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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.

Learn how to calculate the average value of the function f(x) = 3x^2 e^{2x} over the interval [0, 1]. This problem involves calculus concepts such as integration by parts and the evaluation of definite integrals. Suitable for advanced undergraduate students and learners interested in calculus.

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