To find the average value of the function $f(x) = 3x^2 e^{2x}$ on the interval $[0, 1]$, we use the formula for the average value of a function $f$ over the interval $[a, b]$:

$\text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx$

In this case, $a = 0$ and $b = 1$. Thus, we have:

$\text{Average value} = \frac{1}{1-0} \int_0^1 3x^2 e^{2x} \, dx = \int_0^1 3x^2 e^{2x} \, dx$

To solve this integral, we will use integration by parts. Let $u = x^2$ and $dv = 3e^{2x} \, dx$. Then, we have $du = 2x \, dx$ and $v = \frac{3}{2} e^{2x}$.

Using integration by parts formula $\int u \, dv = uv - \int v \, du$:

$\int 3x^2 e^{2x} \, dx = \left( x^2 \cdot \frac{3}{2} e^{2x} \right) - \int \left( \frac{3}{2} e^{2x} \cdot 2x \right) dx$

$= \frac{3}{2} x^2 e^{2x} - 3 \int x e^{2x} \, dx$

Now, we need to integrate $x e^{2x}$. Again, we use integration by parts. Let $u = x$ and $dv = e^{2x} \, dx$. Then, $du = dx$ and $v = \frac{1}{2} e^{2x}$:

$\int x e^{2x} \, dx = \left( x \cdot \frac{1}{2} e^{2x} \right) - \int \left( \frac{1}{2} e^{2x} \cdot 1 \right) dx$

$= \frac{1}{2} x e^{2x} - \frac{1}{2} \int e^{2x} \, dx$

$= \frac{1}{2} x e^{2x} - \frac{1}{2} \cdot \frac{1}{2} e^{2x} = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}$

Putting it all together:

$\int 3x^2 e^{2x} \, dx = \frac{3}{2} x^2 e^{2x} - 3 \left( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \right)$

$= \frac{3}{2} x^2 e^{2x} - \frac{3}{2} x e^{2x} + \frac{3}{4} e^{2x}$

Now, we evaluate this expression from $0$ to $1$:

$\left[ \frac{3}{2} x^2 e^{2x} - \frac{3}{2} x e^{2x} + \frac{3}{4} e^{2x} \right]_0^1$

$= \left( \frac{3}{2} \cdot 1^2 \cdot e^{2 \cdot 1} - \frac{3}{2} \cdot 1 \cdot e^{2 \cdot 1} + \frac{3}{4} e^{2 \cdot 1} \right) - \left( \frac{3}{2} \cdot 0^2 \cdot e^{2 \cdot 0} - \frac{3}{2} \cdot 0 \cdot e^{2 \cdot 0} + \frac{3}{4} e^{2 \cdot 0} \right)$

$= \left( \frac{3}{2} e^2 - \frac{3}{2} e^2 + \frac{3}{4} e^2 \right) - \left( \frac{3}{4} \cdot 1 \right)$

$= \frac{3}{4} e^2 - \frac{3}{4}$

Using a calculator to approximate the values:

$e^2 \approx 7.3891$

$\text{Average value} = \frac{3}{4} \cdot 7.3891 - \frac{3}{4} \approx 5.541825 - 0.75 = 4.791825$

Rounding to three decimal places:

$\text{Average value} \approx 4.792$

The average value of the function $f(x) = 3x^2 e^{2x}$ on the interval $[0, 1]$ is approximately $\boxed{4.792}$.

Would you like more details or have any questions about this problem?

Here are five questions you can ask:

1. How do you use integration by parts?
2. What is the substitution method in integration?
3. How do you find the area under a curve using definite integrals?
4. What are the properties of the exponential function $e^x$?
5. How do you solve integrals involving polynomials and exponentials?

Tip: When dealing with integrals that combine polynomials and exponential functions, integration by parts is often a useful technique.