { \pi } { 6 }\). The finite region \(R\) is bounded by the curve \(y = \cos ^ { 2 } x\), where \(x\) is measured in radians, the positive coordinate axes and the line \(x = \frac { \pi } { 3 }\).
(ii) Use the trapezium rule with two intervals of equal width to estimate the area of \(R\), giving your answer to 3 significant figures. The finite region \(S\) is bounded by the curve \(y = \sin ^ { 2 } x\), where \(x\) is measured in radians, the positive coordinate axes and the line \(x = \frac { \pi } { 3 }\).
(iii) Using your answer to part (b), find an estimate for the area of \(S\).

2 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.

2 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.

9 Fig. 9 shows the cross-section of a straight, horizontal tunnel. The \(x\)-axis from 0 to 6 represents the floor of the tunnel. \begin{figure}[h] \end{figure} With axes as shown, and units in metres, the roof of the tunnel passes through the points shown in the table. The length of the tunnel is 50 m .

1. Use the trapezium rule with 6 strips to estimate the area of cross-section of the tunnel. Hence estimate the volume of earth removed in digging the tunnel.
2. An engineer models the height of the roof of the tunnel using the curve \(y = \frac { 5 } { 81 } \left( 108 x - 54 x ^ { 2 } + 12 x ^ { 3 } - x ^ { 4 } \right)\). This curve is symmetrical about \(x = 3\).
   (A) Show that, according to this model, a vehicle of rectangular cross-section which is 3.6 m wide and 4.4 m high would not be able to pass through the tunnel.
   (B) Use integration to calculate the area of the cross-section given by this model. Hence obtain another estimate of the volume of earth removed in digging the tunnel.

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 \(\lim _ { \mathrm { n } \rightarrow \infty } \left( \mathrm { U } _ { \mathrm { n } } - \mathrm { L } _ { \mathrm { n } } \right) = 0\).

9. 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 of equal width 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).

6. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the curve with equation $$y = \frac { 16 } { x ^ { 2 } } - \frac { x } { 2 } + 1 , \quad x > 0$$ The finite region \(R\), bounded by the lines \(x = 1\), the \(x\)-axis and the curve, is shown shaded in Figure 1. The curve crosses the \(x\)-axis at the point \(( 4,0 )\).

1. Complete the table with the values of \(y\) corresponding to \(x = 2\) and 2.5

   \(x\)	1	1.5	2	2.5	3	3.5	4
   \(y\)	16.5	7.361			1.278	0.556	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 \(R\).

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 \(R\).

（a）Show that the point \(P ( 1,0 )\) lies on \(C\) ．
（b）Find the coordinates of the point \(Q\) ．
（c）Find the area of the shaded region between \(C\) and the line \(P Q\) ．

4 The curve with equation \(y = x ^ { 4 } - 8 x + 9\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-5_410_609_383_721} The point \(( 2,9 )\) lies on the curve.

1. 1. Find \(\int _ { 0 } ^ { 2 } \left( x ^ { 4 } - 8 x + 9 \right) \mathrm { d } x\).
   2. Hence find the area of the shaded region bounded by the curve and the line \(y = 9\).
2. The point \(A ( 1,2 )\) lies on the curve with equation \(y = x ^ { 4 } - 8 x + 9\).
   1. Find the gradient of the curve at the point \(A\).
   2. Hence find an equation of the tangent to the curve at the point \(A\).

（a）Show that the point \(P ( 1,0 )\) lies on \(C\) ．
（b）Find the coordinates of the point \(Q\) ．
（c）Find the area of the shaded region between \(C\) and the line \(P Q\) ．

3 In the diagram, \(A C\) is an arc of a circle, centre \(O\) and radius 6 cm . The line \(B C\) is perpendicular to \(O C\) and \(O A B\) is a straight line. Angle \(A O C = \frac { 1 } { 3 } \pi\) radians. Find the area of the shaded region, giving your answer in terms of \(\pi\) and \(\sqrt { } 3\).

