Question

The value of \[\int_{0}^{1} \left(2x^3 - 3x^2 - x + 1\right)^{\frac{1}{3}} \, dx\]is equal to:

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

The Correct Option is A

The integral to be evaluated is:

$$ I = \int_0^1 \left( 2x^3 - 3x^2 - x + 1 \right)^{1/3} \, dx $$

Applying King's Rule, substitute \( x \) with \( (1 - x) \):

$$ I = \int_0^1 \left[ 2(1 - x)^3 - 3(1 - x)^2 - (1 - x) + 1 \right]^{1/3} \, dx $$

Simplify the integrand:

$$ I = \int_0^1 \left[ 2 - 2x^3 - 6x + 6x^2 - 3 - 3x^2 + 6x - 1 + x + 1 \right]^{1/3} \, dx $$ $$ I = \int_0^1 \left[ -2x^3 + 3x^2 + x - 1 \right]^{1/3} \, dx $$

This simplifies to:

$$ I = - \int_0^1 \left[ 2x^3 - 3x^2 - x + 1 \right]^{1/3} \, dx $$

Therefore, we have the relation:

$$ I = -I $$

Solving for \( I \):

$$ 2I = 0 \Rightarrow I = 0 $$

The value of the integral is \( I = 0 \).