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Home » \cos \theta = \frac{\textadjacent}{\texthypotenuse} = \frac810 = 0.8 - Aurero

\cos \theta = \frac{\textadjacent}{\texthypotenuse} = \frac810 = 0.8 - Aurero

📅 March 15, 2026 👤 scraface

Mar 15, 2026

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📌 P(Y_{(6)} - Y_{(1)} \leq 0.2) = \int_{u=0}^1 \int_{v=u}^{u+0.2} 30(v - u)^4 \, dv \, du

📌 Change variable: let $ s = v - u $, so $ v = u + s $, $ 0 \leq s \leq 0.2 $, and $ s \leq 1 - u $

📌 But since $ u \leq 1 - s \leq 1 $, and $ s \leq 0.2 $, the upper limit $ 1 - u $ is automatically satisfied for $ u \geq 0.8 $, but we integrate $ u $ from 0 to 1.

📌 \int_0^1 \int_0^{\min(0.2, 1 - u)} 30 s^4 \, ds \, du

📌 Split into two regions: $ 0 \leq u \leq 0.8 $, where $ 1 - u \geq 0.2 $, so $ s $ goes to 0.2; for $ u > 0.8 $, $ 1 - u < 0.2 $, so $ s $ goes to $ 1 - u $

📌 = \int_0^{0.8} \int_0^{0.2} 30 s^4 ds \, du + \int_{0.8}^1 \int_0^{1 - u} 30 s^4 ds \, du

📌 First integral:

📌 \int_0^{0.2} 30 \cdot \frac{s^5}{5} ds = 6 \int_0^{0.2} s^5 ds = 6 \cdot \frac{s^6}{6} \Big|_0^{0.2} = (0.2)^6 = \left(\frac{1}{5}\right)^6 = \frac{1}{15625}

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