Evaluate the definite integral. \[ \int_{0}^{2} 4 \mathrm{xe}^{-x^{2}} \mathrm{dx}=[?] \] Round your answer to the nearest thousandth.

To evaluate the definite integral \[ \int_{0}^{2} 4 \mathrm{xe}^{-x^{2}} \mathrm{dx}, \] we can use a substitution. Let \( u = -x^2 \), then \( du = -2x \, dx \), or \( dx = -\frac{du}{2x} \). In terms of \( u \), we also need to change the limits of integration: When \( x = 0 \), \( u = 0 \). When \( x = 2 \), \( u = -4 \). Now substituting into the integral, we get \[ \int_{0}^{2} 4 x e^{-x^2} \, dx = \int_{0}^{-4} 4x e^u \left(-\frac{du}{2x}\right) = -2 \int_{0}^{-4} e^u \, du. \] Next, we reverse the limits of integration and change the sign: \[ -2 \int_{0}^{-4} e^u \, du = 2 \int_{-4}^{0} e^u \, du. \] Now, we can compute the integral of \( e^u \): \[ 2 \left[ e^u \right]_{-4}^{0} = 2 \left( e^0 - e^{-4} \right) = 2 \left( 1 - e^{-4} \right). \] Calculating \( e^{-4} \) (approximately 0.018316) gives us: \[ 2 \left( 1 - 0.018316 \right) = 2 \times 0.981684 \approx 1.963368. \] Rounding to the nearest thousandth, the final result is \[ \int_{0}^{2} 4 \mathrm{xe}^{-x^{2}} \mathrm{dx} \approx 1.963. \]