The value of `int_0^1(2x^3-3x^2-x+1)^(1/3)dx` is equal to

Options

A. `-1`
B. 0
C. 1
D. `1/2`

Answer

To solve the integral \( I = \int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} \, dx \), we will follow these steps:

### Step 1: Identify the function and its roots
We start with the function \( f(x) = 2x^3 - 3x^2 - x + 1 \). To find the roots, we can evaluate \( f(x) \) at specific points. Notably, we check \( f(0) \) and \( f(1) \):

- \( f(0) = 1 \)
- \( f(1) = -1 \)

Since \( f(0) > 0 \) and \( f(1) < 0 \), by the Intermediate Value Theorem, there is at least one root in the interval \( (0, 1) \).

### Step 2: Find the root at \( x = \frac{1}{2} \)
Next, we can check \( f\left(\frac{1}{2}\right) \):

\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = 2 \cdot \frac{1}{8} - 3 \cdot \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{3}{4} - \frac{2}{4} + \frac{4}{4} = 0 \]

Thus, \( x = \frac{1}{2} \) is a root.

### Step 3: Factor the polynomial
Since \( x = \frac{1}{2} \) is a root, we can factor \( f(x) \) as follows:

\[ f(x) = (2x - 1)(Ax^2 + Bx + C) \]

Using polynomial long division or synthetic division, we can find the coefficients \( A \), \( B \), and \( C \). After performing the division, we find:

\[ f(x) = (2x - 1)(x^2 - x - 1) \]

### Step 4: Set up the integral
Now we can rewrite the integral:

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

### Step 5: Use symmetry properties
We can use the property of integrals:

\[ I = \int_0^1 f(x) \, dx = \int_0^1 f(1-x) \, dx \]

Calculating \( f(1-x) \):

\[ f(1-x) = 2(1-x)^3 - 3(1-x)^2 - (1-x) + 1 \]

This simplifies to:

\[ f(1-x) = 2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - (1 - x) + 1 \]

### Step 6: Combine the integrals
Now we can add the two integrals:

\[ 2I = \int_0^1 \left( f(x) + f(1-x) \right) \, dx \]

Since \( f(x) + f(1-x) = 0 \) for all \( x \) in \( [0, 1] \), we conclude that:

\[ 2I = 0 \implies I = 0 \]

### Final Answer
Thus, the value of the integral is:

\[ \int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} \, dx = 0 \]

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