∫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 - AMAZONAWS 📅 March 11, 2026 👤 scraface Mar 11, 2026
![∫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](https://soloferat.biz.id/images/-2-d20-dd--20--2-u-du--20---2-u--ln2---20----12u-ln2----20---122-ln2--121-ln2---20---14-ln2--12-ln2---20----025--05---ln2---20--025--ln2--5--ln2.jpg)
