1.09f Trapezium rule: numerical integration

\includegraphics{figure_2} Figure 2 shows the curve with equation $$y = x^2 \sin\left(\frac{1}{2}x\right), \quad 0 < x \leq 2\pi.$$ The finite region \(R\) bounded by the line \(x = \pi\), the \(x\)-axis, and the curve is shown shaded in Fig 2.

1. Find the exact value of the area of \(R\), by integration. Give your answer in terms of \(\pi\). [7]

The table shows corresponding values of \(x\) and \(y\).

1. Find the value of \(G\). [1]
2. Use the trapezium rule with values of \(x^2 \sin\left(\frac{1}{2}x\right)\)
   1. at \(x = \pi\), \(x = \frac{3\pi}{2}\) and \(x = 2\pi\) to find an approximate value for the area \(R\), giving your answer to 4 significant figures,
   2. at \(x = \pi\), \(x = \frac{5\pi}{4}\), \(x = \frac{3\pi}{2}\), \(x = \frac{7\pi}{4}\) and \(x = 2\pi\) to find an improved approximation for the area \(R\), giving your answer to 4 significant figures.
   [5]

A measure of the effective voltage, \(M\) volts, in an electrical circuit is given by $$M^2 = \int_0^1 V^2 \, dt$$ where \(V\) volts is the voltage at time \(t\) seconds. Pairs of values of \(V\) and \(t\) are given in the following table.

\(t\)	0	0.25	0.5	0.75	1
\(V\)	-48	207	37	-161	-29
\(V^2\)

Use the trapezium rule with five values of \(V^2\) to estimate the value of \(M\). [6]

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]

The speed, \(v\) m s\(^{-1}\), of a lorry at time \(t\) seconds is modelled by $$v = 5(e^{0.1t} - 1) \sin (0.1t), \quad 0 \leq t \leq 30.$$

1. Copy and complete the following table, showing the speed of the lorry at 5 second intervals. Use radian measure for \(0.1t\) and give your values of \(v\) to 2 decimal places where appropriate.

   \(t\)	0	5	10	15	20	25
   \(v\)		1.56	7.23	17.36

   [3]
2. Verify that, according to this model, the lorry is moving more slowly at \(t = 25\) than at \(t = 24.5\). [1]

The distance, \(s\) metres, travelled by the lorry during the first 25 seconds is given by $$s = \int_0^{25} v \, dt.$$

1. Estimate \(s\) by using the trapezium rule with all the values from your table. [4]

\includegraphics{figure_2} Figure 2 shows part of the curve with equation \(y = x^2 + 2\). The finite region \(R\) is bounded by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).

1. Use the trapezium rule with 4 strips of equal width to estimate the area of \(R\). [5]
2. State, with a reason, whether your answer in part \((a)\) is an under-estimate or over-estimate of the area of \(R\). [1]
3. Using integration, find the volume of the solid generated when \(R\) is rotated through \(360°\) about the \(x\)-axis, giving your answer in terms of \(\pi\). [6]

\includegraphics{figure_2} Figure 2 shows the cross-section of a road tunnel and its concrete surround. The curved section of the tunnel is modelled by the curve with equation \(y = 8\sqrt{\sin \frac{\pi x}{10}}\), in the interval \(0 \leq x \leq 10\). The concrete surround is represented by the shaded area bounded by the curve, the \(x\)-axis and the lines \(x = -2\), \(x = 12\) and \(y = 10\). The units on both axes are metres.

1. Using this model, copy and complete the table below, giving the values of \(y\) to 2 decimal places.

   \(x\)	0	2	4	6	8	10
   \(y\)	0	6.13				0

   [2]

The area of the cross-section of the tunnel is given by \(\int_0^{10} y \, dx\).

1. Estimate this area, using the trapezium rule with all the values from your table. [4]
2. Deduce an estimate of the cross-sectional area of the concrete surround. [1]
3. State, with a reason, whether your answer in part \((c)\) over-estimates or under-estimates the true value. [2]

$$f(x) = 2 \sin 2x + x - 2.$$ The root \(\alpha\) of the equation \(f(x) = 0\) lies in the interval \([2, \pi]\). Using the end points of this interval find, by linear interpolation, an approximation to \(\alpha\). [4]

$$f(x) = 2^x + x - 4.$$ The equation \(f(x) = 0\) has a root \(\alpha\) in the interval \([1, 2]\). Use linear interpolation on the values at the end points of this interval to find an approximation to \(\alpha\). [2]

A cyclist and her bicycle have a combined mass of \(100\) kg. She is working at a constant rate of \(80\) W and is moving in a straight line on a horizontal road. The resistance to motion is proportional to the square of her speed. Her initial speed is \(4\) m s\(^{-1}\) and her maximum possible speed under these conditions is \(20\) m s\(^{-1}\). When she is at a distance \(x\) m from a fixed point \(O\) on the road, she is moving with speed \(v\) m s\(^{-1}\) away from \(O\).