5 \includegraphics[max width=\textwidth, alt={}, center]{ea402a1d-3632-4637-9198-2365715b5246-06_323_775_260_685} The diagram shows a triangle \(O A B\) in which angle \(O A B = 90 ^ { \circ }\) and \(O A = 5 \mathrm {~cm}\). The arc \(A C\) is part of a circle with centre \(O\). The arc has length 6 cm and it meets \(O B\) at \(C\). Find the area of the shaded region.

11 \includegraphics[max width=\textwidth, alt={}, center]{11bfe5bd-604c-43e5-81e7-4c1f5676bcbb-4_611_668_1699_737} The diagram shows a sector of a circle with centre \(O\) and radius 20 cm . A circle with centre \(C\) and radius \(x \mathrm {~cm}\) lies within the sector and touches it at \(P , Q\) and \(R\). Angle \(P O R = 1.2\) radians.

1. Show that \(x = 7.218\), correct to 3 decimal places.
2. Find the total area of the three parts of the sector lying outside the circle with centre \(C\).
3. Find the perimeter of the region \(O P S R\) bounded by the \(\operatorname { arc } P S R\) and the lines \(O P\) and \(O R\).

1 Use the trapezium rule with 3 intervals to estimate the value of $$\int _ { 0 } ^ { 3 } \left| 2 ^ { x } - 4 \right| \mathrm { d } x$$

2 Use the trapezium rule with three intervals to find an approximation to $$\int _ { 0 } ^ { 3 } \left| 3 ^ { x } - 10 \right| \mathrm { d } x$$

10 In this question you must show detailed reasoning.
The equation of a curve is \(y = 12 x ^ { 3 } - 24 x ^ { 2 } - 60 x + 72\).
Determine the magnitude of the total area bounded by the curve and the \(x\)-axis.

1. Use Simpson's rule with four strips to estimate the value of the integral

$$\int _ { 0 } ^ { 3 } \mathrm { e } ^ { \cos x } \mathrm {~d} x$$

3 Show that the area of the region bounded by the curve \(y = 3 x ^ { - \frac { 3 } { 2 } }\), the lines \(x = 1 , x = 3\) and the \(x\)-axis is \(6 - 2 \sqrt { 3 }\).

1. The diagram shows the curve \(y = 6 x - x ^ { 2 }\) and the line \(y = 5\). Find the area of the shaded region. You must show detailed reasoning.
(Total 4 marks)
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2 Find the area of the region enclosed by the curve \(y = 2 \sqrt { } x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\).

3 \includegraphics[max width=\textwidth, alt={}, center]{8beee722-7f86-454a-bc36-27e83f1483fd-04_684_455_260_845} The diagram shows the curve \(y = 2 + \mathrm { e } ^ { - 2 x }\). The curve crosses the \(y\)-axis at the point \(A\), and the point \(B\) on the curve has \(x\)-coordinate 1 . The shaded region is bounded by the curve and the line segment \(A B\). Find the exact area of the shaded region.

3 In this question you must show detailed reasoning. Find the exact area of the region enclosed by the curve \(y = \frac { 1 } { x + 2 }\), the two axes and the line \(x = 2.5\).

5. \includegraphics[max width=\textwidth, alt={}, center]{5840974b-b08a-4818-9a59-97b2d3ce9890-1_469_809_1777_484} The diagram shows the curve with equation \(y = x \sqrt { 1 - x } , 0 \leq x \leq 1\).
Use the substitution \(u ^ { 2 } = 1 - x\) to show that the area of the region bounded by the curve and the \(x\)-axis is \(\frac { 4 } { 15 }\).

14. \includegraphics[max width=\textwidth, alt={}, center]{e0c7b6f3-9c28-48bd-a401-635efb2521e3-30_490_992_226_573} The diagram above shows the curve with equation $$y = ( x - 4 ) ^ { 2 } , x \in \mathbb { R }$$ intersected by the straight line with equation \(y = 4\), at the points \(A\) and \(B\). The curve meets the \(y\) axis at the point \(C\). Calculate the exact area of the shaded region, bounded by the curve and the straight line segments \(A B\) and \(B C\).
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