\( \int e ^ { - \frac { x ^ { 2 } } { 2 } } x ^ { 3 } d x \)

To evaluate the integral \[ I = \int e^{-\frac{x^2}{2}} x^3 \, dx, \] we can use integration by parts. We will choose: - \( u = x^3 \) which gives \( du = 3x^2 \, dx \), - \( dv = e^{-\frac{x^2}{2}} \, dx \). To find \( v \), we need to integrate \( dv \): \[ v = \int e^{-\frac{x^2}{2}} \, dx. \] The integral \( \int e^{-\frac{x^2}{2}} \, dx \) does not have a simple closed form, but we can express it in terms of the error function or use it in our integration by parts. Now, we apply integration by parts: \[ I = uv - \int v \, du. \] Substituting our expressions, we have: \[ I = x^3 \int e^{-\frac{x^2}{2}} \, dx - \int \left( \int e^{-\frac{x^2}{2}} \, dx \right) (3x^2) \, dx. \] This integral is still complex, so we can simplify our approach by recognizing a pattern. We can differentiate the integral of the Gaussian function. Instead, we can use the fact that: \[ \int e^{-\frac{x^2}{2}} x^n \, dx \] can be computed using the recurrence relation for the moments of the Gaussian distribution. Specifically, we can use the relation: \[ \int e^{-\frac{x^2}{2}} x^n \, dx = \begin{cases} 0 & \text{if } n \text{ is odd}, \\ (n-1) \int e^{-\frac{x^2}{2}} x^{n-2} \, dx & \text{if } n \text{ is even}. \end{cases} \] Since \( n = 3 \) is odd, we have: \[ \int e^{-\frac{x^2}{2}} x^3 \, dx = 0. \] Thus, the final result is: \[ \int e^{-\frac{x^2}{2}} x^3 \, dx = 0. \]