Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 3

For $f(x)=x^3{e}^{-3x^2/2}/3$, the product rule gives $f^{\prime }(x)=x^2(1-x^2)e^{-3x^2/2}$, so the stationary points occur at $x=-1,0,1$. Its sign is negative for $|x|>1$ and positive for $0<|x|<1$. ConsequentlyThey are also the global extrema, since the function tends to zero at both infinities and is a monotone function on the intervals just identified. At zero the derivative does not change sign. Since $f(x)=x^3/3+O(x^5)$, $(0,0)$ is a stationary inflection point, not an extremum.

The function is odd, with a horizontal tangent line at the origin and tails approaching zero from below on the negative side and above on the positive side. It decreases to its negative minimum at $x=-1$, rises through the stationary inflection point at zero to its positive maximum at $x=1$, then decreases toward zero.

The Gaussian decay makes the integral absolutely convergent, so odd symmetry gives $\boxed{{\int }_{-\infty }^{\infty }f(x)dx=0}$. On the positive half-line substitute $u=3x^2/2$, so $x^{3}dx=2udu/9$. ThusThe last integral is $\Gamma (2)=1$, an instance of the odd Gaussian moments on a half-line.