The value of `int_-3^3(a x^5+b x^3+c x+k)dx` , where a, b, c and k are constants, depends only on

Options

A. a and k
B. a and b
C. a, b and c
D. k

Explanation

To solve the integral \( I = \int_{-3}^{3} (a x^5 + b x^3 + c x + k) \, dx \), we will evaluate the integral term by term and analyze how the constants \( a, b, c, \) and \( k \) affect the result.

### Step-by-Step Solution:

1. **Set up the integral**:
\[ I = \int_{-3}^{3} (a x^5 + b x^3 + c x + k) \, dx \]

2. **Break down the integral**:
We can separate the integral into individual terms:
\[ I = \int_{-3}^{3} a x^5 \, dx + \int_{-3}^{3} b x^3 \, dx + \int_{-3}^{3} c x \, dx + \int_{-3}^{3} k \, dx \]

3. **Evaluate each term**:
- For \( \int_{-3}^{3} a x^5 \, dx \):
\[ \int_{-3}^{3} x^5 \, dx = 0 \quad (\text{since } x^5 \text{ is an odd function}) \]
- For \( \int_{-3}^{3} b x^3 \, dx \):
\[ \int_{-3}^{3} x^3 \, dx = 0 \quad (\text{since } x^3 \text{ is an odd function}) \]
- For \( \int_{-3}^{3} c x \, dx \):
\[ \int_{-3}^{3} x \, dx = 0 \quad (\text{since } x \text{ is an odd function}) \]
- For \( \int_{-3}^{3} k \, dx \):
\[ \int_{-3}^{3} k \, dx = k \cdot (3 - (-3)) = k \cdot 6 = 6k \]

4. **Combine the results**:
Since the integrals of the odd functions are zero, we have:
\[ I = 0 + 0 + 0 + 6k = 6k \]

5. **Conclusion**:
The value of the integral \( I \) depends only on the constant \( k \).

### Final Answer:
The value of \( I \) depends only on \( k \).