The value of \[\int_{0}^{1} \left(2x^3 - 3x^2 - x + 1\right)^{\frac{1}{3}} \, dx\]is equal to:
- 0
- 1
- 2
- -1
The Correct Option is A
Solution and Explanation
The given integral is: \[ I = \int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{1/3} \, dx. \]
Applying the King’s property of definite integrals, replace \(x \to (1 - x)\): \[ I = \int_0^1 \left[2(1 - x)^3 - 3(1 - x)^2 - (1 - x) + 1\right]^{1/3} \, dx. \]
Simplify the expression inside the integral: \[ (1 - x)^3 = 1 - 3x + 3x^2 - x^3, \] \[ (1 - x)^2 = 1 - 2x + x^2. \]
Substitute these into the equation: \[ 2(1 - x)^3 = 2(1 - 3x + 3x^2 - x^3) = 2 - 6x + 6x^2 - 2x^3, \] \[ -3(1 - x)^2 = -3(1 - 2x + x^2) = -3 + 6x - 3x^2, \] \[ -(1 - x) = -1 + x, \quad \text{and } +1 \text{ remains unchanged.} \]
Combine all terms: \[ 2(1 - x)^3 - 3(1 - x)^2 - (1 - x) + 1 = (2 - 6x + 6x^2 - 2x^3) + (-3 + 6x - 3x^2) + (-1 + x) + 1. \]
Simplify: \[ = 2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1. \]
\[ = 2 - 3 - 1 + 1 - 6x + 6x + x + 6x^2 - 3x^2 - 2x^3. \]
Final simplified expression: \[ = -1 + x + 3x^2 - 2x^3. \]
The integral becomes: \[ I = \int_0^1 \left(-1 + x + 3x^2 - 2x^3\right)^{1/3} \, dx. \]
Thus, after applying King’s property, the integral is rewritten as: \[ I = \int_0^1 \left(2(1 - x)^3 - 3(1 - x)^2 - (1 - x) + 1\right)^{1/3} \, dx. \]
Simplify the expression inside the integral: \[ I = \int_0^1 \left[-2x^3 + 3x^2 + x - 1\right]^{1/3} \, dx. \]
Using the property of definite integrals, \(I = \int_0^1 f(x) \, dx = \int_0^1 f(1 - x) \, dx\),
the integral becomes: \[ I = -\int_0^1 \left[2x^3 - 3x^2 - x + 1\right]^{1/3} \, dx. \]
Thus: \[ I = -I. \] Adding \(I\) to both sides: \[ 2I = 0 \quad \Rightarrow \quad I = 0. \]
Therefore, the final answer is: \[ I = 0. \]
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