The trapezium rule Problem set A One question at a time

Use the trapezium rule with 4 strips to estimate \(\int_{0}^{4} \frac{1}{x+3}\,dx\).

• \(h=\dfrac{4-0}{4}=1\)
• x: 0, 1, 2, 3, 4; y: 0.3333, 0.2500, 0.2000, 0.1667, 0.1429
• \(\tfrac12(1)[(0.3333+0.1429)+2(0.2500+0.2000+0.1667)]\) = 0.8548

Use the trapezium rule with 6 strips to estimate \(\int_{1}^{4} \frac{1}{x}\,dx\).

• \(h=\dfrac{4-1}{6}=0.5\)
• x: 1, 1.5, 2, 2.5, 3, 3.5, 4; y: 1.000, 0.6667, 0.5000, 0.4000, 0.3333, 0.2857, 0.2500
• \(\tfrac12(0.5)[(1.000+0.2500)+2(0.6667+0.5000+0.4000+0.3333+0.2857)]\) = 1.405

Use the trapezium rule with 5 strips to estimate \(\int_{0}^{1} 8e^x\,dx\).

• \(h=\dfrac{1-0}{5}=0.2\)
• x: 0, 0.2, 0.4, 0.6, 0.8, 1; y: 8.000, 9.771, 11.93, 14.58, 17.80, 21.75
• \(\tfrac12(0.2)[(8.000+21.75)+2(9.771+11.93+14.58+17.80)]\) = 13.79

Use the trapezium rule with 4 strips to estimate \(\int_{3}^{5} \ln x\,dx\).

• \(h=\dfrac{5-3}{4}=0.5\)
• x: 3, 3.5, 4, 4.5, 5; y: 1.099, 1.253, 1.386, 1.504, 1.609
• \(\tfrac12(0.5)[(1.099+1.609)+2(1.253+1.386+1.504)]\) = 2.749

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

• \(h=\dfrac{4-2}{4}=0.5\)
• x: 2, 2.5, 3, 3.5, 4; y: 2.079, 2.303, 2.485, 2.639, 2.773
• \(\tfrac12(0.5)[(2.079+2.773)+2(2.303+2.485+2.639)]\) = 4.926

The table shows the rate R (litres per second) at which water flows from a tap, measured every 3 seconds. The area under the graph is the volume of water. t: 0, 3, 6, 9, 12 R: 0, 1.8, 2.4, 2.1, 1.1 Estimate the area under the graph.

• \(h=3\)
• \(\tfrac12(3)[(0+1.100)+2(1.800+2.400+2.100)]\) = 20.55
• Answer ≈ 20.6

A sprinter's speed v (m/s) is measured every 0.5 seconds at the start of a race. t: 0, 0.5, 1, 1.5, 2, 2.5 v: 0, 2.9, 4.6, 5.8, 6.7, 7.3 Estimate the distance run in the first 2.5 seconds.

• \(h=0.5\)
• \(\tfrac12(0.5)[(0+7.300)+2(2.900+4.600+5.800+6.700)]\) = 11.82
• Answer ≈ 11.8 m

A pool is filled at the rate \(r=\ln(3t)\) hundred litres per minute, for \(1\le t\le 4\), where t is in minutes. Use the trapezium rule with 6 strips to estimate the volume added, in hundreds of litres, between t = 1 and t = 4.

• \(h=0.5\)
• x: 1, 1.5, 2, 2.5, 3, 3.5, 4; y: 1.099, 1.504, 1.792, 2.015, 2.197, 2.351, 2.485
• \(\tfrac12(0.5)[(1.099+2.485)+2(1.504+1.792+2.015+2.197+2.351)]\) = 5.826

The integral \(\int_{1}^{6} \frac{1}{x}\,dx\) is to be estimated.

1. (a) Use the trapezium rule with 5 strips.

   • \(h=\dfrac{6-1}{5}=1\)
   • x: 1, 2, 3, 4, 5, 6; y: 1.000, 0.5000, 0.3333, 0.2500, 0.2000, 0.1667
   • \(\tfrac12(1)[(1.000+0.1667)+2(0.5000+0.3333+0.2500+0.2000)]\) = 1.867

2. (b) Use the trapezium rule with 10 strips.

   • \(h=\dfrac{6-1}{10}=0.5\)
   • x: 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6; y: 1.000, 0.6667, 0.5000, 0.4000, 0.3333, 0.2857, 0.2500, 0.2222, 0.2000, 0.1818, 0.1667
   • \(\tfrac12(0.5)[(1.000+0.1667)+2(0.6667+0.5000+0.4000+0.3333+0.2857+0.2500+0.2222+0.2000+0.1818)]\) = 1.812

3. (c) Use your calculator to find the value of the integral. Which estimate is closer to it, and why?

   • Calculator: \(\int_{1}^{6}\frac1x\,dx = 1.792\)
   • Differences: 0.0749 for 5 strips, 0.0198 for 10 strips
   • More strips fit the curve more closely.

The integral \(\int_{0}^{2} 3e^x\,dx\) is to be estimated.

1. (a) Use the trapezium rule with 2 strips.

   • \(h=\dfrac{2-0}{2}=1\)
   • x: 0, 1, 2; y: 3.000, 8.155, 22.17
   • \(\tfrac12(1)[(3.000+22.17)+2(8.155)]\) = 20.74

2. (b) Use the trapezium rule with 4 strips.

   • \(h=\dfrac{2-0}{4}=0.5\)
   • x: 0, 0.5, 1, 1.5, 2; y: 3.000, 4.946, 8.155, 13.45, 22.17
   • \(\tfrac12(0.5)[(3.000+22.17)+2(4.946+8.155+13.45)]\) = 19.56

3. (c) Find the value of the integral with your calculator. By how much does each estimate differ from it?

   • Calculator: \(\int_{0}^{2}3e^x\,dx = 19.17\)
   • Differences: 1.57 and 0.398