If `f(0)=0, f(3)=3 and f'(3)=4`, then the value of `int_(0)^(1)xf'' (3x)dx` is equal to

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{0}^{1} x f''(3x) \, dx \).

### Step 1: Change of Variables
We start by making a substitution. Let \( t = 3x \). Then, we have:
- When \( x = 0 \), \( t = 0 \)
- When \( x = 1 \), \( t = 3 \)
- The differential \( dx \) can be expressed as \( dx = \frac{dt}{3} \).

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we get:
\[ I = \int_{0}^{3} \left(\frac{t}{3}\right) f''(t) \left(\frac{dt}{3}\right) = \frac{1}{9} \int_{0}^{3} t f''(t) \, dt. \]

### Step 3: Integration by Parts
Now, we will use integration by parts on the integral \( \int t f''(t) \, dt \). Let:
- \( u = t \) (thus \( du = dt \))
- \( dv = f''(t) \, dt \) (thus \( v = f'(t) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int t f''(t) \, dt = t f'(t) \bigg|_{0}^{3} - \int f'(t) \, dt. \]

### Step 4: Evaluate the Boundary Terms
Now we evaluate \( t f'(t) \) at the limits:
\[ t f'(t) \bigg|_{0}^{3} = 3 f'(3) - 0 \cdot f'(0) = 3 f'(3). \]

### Step 5: Evaluate the Remaining Integral
The remaining integral \( \int f'(t) \, dt \) evaluates to:
\[ \int f'(t) \, dt = f(t) \bigg|_{0}^{3} = f(3) - f(0). \]

### Step 6: Combine Results
Putting it all together, we have:
\[ \int_{0}^{3} t f''(t) \, dt = 3 f'(3) - (f(3) - f(0)). \]
Substituting back into our expression for \( I \):
\[ I = \frac{1}{9} \left( 3 f'(3) - (f(3) - f(0)) \right). \]

### Step 7: Substitute Known Values
From the problem, we know:
- \( f(0) = 0 \)
- \( f(3) = 3 \)
- \( f'(3) = 4 \)

Substituting these values:
\[ I = \frac{1}{9} \left( 3 \cdot 4 - (3 - 0) \right) = \frac{1}{9} \left( 12 - 3 \right) = \frac{1}{9} \cdot 9 = 1. \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{1}. \]