Помогите с интегралом....Заранее спасибо!. Created by Bahaa. algebra-ru

Помогите с интегралом....Заранее спасибо!...

Bahaa @Bahaa

July 2022 1 33 Report

Помогите с интегралом....Заранее спасибо!

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NNNLLL54

Verified answer

$\int _1^{1,5}(1-2x)^3dx=-\frac{1}{2}\cdot \frac{(1-2x)^4}{4}\left |_1^{1,5}=- \frac{1}{8} ((-2)^4-(-1)^4)=$

$=- \frac{1}{8}\cdot (16-1) =- \frac{15}{8} =-1,875$

$P.S.\quad \int\, (ax+b)^{k} \, dx = \frac{1}{k} \cdot \frac{(ax+b)^{k+1}}{k+1} +C$

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