The value of int_0^1 (2x^3 - 3x^2 - x + 1)^{f | vedclass

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(A) Let $I = \int_0^1 (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$.Consider the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.Let $f(x) = (2x^3 - 3x^2 -

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(A) Let $I = \int_0^1 (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$.Consider the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.Let $f(x) = (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}}$.Then $f(1-x) = (2(1-x)^3 - 3(1-x)^2 - (1-x) + 1)^{\frac{1}{3}}$.Expanding the terms: $f(1-x) = (2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - 1 + x + 1)^{\frac{1}{3}}$.$f(1-x) = (2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1)^{\frac{1}{3}}$.$f(1-x) = (-2x^3 + 3x^2 + x - 1)^{\frac{1}{3}} = -(2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} = -f(x)$.Since $f(1-x) = -f(x)$,the integral $I = \int_0^1 f(x) dx$ satisfies $I = -I$,which implies $2I = 0$,so $I = 0$.

The value of $\int_0^1 (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$ is equal to:

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