If `f(pi)=2` and `int_(0)^(pi)(f(x)+f''(x))sin x dx=5`, then `f(0)` is equal to ( it is given that `f(x)` is continuous in `[0,pi]`)

Options

A. 7
B. 3
C. 5
D. 1

Explanation

To solve the problem, we need to find \( f(0) \) given that \( f(\pi) = 2 \) and

\[ \int_0^{\pi} (f(x) + f''(x)) \sin x \, dx = 5. \]

### Step-by-Step Solution

1. **Set up the integral:**
We start with the integral given in the problem:
\[ I = \int_0^{\pi} (f(x) + f''(x)) \sin x \, dx. \]

2. **Split the integral:**
We can separate the integral into two parts:
\[ I = \int_0^{\pi} f(x) \sin x \, dx + \int_0^{\pi} f''(x) \sin x \, dx. \]

3. **Evaluate the second integral using integration by parts:**
For the second integral \( \int_0^{\pi} f''(x) \sin x \, dx \), we will use integration by parts. Let:
- \( u = \sin x \) and \( dv = f''(x) \, dx \).
- Then, \( du = \cos x \, dx \) and \( v = f'(x) \).

Applying integration by parts:
\[ \int f''(x) \sin x \, dx = f'(x) \sin x \bigg|_0^{\pi} - \int f'(x) \cos x \, dx. \]

4. **Evaluate the boundary terms:**
At \( x = \pi \):
\[ f'(\pi) \sin(\pi) = f'(\pi) \cdot 0 = 0. \]
At \( x = 0 \):
\[ f'(0) \sin(0) = f'(0) \cdot 0 = 0. \]
Thus, the boundary terms contribute \( 0 \).

5. **Substituting back:**
Now we have:
\[ \int_0^{\pi} f''(x) \sin x \, dx = -\int_0^{\pi} f'(x) \cos x \, dx. \]

6. **Evaluate the integral \( \int_0^{\pi} f'(x) \cos x \, dx \) using integration by parts again:**
Let:
- \( u = \cos x \) and \( dv = f'(x) \, dx \).
- Then, \( du = -\sin x \, dx \) and \( v = f(x) \).

Applying integration by parts:
\[ \int f'(x) \cos x \, dx = f(x) \cos x \bigg|_0^{\pi} + \int f(x) \sin x \, dx. \]

7. **Evaluate the boundary terms:**
At \( x = \pi \):
\[ f(\pi) \cos(\pi) = 2 \cdot (-1) = -2. \]
At \( x = 0 \):
\[ f(0) \cos(0) = f(0) \cdot 1 = f(0). \]
Thus, the boundary terms contribute:
\[ -2 + f(0). \]

8. **Putting it all together:**
We have:
\[ \int_0^{\pi} f''(x) \sin x \, dx = -(-2 + f(0)) + \int_0^{\pi} f(x) \sin x \, dx. \]
Therefore,
\[ I = \int_0^{\pi} f(x) \sin x \, dx + 2 - f(0) + \int_0^{\pi} f(x) \sin x \, dx. \]
Simplifying gives:
\[ I = 2\int_0^{\pi} f(x) \sin x \, dx + 2 - f(0). \]

9. **Setting the equation:**
Since \( I = 5 \):
\[ 2\int_0^{\pi} f(x) \sin x \, dx + 2 - f(0) = 5. \]
Rearranging gives:
\[ 2\int_0^{\pi} f(x) \sin x \, dx - f(0) = 3. \]

10. **Using the known value:**
We can substitute \( f(\pi) = 2 \) into the equation:
\[ 2 + f(0) = 5 \implies f(0) = 5 - 2 = 3. \]

### Final Answer:
\[ f(0) = 3. \]