Verified answer

Ответ:

[tex]\displaystyle \int\limits^0_ {-1} \frac{xdx}{\sqrt{4-5x} }=-\frac{14}{75}[/tex]

Пошаговое объяснение:

Введем замену

[tex]\sqrt{4-5x} = v[/tex]

[tex]v^2 = 4-5x \\\\ x = \dfrac{4-v^2}{5} \\\\ dx = d\bigg( \dfrac{4-v^2}{5}\bigg ) \\\\ dx = -\dfrac{2}{5} v \; dv[/tex]

Подставим

[tex]\displaystyle \displaystyle \int\limits \frac{xdx}{\sqrt{4-5x} } = \int\limits \left ( \frac{ \dfrac{4-v^2}{5}}{v}\cdot \bigg -\frac{2}{5}\cdot v \right ) dv =-\frac{2}{25} \int\limits (4-v^2) \, du = \\\\\\ =- \frac{2}{25}\left( 4v - \frac{v^3}{3} \right) = -\frac{2}{25} \bigg(4\sqrt{4-5x} - \frac{\sqrt{(4-5x)^3}}{3} \bigg) +C[/tex]

Найдем определенный интеграл

[tex]\displaystyle \int\limits^0_ {-1} \frac{xdx}{\sqrt{4-5x} }= -\frac{2}{25} \bigg(4\sqrt{4-5x} - \frac{\sqrt{(4-5x)^3}}{3} \bigg) \Bigg |^0_{-1} = \\\\\\\ =-\frac{2}{25}\Bigg ( 4\cdot 2-\frac{8}{3} - \bigg ( 4\cdot 3 - \frac{27}{3} \bigg ) \Bigg ) = -\frac{2}{25} \bigg(8 - 2\frac{2}{3} -3 \bigg) =- \frac{2}{25} \cdot \frac{7}{3} =\boxed{- \frac{14}{75} }[/tex]

#SPJ1