To solve the problem, we start with the function given:
\[ f(x) = x + \int_0^1 x^2 z f(z) \, dz + \int_0^1 x z^2 f(z) \, dz \]
### Step 1: Simplify the function \( f(x) \)
We can factor out \( x^2 \) and \( x \) from the integrals:
1. Let \( \lambda = \int_0^1 z f(z) \, dz \)
2. Let \( \alpha = \int_0^1 z^2 f(z) \, dz \)
Thus, we can rewrite \( f(x) \):
\[ f(x) = x + x^2 \lambda + x \alpha \]
### Step 2: Combine terms
Now, we can combine the terms:
\[ f(x) = x + x^2 \lambda + x \alpha = x + x \alpha + x^2 \lambda = x(1 + \alpha) + x^2 \lambda \]
### Step 3: Analyze the limit given in the problem
We need to evaluate:
\[ \lim_{x \to \infty} \frac{f(x)}{ax^2 + 7x} = \frac{20}{119} \]
Substituting \( f(x) \):
\[ \lim_{x \to \infty} \frac{x(1 + \alpha) + x^2 \lambda}{ax^2 + 7x} \]
### Step 4: Factor out \( x^2 \)
Dividing the numerator and denominator by \( x^2 \):
\[ \lim_{x \to \infty} \frac{\frac{1 + \alpha}{x} + \lambda}{a + \frac{7}{x}} \]
As \( x \to \infty \), the terms \( \frac{1 + \alpha}{x} \) and \( \frac{7}{x} \) approach 0:
\[ \lim_{x \to \infty} \frac{\lambda}{a} = \frac{20}{119} \]
### Step 5: Solve for \( a \)
From the limit, we have:
\[ \frac{\lambda}{a} = \frac{20}{119} \]
This implies:
\[ \lambda = \frac{20}{119} a \]
### Step 6: Find \( \lambda \)
Now, we need to find \( \lambda \):
Recall that:
\[ \lambda = \int_0^1 z f(z) \, dz \]
Substituting \( f(z) \):
\[ f(z) = z + z^2 \lambda + z \alpha \]
Thus,
\[ \lambda = \int_0^1 z(z + z^2 \lambda + z \alpha) \, dz \]
Calculating the integrals:
1. \( \int_0^1 z^2 \, dz = \frac{1}{3} \)
2. \( \int_0^1 z^3 \, dz = \frac{1}{4} \)
So,
\[ \lambda = \int_0^1 z^2 \, dz + \lambda \int_0^1 z^3 \, dz + \alpha \int_0^1 z^2 \, dz \]
This gives:
\[ \lambda = \frac{1}{3} + \frac{\lambda}{4} + \frac{\alpha}{3} \]
### Step 7: Rearranging the equation
Rearranging gives:
\[ \lambda - \frac{\lambda}{4} = \frac{1}{3} + \frac{\alpha}{3} \]
Multiplying through by 12 to eliminate fractions:
\[ 12\lambda - 3\lambda = 4 + 4\alpha \]
Thus:
\[ 9\lambda = 4 + 4\alpha \]
### Step 8: Solve the equations
Now we have two equations:
1. \( \lambda = \frac{20}{119} a \)
2. \( 9\lambda = 4 + 4\alpha \)
Substituting \( \lambda \) from the first equation into the second:
\[ 9 \left(\frac{20}{119} a\right) = 4 + 4\alpha \]
This simplifies to:
\[ \frac{180}{119} a = 4 + 4\alpha \]
### Step 9: Solve for \( a \)
We can express \( \alpha \) in terms of \( a \) and substitute back to find \( a \). After solving the equations, we find:
\[ a = 4 \]
### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{4} \]
