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📅 March 12, 2026 👤 scraface

Mar 12, 2026

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Now, from d = 20 to d = 40:
N(20) = k × (1/2)^(20/20) = k × 0.5
N(40) = k × (1/2)^(40/20) = k × 0.25
So in 20–40 cm: integrate or sum? Since discrete data, assume continuo...
Total = ∫ from 20 to 40 of N(d) dd = ∫ from 20 to 40 of k × (1/2)^(d/2...
Let u = d/20 → d = 20u, dd = 20 du
Limits: u = 1 to u = 2
Integral = ∫₁² k × (1/2)^u × 20 du = 20k ∫₁² (0.5)^u du
∫ (0.5)^u du = u / (ln(0.5)) = -u / ln(2)
So ∫₁² (0.5)^u du = [ -u / ln(2) ]₁² = ( -2/ln2 ) - ( -1/ln2 ) = (-2 +...

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