Let `inte^(x).x^(2)dx=f(x)e^(x)+C` (where, C is the constant of integration). The range of f(x) as `x in R` is `[a, oo)`. The value of `(a)/(4)` is

Answer

To solve the problem, we need to analyze the given integral and find the function \( f(x) \) such that the range of \( f(x) \) is \([a, \infty)\). We will then determine the value of \( \frac{a}{4} \).

### Step-by-Step Solution:

1. **Start with the given integral**:
\[ \int e^x x^2 \, dx = f(x) e^x + C \]

2. **Use integration by parts**:
We will apply integration by parts where we let:
- \( u = x^2 \) and \( dv = e^x \, dx \)
- Then, \( du = 2x \, dx \) and \( v = e^x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int e^x x^2 \, dx = x^2 e^x - \int e^x (2x) \, dx \]

3. **Apply integration by parts again**:
Now we need to integrate \( \int e^x (2x) \, dx \):
- Let \( u = 2x \) and \( dv = e^x \, dx \)
- Then, \( du = 2 \, dx \) and \( v = e^x \)

Applying integration by parts again:
\[ \int e^x (2x) \, dx = 2x e^x - \int 2 e^x \, dx = 2x e^x - 2 e^x \]

4. **Combine the results**:
Substitute back into the equation:
\[ \int e^x x^2 \, dx = x^2 e^x - (2x e^x - 2 e^x) = x^2 e^x - 2x e^x + 2 e^x \]
Simplifying gives:
\[ \int e^x x^2 \, dx = e^x (x^2 - 2x + 2) + C \]

5. **Identify \( f(x) \)**:
From the integral, we can identify:
\[ f(x) = x^2 - 2x + 2 \]

6. **Determine the range of \( f(x) \)**:
The function \( f(x) = x^2 - 2x + 2 \) is a quadratic function. To find its minimum value, we can complete the square:
\[ f(x) = (x - 1)^2 + 1 \]
The minimum value occurs at \( x = 1 \):
\[ f(1) = (1 - 1)^2 + 1 = 1 \]

Therefore, the range of \( f(x) \) is \([1, \infty)\).

7. **Find \( a \)**:
From the range \([a, \infty)\), we have \( a = 1 \).

8. **Calculate \( \frac{a}{4} \)**:
\[ \frac{a}{4} = \frac{1}{4} = 0.25 \]

### Final Answer:
The value of \( \frac{a}{4} \) is \( 0.25 \).