1. Show that $$v \frac{dv}{dx} = \frac{8000 - v^3}{10000v}.$$ [5]
2. Find the distance she travels as her speed increases from \(4\) m s\(^{-1}\) to \(8\) m s\(^{-1}\). [5]
3. Use the trapezium rule, with 2 intervals, to estimate how long it takes for her speed to increase from \(4\) m s\(^{-1}\) to \(8\) m s\(^{-1}\). [4]

\includegraphics{figure_8} The diagram shows the curve \(y = 1.25^x\).

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

Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O. \includegraphics{figure_11}

1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section. [4]
2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. [2]

The curve of the roof may be modelled by \(y = -0.013x^3 + 0.16x^2 - 0.082x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.

1. Use integration to find the area of the cross-section according to this model. [4]
2. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\). [2]

\includegraphics{figure_12} A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water.

1. Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough. Hence estimate the volume of water in the trough. [5]
2. A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation \(y = 8x^3 - 3x^2 - 0.5x - 0.15\), for \(0 \leq x \leq 0.5\). Calculate \(\int_0^{0.5} (8x^3 - 3x^2 - 0.5x - 0.15) \, \text{d}x\) and state what this represents. Hence find the volume of water in the trough as given by this model. [7]

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]

Oskar is designing a building. Fig. 12 shows his design for the end wall and the curve of the roof. The units for \(x\) and \(y\) are metres. \includegraphics{figure_12}

1. Use the trapezium rule with 5 strips to estimate the area of the end wall of the building. [4]
2. Oskar now uses the equation \(y = -0.001x^3 - 0.025x^2 + 0.6x + 9\), for \(0 \leq x \leq 15\), to model the curve of the roof.
   1. Calculate the difference between the height of the roof when \(x = 12\) given by this model and the data shown in Fig. 12. [2]
   2. Use integration to find the area of the end wall given by this model. [4]

Fig. 9 shows the cross-section of a straight, horizontal tunnel. The \(x\)-axis from 0 to 6 represents the floor of the tunnel. \includegraphics{figure_9} With axes as shown, and units in metres, the roof of the tunnel passes through the points shown in the table.

\(x\)	0	1	2	3	4	5	6
\(y\)	0	4.0	4.9	5.0	4.9	4.0	0

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. [4]
2. An engineer models the height of the roof of the tunnel using the curve \(y = \frac{x}{81}(108x - 54x^2 + 12x^3 - x^4)\). This curve is symmetrical about \(x = 3\).
   1. 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. [2]
   2. 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. [5]

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 \(R\). [6]
2. Use integration to find the exact area of \(R\) in the form \(a + b\sqrt{2}\). [5]
3. Find the percentage error in the estimate made in part (a). [2]

\includegraphics{figure_2} The diagram shows the curve with equation \(y = 2^x\). Use the trapezium rule with four intervals, each of width 1, to estimate the area of the shaded region bounded by the curve, the \(x\)-axis and the lines \(x = -2\) and \(x = 2\). [4]

\includegraphics{figure_4} The diagram shows the curve with equation \(y = (x - \log_{10} x)^2\), \(x > 0\).

1. Copy and complete the table below for points on the curve, giving the \(y\) values to 2 decimal places.

   \(x\)	2	3	4	5	6
   \(y\)	2.89	6.36

   [2]

The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 6\).

1. Use the trapezium rule with all the values in your table to estimate the area of the shaded region. [3]
2. State, with a reason, whether your answer to part (b) is an under-estimate or an over-estimate of the true area. [2]

The diagram shows the curve with equation \(y = \ln(6x)\). \includegraphics{figure_1}

1. State the \(x\)-coordinate of the point of intersection of the curve with the \(x\)-axis. [1]
2. Find \(\frac{dy}{dx}\). [2]
3. Use Simpson's rule with 6 strips (7 ordinates) to find an estimate for \(\int_1^7 \ln(6x) \, dx\), giving your answer to three significant figures. [4]

\includegraphics{figure_8} The diagram shows part of the curve \(y = \ln(5 - x^2)\) which meets the \(x\)-axis at the point \(P\) with coordinates \((2, 0)\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(PQ\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\). [5]
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures. [4]
3. Deduce an approximation to the area of the region \(B\). [2]

\includegraphics{figure_8} The diagram shows the curve with equation \(y = x^8 e^{-x^2}\). The curve has maximum points at \(P\) and \(Q\). The shaded region \(A\) is bounded by the curve, the line \(y = 0\) and the line through \(Q\) parallel to the \(y\)-axis. The shaded region \(B\) is bounded by the curve, the line \(y = 0\) and the line \(PQ\).

1. Show by differentiation that the \(x\)-coordinate of \(Q\) is 2. [5]
2. Use Simpson's rule with 4 strips to find an approximation to the area of region \(A\). Give your answer correct to 3 decimal places. [4]
3. Deduce an approximation to the area of region \(B\). [2]