The value of _0^1(2 x^3-3 x^2-x+1)^(1)/(3) ~d x is equal to :

Question

The value of $\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} \mathrm{~d} x$ is equal to :

Accepted Answer

$\begin{aligned} & I=\int_0^1\left(2 x^3-3 x^2-x+1ight)^{1 / 3} d x \\ & I=\int_0^1\left((2 x-1)\left(x^2-x-1ight)ight)^{1 / 3} d x \\ & I=\int_0^1\left[(2(1-x)-1)\left((1-x)^2-(1-x)-1ight)ight]^{1 / 3} d x \\ & I=\int_0^1\left((1-2 x)\left(x^2-x-1ight)ight)^{1 / 3} d x\end{aligned}$

$\begin{aligned} & I=-\int_0^1\left((2 x-1)\left(x^2-x-1ight)ight)^{1 / 3} d x \\ & I=-I \\ & 2 I=0 \\ & I=0\end{aligned}$

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