The value of intlimits_0^1 {sqrt[3]{{2{x^3} - | vedclass

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(B) Let $I = \int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx}$.Using the property $\int\limits_0^a {f(x)\,dx} = \int\limits_0^a {f(a - x)\,dx

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(B) Let $I = \int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx}$.Using the property $\int\limits_0^a {f(x)\,dx} = \int\limits_0^a {f(a - x)\,dx}$,we have:$I = \int\limits_0^1 {\sqrt[3]{{2{{(1 - x)}^3} - 3{{(1 - x)}^2} - (1 - x) + 1}}\,dx}$Expanding the terms inside the cube root:$2{(1 - x)^3} = 2(1 - 3x + 3{x^2} - {x^3}) = 2 - 6x + 6{x^2} - 2{x^3}$$-3{(1 - x)^2} = -3(1 - 2x + {x^2}) = -3 + 6x - 3{x^2}$$-(1 - x) + 1 = -1 + x + 1 = x$Summing these: $(2 - 3 - 1 + 1) + (-6x + 6x + x) + (6{x^2} - 3{x^2}) + (-2{x^3}) = -1 + x + 3{x^2} - 2{x^3}$Thus,$I = \int\limits_0^1 {\sqrt[3]{{ - 2{x^3} + 3{x^2} + x - 1}}\,dx} = \int\limits_0^1 { - \sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} = -I$.Therefore,$2I = 0$,which implies $I = 0$.

The value of $\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ is

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