If `f(x) = |(0,x^2 - sin x, cos x - 2),(sin x - x^2 ,0, 1 - 2x),(2 - cos x, 2x - 1 , 0)|`
then `int f(x) dx` is equal to

Options

A. `(x^3)/(3) - x^2 sin x + sin 2x `
B. `(x^3)/3 - x^2 sin x - cos 2x`
C. `(x^3)/3 - x^2 cos x - cos 2x`
D. c

Answer

To solve the problem, we need to evaluate the integral of the determinant function \( f(x) \) defined as:

\[ f(x) = \begin{vmatrix} 0 & x^2 - \sin x & \cos x - 2 \\ \sin x - x^2 & 0 & 1 - 2x \\ 2 - \cos x & 2x - 1 & 0 \end{vmatrix} \]

### Step 1: Expand the Determinant
We will expand the determinant using the first row. The determinant can be expressed as:

\[ f(x) = 0 \cdot D_1 - (x^2 - \sin x) \cdot D_2 + (\cos x - 2) \cdot D_3 \]

Where \( D_1, D_2, D_3 \) are the determinants of the 2x2 matrices obtained by removing the corresponding row and column.

Since the first term is multiplied by zero, it contributes nothing. We will focus on the second and third terms.

### Step 2: Calculate \( D_2 \)
For \( D_2 \):

\[ D_2 = \begin{vmatrix} \sin x - x^2 & 1 - 2x \\ 2 - \cos x & 0 \end{vmatrix} \]

Calculating this determinant:

\[ D_2 = (\sin x - x^2) \cdot 0 - (1 - 2x)(2 - \cos x) = -(1 - 2x)(2 - \cos x) \]

### Step 3: Calculate \( D_3 \)
For \( D_3 \):

\[ D_3 = \begin{vmatrix} \sin x - x^2 & 0 \\ 2x - 1 & 2 - \cos x \end{vmatrix} \]

Calculating this determinant:

\[ D_3 = (\sin x - x^2)(2 - \cos x) - 0 \cdot (2x - 1) = (\sin x - x^2)(2 - \cos x) \]

### Step 4: Substitute Back into \( f(x) \)
Now substituting \( D_2 \) and \( D_3 \) back into the expression for \( f(x) \):

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

### Step 5: Simplify \( f(x) \)
After simplification, we can see that the terms will cancel out, leading to:

\[ f(x) = 0 \]

### Step 6: Integrate \( f(x) \)
Now we need to find the integral:

\[ \int f(x) \, dx = \int 0 \, dx = C \]

Where \( C \) is a constant of integration.

### Final Answer
Thus, the integral \( \int f(x) \, dx \) is equal to a constant \( C \).