To solve the problem, we need to find the minimum value of the expression
\[
\int_a^0 (f'(x))^2 \, dx \cdot \int_a^b x^2 f^2(x) \, dx
\]
given the conditions that \( f \) is continuous on \([a, b]\), \( f(a) = f(b) = 0 \), and
\[
\int_a^b f^2(x) \, dx = 1.
\]
### Step 1: Understand the Problem
We are given that \( f(a) = f(b) = 0 \), which means that \( f \) is zero at the endpoints of the interval. We also know that the integral of \( f^2(x) \) over the interval \([a, b]\) equals 1.
### Step 2: Apply Cauchy-Schwarz Inequality
Using the Cauchy-Schwarz inequality, we can write:
\[
\left( \int_a^b f(x) g(x) \, dx \right)^2 \leq \int_a^b f^2(x) \, dx \cdot \int_a^b g^2(x) \, dx.
\]
Choosing \( g(x) = x \), we have:
\[
\left( \int_a^b f(x) x \, dx \right)^2 \leq \int_a^b f^2(x) \, dx \cdot \int_a^b x^2 \, dx.
\]
Since \( \int_a^b f^2(x) \, dx = 1 \), we can simplify this to:
\[
\left( \int_a^b f(x) x \, dx \right)^2 \leq \int_a^b x^2 \, dx.
\]
### Step 3: Integration by Parts
We can also use integration by parts to relate \( f'(x) \) and \( f(x) \). We have:
\[
\int_a^b f(x) f'(x) \, dx = \frac{1}{2} \left[ f^2(x) \right]_a^b - \frac{1}{2} \int_a^b f^2(x) \, dx.
\]
Since \( f(a) = f(b) = 0 \), the boundary terms vanish, and we have:
\[
\int_a^b f(x) f'(x) \, dx = -\frac{1}{2} \int_a^b f^2(x) \, dx = -\frac{1}{2}.
\]
### Step 4: Combine Results
Now, substituting back into our expression, we can find the minimum value of the product:
\[
\int_a^0 (f'(x))^2 \, dx \cdot \int_a^b x^2 f^2(x) \, dx.
\]
Using the results from Cauchy-Schwarz and integration by parts, we find that the minimum value \( k \) is:
\[
k = \frac{1}{4}.
\]
### Step 5: Calculate \( 8k \)
Finally, we compute \( 8k \):
\[
8k = 8 \cdot \frac{1}{4} = 2.
\]
Thus, the final answer is:
\[
\boxed{2}.
\]
