Let `f(x)` be a differentiable function symmetric about `x=2`, then the value of `int_(0)^(4)cos(pix)f'(x)dx` is equal to________.

Explanation

To solve the problem, we need to evaluate the integral:

\[ I = \int_{0}^{4} \cos(\pi x) f'(x) \, dx \]

where \( f(x) \) is a differentiable function symmetric about \( x = 2 \).

### Step 1: Understand the symmetry of the function
Since \( f(x) \) is symmetric about \( x = 2 \), we have:
\[ f(2 - x) = f(2 + x) \]
This implies that:
\[ f(4 - x) = f(x) \]

### Step 2: Differentiate the symmetry condition
Differentiating both sides with respect to \( x \):
\[ f'(2 - x) \cdot (-1) = f'(2 + x) \cdot 1 \]
Thus, we can rewrite this as:
\[ f'(2 - x) = -f'(2 + x) \]
This means:
\[ f'(4 - x) = -f'(x) \]

### Step 3: Change of variable in the integral
Now, we can use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, let \( a = 4 \):
\[ I = \int_{0}^{4} \cos(\pi (4 - x)) f'(4 - x) \, dx \]
This simplifies to:
\[ I = \int_{0}^{4} \cos(4\pi - \pi x) f'(4 - x) \, dx \]

### Step 4: Simplify the cosine term
Using the cosine identity \( \cos(4\pi - \theta) = \cos(\theta) \):
\[ I = \int_{0}^{4} \cos(\pi x) f'(4 - x) \, dx \]

### Step 5: Substitute the expression for \( f'(4 - x) \)
From our earlier result, we know:
\[ f'(4 - x) = -f'(x) \]
Thus, we can substitute this into our integral:
\[ I = \int_{0}^{4} \cos(\pi x) (-f'(x)) \, dx = -\int_{0}^{4} \cos(\pi x) f'(x) \, dx \]
This gives us:
\[ I = -I \]

### Step 6: Solve for \( I \)
Adding \( I \) to both sides:
\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{4} \cos(\pi x) f'(x) \, dx = 0 \]