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.