The trapezium rule with data, and its error Problem set A Worked solutions

The graph shows part of \(y=f(x)\). The table gives some values. x: 1, 1.5, 2, 2.5, 3, 3.5 y: 2.4, 3.8, 4.5, 4.1, 3, 1.6 Estimate the area between the graph and the x-axis from x = 1 to x = 3.5.

• \(h=0.5\)
• \(\tfrac12(0.5)[(2.400+1.600)+2(3.800+4.500+4.100+3.000)]\) = 8.700

Use the trapezium rule with 4 strips to estimate \(\int_{1}^{5} \sqrt{2x-1}\,dx\).

• \(h=\dfrac{5-1}{4}=1\)
• x: 1, 2, 3, 4, 5; y: 1.000, 1.732, 2.236, 2.646, 3.000
• \(\tfrac12(1)[(1.000+3.000)+2(1.732+2.236+2.646)]\) = 8.614

Use the trapezium rule with 5 strips to estimate \(\int_{0}^{3} \sqrt{5x+1}\,dx\).

• \(h=\dfrac{3-0}{5}=0.6\)
• x: 0, 0.6, 1.2, 1.8, 2.4, 3; y: 1.000, 2.000, 2.646, 3.162, 3.606, 4.000
• \(\tfrac12(0.6)[(1.000+4.000)+2(2.000+2.646+3.162+3.606)]\) = 8.348

The trapezium rule with 3 strips is used to estimate the area under \(y=\frac{12}{x+1}\) from x = 0 to x = 3.

1. (a) Find the estimate.

   • \(h=1\)
   • x: 0, 1, 2, 3; y: 12.00, 6.000, 4.000, 3.000
   • \(\tfrac12(1)[(12.00+3.000)+2(6.000+4.000)]\) = 17.50

2. (b) Is it an overestimate or an underestimate? Give a reason.

   • Look at how the curve bends.
   • The curve is concave up, so the chords lie above it and the estimate is too large.

The width of a lake is measured every 20 m along its length: 0, 45, 68, 82, 74 and 0 m. Estimate the area of the lake.

• \(h=20\)
• \(\tfrac12(20)[(0+0)+2(45.00+68.00+82.00+74.00)]\) = 5380
• Area ≈ 5380 m²

A bridge arch is 12 m wide. Its height above the ground is measured every 3 m from one end: 0, 2.8, 4.1, 3.3 and 0 m. The opening under the arch is the area between the arch and the ground.

1. (a) Estimate the cross-sectional area.

   • \(h=3\)
   • \(\tfrac12(3)[(0+0)+2(2.800+4.100+3.300)]\) = 30.60

2. (b) The roof curves outwards between the measured points. Is the real area larger or smaller than your estimate? Explain.

   • The roof curves outwards (concave down).
   • The straight chords lie below the roof, so the trapeziums cover less than the real area.
   • The real area is larger than the estimate.

A field is drawn on a map. Some points on its edges are marked. Distances are in metres. Estimate the area of the field. Top edge: (20, 150), (40, 172), (60, 160), (80, 181), (100, 168). Bottom edge: (20, 62), (40, 58), (60, 66), (80, 61), (100, 64).

• Top: \(\tfrac12(20)[(150.0+168.0)+2(172.0+160.0+181.0)]\) = 13440
• Bottom: \(\tfrac12(20)[(62.00+64.00)+2(58.00+66.00+61.00)]\) = 4960
• 13440 − 4960 = 8480 ≈ 8480 m²

The integral \(\int_{0}^{2} (x^3+x)\,dx\) is estimated with the trapezium rule using 4 strips.

1. (a) Find the estimate.

   • \(h=0.5\)
   • x: 0, 0.5, 1, 1.5, 2; y: 0, 0.6250, 2.000, 4.875, 10.00
   • \(\tfrac12(0.5)[(0+10.00)+2(0.6250+2.000+4.875)]\) = 6.250

2. (b) Use your calculator to find the exact value of the integral.

   • Calculator: \(\int_{0}^{2}(x^3+x)\,dx = 6.000\)

3. (c) Find the percentage error of the estimate.

   • \(\dfrac{|6.250-6.000|}{6.000}\times100 = 4.17\%\)
   • The estimate is an overestimate.

A table gives some values of \(y=f(x)\). x: 0, 2, 4, 6, 8 y: 5, 7.5, k, 6.5, 3 The trapezium rule with 4 strips gives an estimate of 54.4 for the area under the graph. Find the value of k.

• \(h=2\), so \(\tfrac12(2)[(5+3)+2(k+7.5+6.5)]=54.4\)
• 2k + 36.000 = 54.4
• k = 9.2