If `f (x) = k,` where k is constant then `int k dx =0`

Answer

To solve the problem, we need to evaluate the integral of a constant function \( f(x) = k \), where \( k \) is a constant. The statement claims that the integral of \( k \) with respect to \( x \) is equal to 0, which we will verify step by step.

### Step-by-Step Solution:

1. **Identify the integral**: We start with the integral of the constant \( k \):
\[ \int k \, dx \]

2. **Factor out the constant**: Since \( k \) is a constant, we can factor it out of the integral:
\[ \int k \, dx = k \int 1 \, dx \]

3. **Integrate \( 1 \)**: The integral of \( 1 \) with respect to \( x \) is:
\[ \int 1 \, dx = x + C \]
where \( C \) is the constant of integration.

4. **Combine the results**: Now we substitute back into our equation:
\[ k \int 1 \, dx = k(x + C) = kx + kC \]

5. **Conclusion**: The result of the integral is:
\[ \int k \, dx = kx + C \]
This shows that the integral of a constant \( k \) is not equal to zero but rather \( kx + C \).

### Final Result:
Thus, the statement that \( \int k \, dx = 0 \) is false. The correct result is:
\[ \int k \, dx = kx + C \]