∫₂⁴ 2^(-d/20) dd = 20 ∫₁² 2^(-u) du = 20 [ -2^(-u) / ln2 ]₁² = 20 [ ( -1/(2^u ln2) ) ]₁² = 20 [ -1/(2^2 ln2) + 1/(2^1 ln2) ] = 20 [ -1/(4 ln2) + 1/(2 ln2) ] = 20 [ ( -0.25 + 0.5 ) / ln2 ] = 20 × (0.25 / ln2) = 5 / ln2 - Nelissen Grade advocaten
Feb 28, 2026
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