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The given problem is a double integral:

\[
\int_{-1}^{1} \left( \int_0^y xye^{x^2} \, dx \right) dy
\]

### Step-by-step solution:

1. **Inner Integral**: We first evaluate the inner integral with respect to \( x \), while treating \( y \) as a constant:

\[
I_1 = \int_0^y xye^{x^2} \, dx
\]

To solve this, we perform substitution. Let \( u = x^2 \), so \( du = 2x \, dx \), or \( dx = \frac{du}{2x} \).

Thus, the integral becomes:

\[
I_1 = y \int_0^y xe^{x^2} \, dx = \frac{y}{2} \int_0^{y^2} e^u \, du
\]

The integral of \( e^u \) is simply \( e^u \). Substituting the limits:

\[
I_1 = \frac{y}{2} \left( e^{y^2} - e^0 \right) = \frac{y}{2} \left( e^{y^2} - 1 \right)
\]

2. **Outer Integral**: Now we need to integrate this result with respect to \( y \):

\[
I_2 = \int_{-1}^{1} \frac{y}{2} \left( e^{y^2} - 1 \right) \, dy
\]

This can be broken into two parts:

\[
I_2 = \frac{1}{2} \int_{-1}^{1} y e^{y^2} \, dy - \frac{1}{2} \int_{-1}^{1} y \, dy
\]

- The second integral, \( \int_{-1}^{1} y \, dy \), is zero because it's an odd function over a symmetric interval.
- For the first integral, perform substitution \( u = y^2 \), so \( du = 2y \, dy \). Changing limits, when \( y = -1 \), \( u = 1 \), and when \( y = 1 \), \( u = 1 \):

\[
\int_{-1}^{1} y e^{y^2} \, dy = 0
\]

Thus, the entire integral evaluates to:

\[
I_2 = 0
\]

### Final Answer:

\[
\int_{-1}^{1} \left( \int_0^y xye^{x^2} \, dx \right) dy = 0
\]

---

Would you like further clarification or more details on any specific step?

Here are five related questions to deepen your understanding:

1. How does the substitution method simplify integrals involving exponential functions?
2. Why is the integral of an odd function over a symmetric interval zero?
3. Can you apply this method to other integrals involving products of \( e^{x^2} \) and polynomials?
4. How would the result change if the limits of integration were different?
5. What is the significance of the exponential function in this context?

**Tip**: When dealing with complex integrals, breaking them into smaller, manageable parts often simplifies the solution.

Evaluate the double integral: \( \int_{-1}^{1} \left( \int_0^y xye^{x^2} \, dx \right) dy \)

Learn how to evaluate the double integral \( \int_{-1}^{1} \left( \int_0^y xye^{x^2} \, dx \right) dy \). This problem uses key calculus concepts, including substitution in integration, the exponential function, and the odd function theorem. Suitable for college-level students studying calculus.

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