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To solve the definite integral 

\[
\int_{0}^{1} x^2 e^{-x^3} \, dx,
\]

we can use the substitution method.

### Step 1: Substitution
Let \( u = x^3 \). Then, differentiate both sides with respect to \( x \):

\[
du = 3x^2 \, dx \quad \Rightarrow \quad \frac{du}{3} = x^2 \, dx.
\]

### Step 2: Change of Limits
When \( x = 0 \), \( u = 0^3 = 0 \).

When \( x = 1 \), \( u = 1^3 = 1 \).

So, the limits of integration in terms of \( u \) are from 0 to 1.

### Step 3: Rewrite the Integral
Substitute into the integral:

\[
\int_{0}^{1} x^2 e^{-x^3} \, dx = \int_{0}^{1} e^{-u} \cdot \frac{du}{3}.
\]

This simplifies to:

\[
\frac{1}{3} \int_{0}^{1} e^{-u} \, du.
\]

### Step 4: Integrate
The integral of \( e^{-u} \) with respect to \( u \) is \( -e^{-u} \):

\[
\frac{1}{3} \left[-e^{-u}\right]_{0}^{1}.
\]

### Step 5: Evaluate the Definite Integral
Now, evaluate this expression at the limits 0 and 1:

\[
\frac{1}{3} \left[-e^{-1} - (-e^{0})\right] = \frac{1}{3} \left[-\frac{1}{e} + 1\right].
\]

So the final answer is:

\[
\frac{1}{3} \left(1 - \frac{1}{e}\right) = \frac{1}{3} - \frac{1}{3e}.
\]

This is the value of the definite integral.

Would you like any more details or explanations on any part of this process? 

Here are some relative questions for further practice:
1. How would you approach a definite integral where the exponent is a different function of \(x\)?
2. What would change in the solution if the integral was from \(1\) to \(2\) instead of \(0\) to \(1\)?
3. Can this method be used for any integral of the form \( \int x^n e^{-x^m} dx \)?
4. How does the substitution method simplify the integration process?
5. What if the exponent in the original integral was positive? How would the integration change?

**Tip:** When facing complex exponents or logarithmic expressions in integrals, substitution is often a powerful tool to simplify the problem.

Learn how to solve the definite integral ∫₀¹ x² e⁻ᵘ du using the substitution method. Step-by-step solution with detailed integration techniques. Suitable for college-level students and those interested in advanced calculus concepts.

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