The trapezium rule with data, and its error Problem set B One question at a time

The graph shows part of \(y=f(x)\). The table gives some values. x: 0, 0.6, 1.2, 1.8, 2.4, 3 y: 1.8, 3.2, 3.9, 3.5, 2.6, 1.1 Estimate the area between the graph and the x-axis from x = 0 to x = 3.

• \(h=0.6\)
• \(\tfrac12(0.6)[(1.800+1.100)+2(3.200+3.900+3.500+2.600)]\) = 8.790

Use the trapezium rule with 4 strips to estimate \(\int_{3}^{7} \sqrt{4x-11}\,dx\).

• \(h=\dfrac{7-3}{4}=1\)
• x: 3, 4, 5, 6, 7; y: 1.000, 2.236, 3.000, 3.606, 4.123
• \(\tfrac12(1)[(1.000+4.123)+2(2.236+3.000+3.606)]\) = 11.40

Use the trapezium rule with 5 strips to estimate \(\int_{0}^{2} \sqrt{7x+2}\,dx\).

• \(h=\dfrac{2-0}{5}=0.4\)
• x: 0, 0.4, 0.8, 1.2, 1.6, 2; y: 1.414, 2.191, 2.757, 3.225, 3.633, 4.000
• \(\tfrac12(0.4)[(1.414+4.000)+2(2.191+2.757+3.225+3.633)]\) = 5.805

The trapezium rule with 3 strips is used to estimate the area under \(y=5\ln x\) from x = 1 to x = 4.

1. (a) Find the estimate.

   • \(h=1\)
   • x: 1, 2, 3, 4; y: 0, 3.466, 5.493, 6.931
   • \(\tfrac12(1)[(0+6.931)+2(3.466+5.493)]\) = 12.42

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

   • Look at how the curve bends.
   • The curve is concave down, so the chords lie below it and the estimate is too small.

The width of a pond is measured every 5 m along its length: 0, 12, 19, 23, 17 and 0 m. Estimate the area of the pond.

• \(h=5\)
• \(\tfrac12(5)[(0+0)+2(12.00+19.00+23.00+17.00)]\) = 355.0
• Area ≈ 355 m²

A greenhouse is 10 m wide. Its height above the ground is measured every 2.5 m across the width: 0, 2.2, 3.0, 2.2 and 0 m. The cross-section is the area between the roof and the ground.

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

   • \(h=2.5\)
   • \(\tfrac12(2.5)[(0+0)+2(2.200+3.000+2.200)]\) = 18.50

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 lake is drawn on a map. Some points on its shores are marked. Distances are in metres. Estimate the area of the lake. Top edge: (10, 84), (25, 96), (40, 91), (55, 103), (70, 95). Bottom edge: (10, 30), (25, 26), (40, 33), (55, 28), (70, 31).

• Top: \(\tfrac12(15)[(84.00+95.00)+2(96.00+91.00+103.0)]\) = 5692.5
• Bottom: \(\tfrac12(15)[(30.00+31.00)+2(26.00+33.00+28.00)]\) = 1762.5
• 5692.5 − 1762.5 = 3930 ≈ 3930 m²

The integral \(\int_{0}^{3} (x^2+1)\,dx\) is estimated with the trapezium rule using 3 strips.

1. (a) Find the estimate.

   • \(h=1\)
   • x: 0, 1, 2, 3; y: 1.000, 2.000, 5.000, 10.00
   • \(\tfrac12(1)[(1.000+10.00)+2(2.000+5.000)]\) = 12.50

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

   • Calculator: \(\int_{0}^{3}(x^2+1)\,dx = 12.00\)

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

   • \(\dfrac{|12.50-12.00|}{12.00}\times100 = 4.17\%\)
   • The estimate is an overestimate.

A table gives some values of \(y=f(x)\). x: 0, 1.5, 3, 4.5, 6 y: 4, 6.5, k, 7.5, 3.5 The trapezium rule with 4 strips gives an estimate of 39.225 for the area under the graph. Find the value of k.

• \(h=1.5\), so \(\tfrac12(1.5)[(4+3.5)+2(k+6.5+7.5)]=39.225\)
• 1.5k + 26.625 = 39.225
• k = 8.4