The value of ∫ _ 0 ^ 1 (2 x ^ 3 - 3 x ^ 2 - x + 1)(1)/(3)dx is...

$I = \mathop{∫}_{0}^{1}\left(2{x}^{3} - 3{x}^{2} - x + 1\right)\dfrac{1}{3}dx$

Using $\mathop{∫}_{0}^{2a}f\left(x\right)dx = 0$ when$f\left(2a - x\right) = - f\left(x\right)$

Now, finding $f\left(1 - x\right)$ we get,

$⇒ I = \mathop{∫}_{0}^{1}\left(2{\left(1 - x\right)}^{3} - 3{\left(1 - x\right)}^{2} - \left(1 - x\right) + 1\right)\dfrac{1}{3}dx$

$⇒ I = \mathop{∫}_{0}^{1}\left(2\left(1 - {x}^{3} - 3x + 3{x}^{2}\right) - 3{\left(1 + {x}^{2} - 2x\right)}^{2} - \left(1 - x\right) + 1\right)\dfrac{1}{3}dx$

$⇒ I = \mathop{∫}_{0}^{1}\left(2 - 2{x}^{3} - 6x + 6{x}^{2} - 3 - 3{x}^{2} + 6x - 1 + x + 1\right)\dfrac{1}{3}dx$

$⇒ I = \mathop{∫}_{0}^{1}\left(- 2{x}^{3} + 3{x}^{2} + x - 1\right)\dfrac{1}{3}dx$

$⇒ I = - I$

$⇒ f\left(1 - x\right) = - f\left(x\right)$

$∴ I = 0$