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Gauss 2-point and 3-point: log(x+1)+sin 2x

Short answer

With Gauss-Legendre quadrature after mapping x=0.6t+0.8: 2-point gives ≈1.6056, 3-point gives ≈1.6141.

Evaluate the integral I = ∫ from 0.2 to 1.4 of (log(x + 1) + sin 2x) dx using Gaussian 2 point and 3 point formula. [limits partly illegible in the scan]

Assumptions: log means the natural logarithm ln, sin 2x uses radians, and the limits are 0.2 to 1.4 (partly illegible in the scan).

Change of limits

x = [(b−a)t + (b+a)]/2 = [(1.4−0.2)t + (1.4+0.2)]/2 = 0.6t + 0.8, dx = 0.6 dt.

I = 0.6 ∫₋₁¹ F(t) dt, where F(t) = ln(1.8 + 0.6t) + sin(1.6 + 1.2t).

2-point Gauss–Legendre (t = ±0.577350, w = 1, 1)

tweightx = 0.6t+0.8f(x)w·f(x)
-0.57741.00000.45361.16181.1618
0.57741.00001.14641.51431.5143
I₂ = 0.6 × (2.6761) = 1.6056

3-point Gauss–Legendre (t = 0, ±0.774597; w = 8/9, 5/9, 5/9)

tweightx = 0.6t+0.8f(x)w·f(x)
-0.77460.55560.33520.91050.5058
0.00000.88890.80001.58741.4110
0.77460.55561.26481.39200.7734
I₃ = 0.6 × (2.6902) = 1.6141

Answer: 2-point result ≈ 1.6056; 3-point result ≈ 1.6141 (exact value 1.6140; the 3-point result is closer).

✓ Answer checked by Claude Opus 5.5 (high effort)

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