The value of `int_(-3)^(3)(ax^()+bx^(3)+cx+k)dx`, where a,b,c,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 \( \int_{-3}^{3} (ax + bx^3 + cx + k) \, dx \), we will follow these steps:

### Step 1: Break down the integral
We can separate the integral into individual parts:
\[ \int_{-3}^{3} (ax + bx^3 + cx + k) \, dx = \int_{-3}^{3} ax \, dx + \int_{-3}^{3} bx^3 \, dx + \int_{-3}^{3} cx \, dx + \int_{-3}^{3} k \, dx \]

### Step 2: Evaluate each integral
1. **For \( \int_{-3}^{3} ax \, dx \)**:
\[ \int ax \, dx = \frac{a x^2}{2} \Big|_{-3}^{3} = \frac{a(3^2)}{2} - \frac{a(-3^2)}{2} = \frac{9a}{2} - \frac{9a}{2} = 0 \]

2. **For \( \int_{-3}^{3} bx^3 \, dx \)**:
\[ \int bx^3 \, dx = \frac{b x^4}{4} \Big|_{-3}^{3} = \frac{b(3^4)}{4} - \frac{b(-3^4)}{4} = \frac{81b}{4} - \frac{81b}{4} = 0 \]

3. **For \( \int_{-3}^{3} cx \, dx \)**:
\[ \int cx \, dx = \frac{c x^2}{2} \Big|_{-3}^{3} = \frac{c(3^2)}{2} - \frac{c(-3^2)}{2} = \frac{9c}{2} - \frac{9c}{2} = 0 \]

4. **For \( \int_{-3}^{3} k \, dx \)**:
\[ \int k \, dx = kx \Big|_{-3}^{3} = k(3) - k(-3) = 3k + 3k = 6k \]

### Step 3: Combine the results
Now, we combine the results of all the integrals:
\[ \int_{-3}^{3} (ax + bx^3 + cx + k) \, dx = 0 + 0 + 0 + 6k = 6k \]

### Conclusion
The value of the integral depends only on \( k \).