If `int _(0)^(2)(3x ^(2) -3x +1) cos (x ^(3) -3x ^(2)+4x -2) dx = a sin (b),` where a and b are positive integers. Find the value of `(a+b)/2.`

Answer

To solve the integral \( I = \int_{0}^{2} (3x^2 - 3x + 1) \cos(x^3 - 3x^2 + 4x - 2) \, dx \), we can use the property of definite integrals which states that:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

In our case, \( a = 0 \) and \( b = 2 \). Thus, we can write:

\[ I = \int_{0}^{2} (3(2-x)^2 - 3(2-x) + 1) \cos((2-x)^3 - 3(2-x)^2 + 4(2-x) - 2) \, dx \]

### Step 1: Substitute \( 2 - x \) into the integral

Calculating \( 3(2-x)^2 - 3(2-x) + 1 \):

\[ 3(2-x)^2 = 3(4 - 4x + x^2) = 12 - 12x + 3x^2 \]
\[ -3(2-x) = -6 + 3x \]
Combining these gives:

\[ 3(2-x)^2 - 3(2-x) + 1 = (12 - 12x + 3x^2) + (-6 + 3x) + 1 = 3x^2 - 9x + 7 \]

### Step 2: Substitute into the cosine term

Now, we calculate \( (2-x)^3 - 3(2-x)^2 + 4(2-x) - 2 \):

\[ (2-x)^3 = 8 - 12x + 6x^2 - x^3 \]
\[ -3(2-x)^2 = -3(4 - 4x + x^2) = -12 + 12x - 3x^2 \]
\[ 4(2-x) = 8 - 4x \]
Combining these gives:

\[ (2-x)^3 - 3(2-x)^2 + 4(2-x) - 2 = (8 - 12x + 6x^2 - x^3) + (-12 + 12x - 3x^2) + (8 - 4x) - 2 \]
\[ = 8 - 12 + 8 - 2 + (-x^3 + 3x^2 - 4x) = -x^3 + 3x^2 + 0x + 0 = -x^3 + 3x^2 \]

### Step 3: Combine the two parts

Now we can write:

\[ I = \int_{0}^{2} (3x^2 - 9x + 7) \cos(-x^3 + 3x^2) \, dx \]

Using the property of cosine, \( \cos(-\theta) = \cos(\theta) \):

\[ I = \int_{0}^{2} (3x^2 - 9x + 7) \cos(x^3 - 3x^2) \, dx \]

### Step 4: Add the two integrals

Now we have:

\[ 2I = \int_{0}^{2} \left( (3x^2 - 3x + 1) + (3x^2 - 9x + 7) \right) \cos(x^3 - 3x^2) \, dx \]
\[ = \int_{0}^{2} (6x^2 - 12x + 8) \cos(x^3 - 3x^2) \, dx \]

### Step 5: Factor out constants

Factor out 2:

\[ = 2 \int_{0}^{2} (3x^2 - 6x + 4) \cos(x^3 - 3x^2) \, dx \]

### Step 6: Change of variable

Let \( t = x^3 - 3x^2 + 4x - 2 \). Then:

\[ dt = (3x^2 - 6x + 4) \, dx \]

Thus, the integral becomes:

\[ I = \int_{t_0}^{t_2} \cos(t) \, dt \]
where \( t_0 = 0 \) and \( t_2 = 2 \).

### Step 7: Evaluate the integral

The integral of \( \cos(t) \) is \( \sin(t) \):

\[ I = \left[ \sin(t) \right]_{0}^{2} = \sin(2) - \sin(0) = \sin(2) \]

### Step 8: Relate to the form \( a \sin(b) \)

From the problem, we have \( I = a \sin(b) \). Here, \( a = 1 \) and \( b = 2 \).

### Final Calculation

Now, we need to find \( \frac{a + b}{2} \):

\[ \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \]

Thus, the final answer is:

\[ \boxed{2} \]