We are asked to evaluate the iterated integral $\int_{0}^{2} \int_{-x}^{x} e^{-x^2} dy dx$.

1. Problem Description

We are asked to evaluate the iterated integral $\int_{0}^{2} \int_{-x}^{x} e^{-x^2} dy dx$.

2. Solution Steps

First, we evaluate the inner integral with respect to $y$:

$\int_{-x}^{x} e^{-x^2} dy = e^{-x^2} \int_{-x}^{x} dy = e^{-x^2} [y]_{-x}^{x} = e^{-x^2} (x - (-x)) = e^{-x^2} (2x) = 2xe^{-x^2}$

Now, we evaluate the outer integral with respect to $x$:

$\int_{0}^{2} 2xe^{-x^2} dx$

Let $u = -x^2$. Then $du = -2x dx$, so $-du = 2x dx$.

When $x = 0$, $u = -0^2 = 0$.

When $x = 2$, $u = -2^2 = -4$.

Thus, the integral becomes:

$\int_{0}^{-4} e^u (-du) = -\int_{0}^{-4} e^u du = \int_{-4}^{0} e^u du = [e^u]_{-4}^{0} = e^0 - e^{-4} = 1 - e^{-4} = 1 - \frac{1}{e^4}$

3. Final Answer

$1 - e^{-4}$