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Mathematics Question on integral

The value of ∫01(2x3−3x2−x+1)13 dx\int_{0}^{1} \left(2x^3 - 3x^2 - x + 1\right)^{\frac{1}{3}} \, dxis equal to:

A

0

B

1

C

2

D

-1

Answer

0

Explanation

Solution

The given integral is: I=∫01(2x3−3x2−x+1)1/3 dx.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→(1−x)x \to (1 - x): I=∫01[2(1−x)3−3(1−x)2−(1−x)+1]1/3 dx.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+3x2−x3,(1 - x)^3 = 1 - 3x + 3x^2 - x^3, (1−x)2=1−2x+x2.(1 - x)^2 = 1 - 2x + x^2.

Substitute these into the equation: 2(1−x)3=2(1−3x+3x2−x3)=2−6x+6x2−2x3,2(1 - x)^3 = 2(1 - 3x + 3x^2 - x^3) = 2 - 6x + 6x^2 - 2x^3, −3(1−x)2=−3(1−2x+x2)=−3+6x−3x2,-3(1 - x)^2 = -3(1 - 2x + x^2) = -3 + 6x - 3x^2, −(1−x)=−1+x,and +1 remains unchanged.-(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+6x2−2x3)+(−3+6x−3x2)+(−1+x)+1.2(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+6x2−2x3−3+6x−3x2−1+x+1.= 2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1.
=2−3−1+1−6x+6x+x+6x2−3x2−2x3.= 2 - 3 - 1 + 1 - 6x + 6x + x + 6x^2 - 3x^2 - 2x^3.
Final simplified expression: =−1+x+3x2−2x3.= -1 + x + 3x^2 - 2x^3.

The integral becomes: I=∫01(−1+x+3x2−2x3)1/3 dx.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=∫01(2(1−x)3−3(1−x)2−(1−x)+1)1/3 dx.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=∫01[−2x3+3x2+x−1]1/3 dx.I = \int_0^1 \left[-2x^3 + 3x^2 + x - 1\right]^{1/3} \, dx.
Using the property of definite integrals, I=∫01f(x) dx=∫01f(1−x) dxI = \int_0^1 f(x) \, dx = \int_0^1 f(1 - x) \, dx,
the integral becomes: I=−∫01[2x3−3x2−x+1]1/3 dx.I = -\int_0^1 \left[2x^3 - 3x^2 - x + 1\right]^{1/3} \, dx.

Thus: I=−I.I = -I. Adding II to both sides: 2I=0⇒I=0.2I = 0 \quad \Rightarrow \quad I = 0.
Therefore, the final answer is: I=0.I = 0.