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Integration

4.6 Numerical Integration

Focus on trapezoidal, midpoint, and simpson rules, error bounds, choose a sampling method.

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Ideas in this lesson

01Trapezoidal, midpoint, and Simpson rulesWorked steps · five questions02Error boundsWorked steps · five questions03Choose a sampling methodWorked steps · five questions
Idea 1 of 3

Trapezoidal, midpoint, and Simpson rules

Use the appropriate weights for the chosen rule. Simpson’ss Rule requires an even number of equal subintervals.

Worked example 1

Approximate ∫01x2 dx\int_{0}^{1} x^{2}\,dx with n=2n = 2 by the Trapezoidal Rule.
  1. h=12h = \dfrac{1}{2}; xxᵢ =i2= \dfrac{i}{2}
  2. Tn=h2[f(0)+2∑f(xT_{n} = \dfrac{h}{2}[f(0) + 2\sum f(xᵢ)+f(1)]) + f(1)]
  3. T2=13+124=38≈0.38T_2 = \dfrac{1}{3} + \dfrac{1}{24} = \dfrac{3}{8} \approx 0.38

Answer: 38≈0.38\dfrac{3}{8} \approx 0.38

Worked example 2

Approximate ∫01x2 dx\int_{0}^{1} x^{2}\,dx with n=4n = 4 by Simpson’ss Rule.
  1. h=14h = \dfrac{1}{4}; xxᵢ =i4= \dfrac{i}{4}
  2. Sn=h3[f(0)+4∑f(xodd)+2∑f(xeven)+f(1)]S_{n} = \dfrac{h}{3}[f(0) + 4\sum f(x_{\mathrm{odd}}) + 2\sum f(x_{\mathrm{even}}) + f(1)]
  3. S4=13=13≈0.33S_4 = \dfrac{1}{3} = \dfrac{1}{3} \approx 0.33

Answer: 13≈0.33\dfrac{1}{3} \approx 0.33

Idea 2 of 3

Error bounds

A derivative bound controls the approximation error. A zero fourth derivative makes Simpson’ss Rule exact for a quadratic.

Worked example 1

For ∫01x2 dx\int_{0}^{1} x^{2}\,dx with n=4n = 4, find the Trapezoidal error bound.
  1. f′′=2f'' = 2; ff⁽4^{4}⁾ =0= 0; M2=2M_{2} = 2; M4=0M_{4} = 0
  2. ∣ET∣≤2(1−0)3192=196≈0.01|E_T| \le \dfrac{2(1 - 0)^{3}}{192} = \dfrac{1}{96} \approx 0.01

Answer: 196≈0.01\dfrac{1}{96} \approx 0.01

Worked example 2

For ∫01x2 dx\int_{0}^{1} x^{2}\,dx with n=6n = 6, find the Midpoint error bound.
  1. f′′=2f'' = 2; ff⁽4^{4}⁾ =0= 0; M2=2M_{2} = 2; M4=0M_{4} = 0
  2. ∣EM∣≤2(1−0)3864=1432≈0.00231|E_M| \le \dfrac{2(1 - 0)^{3}}{864} = \dfrac{1}{432} \approx 0.00231

Answer: 1432≈0.00231\dfrac{1}{432} \approx 0.00231

Idea 3 of 3

Choose a sampling method

Left, right, and midpoint sums use different heights on the same partition; a larger nn alone is not an exact answer.

Worked example 1

Approximate ∫02x2 dx\int_{0}^{2} x^{2}\,dx with 33 equal left rectangles.
xx000.66666666670.66666666671.3333333331.333333333
f(x)f(x)000.44444444440.44444444441.7777777781.777777778
  1. Δx=23≈0.67\Delta x = \dfrac{2}{3} \approx 0.67
  2. Sample points: xxᵢ =0,≈0.67,≈1.33= 0, \approx 0.67, \approx 1.33
  3. L3=23[02+≈0.672+≈1.332]=4027≈1.48L_3 = \dfrac{2}{3}[0^{2} + \approx 0.67^{2} + \approx 1.33^{2}] = \dfrac{40}{27} \approx 1.48

Answer: 4027≈1.48\dfrac{40}{27} \approx 1.48

Worked example 2

Approximate ∫03x2 dx\int_{0}^{3} x^{2}\,dx with 44 equal right rectangles.
xx0.750.751.51.52.252.2533
f(x)f(x)0.56250.56252.252.255.06255.062599
  1. Δx=34≈0.75\Delta x = \dfrac{3}{4} \approx 0.75
  2. Sample points: xxᵢ =0.75,1.5,2.25,3= 0.75, 1.5, 2.25, 3
  3. R4=34[0.752+1.52+2.252+32]=40532≈12.66R_4 = \dfrac{3}{4}[0.75^{2} + 1.5^{2} + 2.25^{2} + 3^{2}] = \dfrac{405}{32} \approx 12.66

Answer: 40532≈12.66\dfrac{405}{32} \approx 12.66

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §4.6,p4.6, p. 305305. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.