I = intlimits_0^1 sqrt[3]{2x^3 - 3x^2 - x + 1 | vedclass

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

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(B) Let $I = \int\limits_0^1 f(x) \, dx$,where $f(x) = \sqrt[3]{2x^3 - 3x^2 - x + 1}$.Using the property $\int\limits_a^b f(x) \, dx = \int\limits_a^b f(a+b-x) \, dx$,we have $I = \int\limits_0^1 f(1-x) \, dx$.Calculate $f(1-x) = \sqrt[3]{2(1-x)^3 - 3(1-x)^2 - (1-x) + 1}$.$f(1-x) = \sqrt[3]{2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - 1 + x + 1}$.$f(1-x) = \sqrt[3]{2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1}$.$f(1-x) = \sqrt[3]{-2x^3 + 3x^2 + x - 1} = -\sqrt[3]{2x^3 - 3x^2 - x + 1} = -f(x)$.Therefore,$I = \int\limits_0^1 -f(x) \, dx = -I$.$2I = 0 \implies I = 0$.

$I = \int\limits_0^1 \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx$ is equal to

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