Let `R to R ` be a differentiable functions such that its derivative f' is continuous and `f(pi) = - 6` . If F : `[0,pi] to R ` is defined by `F(x) int_(0)^(x) f(t) dt , and int_(0)^(x) f(t) dt and if int_(0)^(pi) (f'(x) +F(x))cos xdx = 2 `
then the value of f(0) is _________

Answer

To solve the problem, we will follow these steps systematically:

### Step 1: Understand the given information
We are given a differentiable function \( f: \mathbb{R} \to \mathbb{R} \) such that its derivative \( f' \) is continuous, and we know that \( f(\pi) = -6 \). We also have a function \( F \) defined as:
\[ F(x) = \int_0^x f(t) \, dt \]

### Step 2: Set up the integral equation
We need to evaluate the integral:
\[ \int_0^\pi (f'(x) + F(x)) \cos x \, dx = 2 \]

### Step 3: Break down the integral
We can split the integral into two parts:
\[ \int_0^\pi f'(x) \cos x \, dx + \int_0^\pi F(x) \cos x \, dx = 2 \]

### Step 4: Evaluate the first integral using integration by parts
For the first integral, we can use integration by parts. Let:
- \( u = \cos x \) so that \( du = -\sin x \, dx \)
- \( dv = f'(x) \, dx \) so that \( v = f(x) \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int_0^\pi f'(x) \cos x \, dx = \left[ f(x) \cos x \right]_0^\pi - \int_0^\pi f(x)(-\sin x) \, dx \]
Evaluating the boundary term:
\[ \left[ f(x) \cos x \right]_0^\pi = f(\pi) \cos(\pi) - f(0) \cos(0) = -6(-1) - f(0)(1) = 6 - f(0) \]
Thus, we have:
\[ \int_0^\pi f'(x) \cos x \, dx = 6 - f(0) + \int_0^\pi f(x) \sin x \, dx \]

### Step 5: Substitute back into the integral equation
Now we substitute this back into our split integral:
\[ 6 - f(0) + \int_0^\pi f(x) \sin x \, dx + \int_0^\pi F(x) \cos x \, dx = 2 \]
This simplifies to:
\[ \int_0^\pi F(x) \cos x \, dx + \int_0^\pi f(x) \sin x \, dx = 2 + f(0) - 6 \]
\[ \int_0^\pi F(x) \cos x \, dx + \int_0^\pi f(x) \sin x \, dx = f(0) - 4 \]

### Step 6: Evaluate \( \int_0^\pi F(x) \cos x \, dx \)
We can evaluate \( \int_0^\pi F(x) \cos x \, dx \) using integration by parts again. Let:
- \( u = F(x) \) so that \( du = f(x) \, dx \)
- \( dv = \cos x \, dx \) so that \( v = \sin x \)

Thus,
\[ \int_0^\pi F(x) \cos x \, dx = \left[ F(x) \sin x \right]_0^\pi - \int_0^\pi f(x) \sin x \, dx \]
Evaluating the boundary term:
\[ \left[ F(x) \sin x \right]_0^\pi = F(\pi) \sin(\pi) - F(0) \sin(0) = 0 - 0 = 0 \]
So we have:
\[ \int_0^\pi F(x) \cos x \, dx = -\int_0^\pi f(x) \sin x \, dx \]

### Step 7: Substitute back into the equation
Substituting this into our previous equation gives:
\[ -\int_0^\pi f(x) \sin x \, dx + \int_0^\pi f(x) \sin x \, dx = f(0) - 4 \]
This simplifies to:
\[ 0 = f(0) - 4 \]
Thus, we find:
\[ f(0) = 4 \]

### Final Answer
The value of \( f(0) \) is \( \boxed{4} \).