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📅 May 15, 2026 👤 scraface

May 15, 2026

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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 +...
So integral = 20k × ( -1/ln2 ) = -20k / ln2
ln2 ≈ 0.6931
k ≈ 636.39
So -20 × 636.39 / 0.6931 ≈ -12727.8 / 0.6931 ≈ -18368. But this is neg...
Wait: correct formula:
∫ a^u du = u / (ln a), but a = 0.5, ln(0.5) < 0
So ∫ (0.5)^u du = u / (ln 0.5) = u / (-ln 2) = -u / ln 2

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