$I=\frac{1}{3}\int^1_0 x^3\sqrt{1-x^3}dx^3$
Đặt $ t=\sqrt{1-x^3}\Rightarrow x^3=1-t^2$
$\Rightarrow I=\frac{-2}{3}\int^0_1(1-t^2)t^2dt=\frac{2}{3}\int^1_0(t^2-t^4)dt$
$=\frac{2}{3}(\frac{t^3}{3}-\frac{t^5}{5})|^{1}_{0}=\frac{4}{45}$
Thẻ
Lượt xem