If `int_(1)^(2) {K^(2) + (4-4K)x+4x^(3)} dx le 12`, then which one of the following is correct ?

Options

A. `K = 3`
B. `0 le K le 3`
C. `K le 4`
D. `K = 0`

Explanation

To solve the problem, we need to evaluate the definite integral and find the values of \( K \) such that the integral is less than or equal to 12.

### Step-by-Step Solution:

1. **Set Up the Integral**:
We start with the integral:
\[ \int_{1}^{2} \left( K^2 + (4 - 4K)x + 4x^3 \right) dx \]

2. **Integrate Term by Term**:
We can integrate each term separately:
- The integral of \( K^2 \) from 1 to 2 is:
\[ K^2 \int_{1}^{2} dx = K^2 [x]_{1}^{2} = K^2 (2 - 1) = K^2 \]
- The integral of \( (4 - 4K)x \) is:
\[ (4 - 4K) \int_{1}^{2} x \, dx = (4 - 4K) \left[ \frac{x^2}{2} \right]_{1}^{2} = (4 - 4K) \left( \frac{4}{2} - \frac{1}{2} \right) = (4 - 4K) \cdot \frac{3}{2} = 6 - 6K \]
- The integral of \( 4x^3 \) is:
\[ 4 \int_{1}^{2} x^3 \, dx = 4 \left[ \frac{x^4}{4} \right]_{1}^{2} = [x^4]_{1}^{2} = 16 - 1 = 15 \]

3. **Combine the Results**:
Now, we combine all the results from the integrations:
\[ K^2 + (6 - 6K) + 15 \]
Simplifying this gives:
\[ K^2 - 6K + 21 \]

4. **Set Up the Inequality**:
We need to satisfy the inequality:
\[ K^2 - 6K + 21 \leq 12 \]

5. **Rearranging the Inequality**:
Rearranging gives:
\[ K^2 - 6K + 9 \leq 0 \]

6. **Factoring the Quadratic**:
The quadratic can be factored as:
\[ (K - 3)^2 \leq 0 \]

7. **Finding the Solution**:
The only solution to this inequality is:
\[ K - 3 = 0 \implies K = 3 \]

### Conclusion:
Thus, the value of \( K \) that satisfies the given condition is \( K = 3 \).