Trapezium rule for applications

\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. 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]