If `f(x)=|{:(0,x^2-sinx,cosx-2),(sinx-x^2,0,1-2x),(2-cosx,2x-1,0):}|` then `intf(x)` dx is equal to

Options

A. `x^3/3-x^2sinx+sin2x+C`
B. `x^3/3-x^2sinx-cos2x+C`
C. `x^3/3-x^2cosx-cos2x+C`
D. None of the above

Answer

To solve the problem, we need to evaluate the function \( f(x) \) given by the determinant:

\[ f(x) = \left| \begin{array}{ccc} 0 & x^2 - \sin x & \cos x - 2 \\ \sin x - x^2 & 0 & 1 - 2x \\ 2 - \cos x & 2x - 1 & 0 \end{array} \right| \]

### Step 1: Expand the Determinant

We will expand the determinant using the first row. The first row has a zero, which simplifies our calculation:

\[ f(x) = 0 \cdot \text{(minor)} - (x^2 - \sin x) \cdot \left| \begin{array}{cc} \sin x - x^2 & 1 - 2x \\ 2 - \cos x & 0 \end{array} \right| + (\cos x - 2) \cdot \left| \begin{array}{cc} \sin x - x^2 & 0 \\ 2 - \cos x & 2x - 1 \end{array} \right| \]

### Step 2: Calculate the 2x2 Determinants

1. For the first determinant:

\[ \left| \begin{array}{cc} \sin x - x^2 & 1 - 2x \\ 2 - \cos x & 0 \end{array} \right| = 0 \cdot (\sin x - x^2) - (1 - 2x)(2 - \cos x) = -(1 - 2x)(2 - \cos x) \]

2. For the second determinant:

\[ \left| \begin{array}{cc} \sin x - x^2 & 0 \\ 2 - \cos x & 2x - 1 \end{array} \right| = (\sin x - x^2)(2x - 1) - 0 = (\sin x - x^2)(2x - 1) \]

### Step 3: Substitute Back into \( f(x) \)

Now substituting these back into the expression for \( f(x) \):

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

### Step 4: Simplify \( f(x) \)

Notice that both terms involve similar expressions. However, we can observe that if we expand and simplify, we will find that they cancel each other out.

After careful simplification, we find that:

\[ f(x) = 0 \]

### Step 5: Integrate \( f(x) \)

Now we need to integrate \( f(x) \):

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

where \( C \) is the constant of integration.

### Final Answer

Thus, the value of \( \int f(x) \, dx \) is:

\[ C \]