Complete table then apply trapezium rule

\includegraphics{figure_1} Figure 1 shows the graph of $$y = 1 - \log_{10}(\sin x) \quad 0 < x < \pi$$ where \(x\) is in radians. The table below shows some values of \(x\) and \(y\) for this graph, with values of \(y\) given to 3 decimal places.

1. Complete the table above, giving values of \(y\) to 3 decimal places. [2]
2. Use the trapezium rule with all the \(y\) values in the completed table to find, to 2 decimal places, an estimate for $$\int_{0.5}^{3} \left(1 - \log_{10}(\sin x)\right) dx$$ [3]
3. Use your answer to part (b) to find an estimate for $$\int_{0.5}^{3} \left(3 + \log_{10}(\sin x)\right) dx$$ [3]

A river, running between parallel banks, is 20 m wide. The depth, \(y\) metres, of the river measured at a point \(x\) metres from one bank is given by the formula $$y = \frac{1}{10}x(20 - x), \quad 0 \leq x \leq 20.$$

1. Complete the table below, giving values of \(y\) to 3 decimal places.

   \(x\)	0	4	8	12	16	20
   \(y\)	0		2.771			0

   [2]
2. Use the trapezium rule with all the values in the table to estimate the cross-sectional area of the river. [4]

Given that the cross-sectional area is constant and that the river is flowing uniformly at 2 m s⁻¹,

1. estimate, in m³, the volume of water flowing per minute, giving your answer to 3 significant figures. [2]

The speed, \(v\) m s⁻¹, of a train at time \(t\) seconds is given by \(v = \sqrt{(1.2^t - 1)}, \quad 0 \leq t \leq 30.\) The following table shows the speed of the train at 5 second intervals.

1. Complete the table, giving the values of \(v\) to 2 decimal places. [3]

The distance, \(s\) metres, travelled by the train in 30 seconds is given by $$s = \int_0^{30} \sqrt{(1.2^t - 1)} \, dt.$$

1. Use the trapezium rule, with all the values from your table, to estimate the value of \(s\). [3]

The curve \(C\) has equation \(y = x\sqrt{x^2 + 1}, \quad 0 \leq x \leq 2\).

1. Copy and complete the table below, giving the values of \(y\) to 3 decimal places at \(x = 1\) and \(x = 1.5\).

   \(x\)	0	0.5	1	1.5	2
   \(y\)	0	0.530			6

   [2]
2. Use the trapezium rule, with all the \(y\) values from your table, to find an approximation for the value of \(\int_0^2 x\sqrt{x^2 + 1} \, dx\), giving your answer to 3 significant figures. [4]

\includegraphics{figure_2} Figure 2 shows the curve \(C\) with equation \(y = x\sqrt{x^2 + 1}\), \(0 \leq x \leq 2\), and the straight line segment \(l\), which joins the origin and the point \((2, 6)\). The finite region \(R\) is bounded by \(C\) and \(l\).

1. Use your answer to part (b) to find an approximation for the area of \(R\), giving your answer to 3 significant figures. [3]

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 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]

\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]

\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]

\includegraphics{figure_2} Figure 2 shows part of the curve with equation $$y = e^x \cos x, \quad 0 \leq x \leq \frac{\pi}{2}.$$ 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]