Bài 106155

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Tính tích phân:

$I = \int\limits_0^1 {x^5\sqrt {1 - x^3} dx}$

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$I=\frac{1}{3}\int^1_0 x^3\sqrt{1-x^3}dx^3$

Đặt $t=\sqrt{1-x^3}\Rightarrow x^3=1-t^2$

$\Rightarrow I=\frac{-2}{3}\int^0_1(1-t^2)t^2dt=\frac{2}{3}\int^1_0(t^2-t^4)dt$

$=\frac{2}{3}(\frac{t^3}{3}-\frac{t^5}{5})|^{1}_{0}=\frac{4}{45}$

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