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

Question

The value of $\int _{0}^{1}{(2{x}^{3}-3{x}^{2}-x+1)}^{\frac{1}{3}}dx$ is equal to :

Accepted Answer

$\begin{gathered} I={\int }_{1}^{0}{(2{x}^{3}-3{x}^{2}-x+1)}^{1/3}dx \ I={\int }_{1}^{0}{((2x-1)({x}^{2}-x-1))}^{1/3}dx \ I={\int }_{1}^{0}{[(2(1-x)-1)((1-x{)}^{2}-(1-x)-1)]}^{1/3}dx \ I={\int }_{1}^{0}{((1-2x)({x}^{2}-x-1))}^{1/3}dx \end{gathered}$

$\begin{gathered} I=-{\int }_{1}^{0}{((2x-1)({x}^{2}-x-1))}^{1/3}dx \ I=-I \ 2I=0 \ I=0 \end{gathered}$

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