Calcular a integral definida $I\ =\ \displaystyle\int_0^1 \dfrac{y^2}{\sqrt{4 - 3y}}\ dy$.

Seja $4 - 3y = x$, $y = \dfrac{4 - x}{3}$ e $dx = -3dy$.

 

$I\ =\ -\dfrac{1}{27}\displaystyle\int_4^1 \dfrac{16 - 8x + x^2}{\sqrt{x}}\ dx\ =\ -\dfrac{1}{27}\left.\left(32\sqrt{x} - \dfrac{16}{3}\sqrt{x^3} + \dfrac{2}{5}\sqrt{x^5}\right)\right|_4^1 =$

 

$=\ -\dfrac{32 - 64 - \dfrac{16}{3} + \dfrac{128}{3} + \dfrac{2}{5} - \dfrac{64}{5}}{27} = \fbox{$\dfrac{106}{405}$}$