Delta function!

With $\delta(x^2 - 2) \; = \; \tfrac{1}{2\sqrt{2}}\big[\delta(x-\sqrt{2}) + \delta(x + \sqrt{2})\big]$ so that $\int_\mathbf{R} f(x) \delta(x^2 - 2)\,dx \; = \; \tfrac{1}{2\sqrt{2}}\big[f(\sqrt{2}) + f(-\sqrt{2})\big]$ we see that $\int_\mathbb{R} e^{-x^2}\delta(x^2-2)\,dx \; = \; \frac{e^{-2}}{\sqrt{2}}$ making the answer $1 + 2 + 1 + 2 + 2 +2= \boxed{10}$ .

Change of variables approach:

First, to avoid minor complications later, use the symmetry property of the integrand (i.e. $f(-x) = f(x)$ ) to rewrite the integral as $\displaystyle\int_{-\infty}^{\infty} \! e^{-x^2} \delta(x^2 - 2) \, \, \mathrm{d}x = 2 \displaystyle\int_{0}^{\infty} \! e^{-x^2} \delta(x^2 - 2) \, \, \mathrm{d}x$

Now, apply a change of variables. $y \equiv x^2 \Rightarrow \mathrm{d}y = 2x \, \mathrm{d}x \, \, \mathrm{or} \, \, \frac{\mathrm{d}y}{2\sqrt{y}} = \mathrm{d}x$

Simple substitution gives $2 \displaystyle\int_{0}^{\infty} \! e^{-x^2} \delta(x^2 - 2) \, \, \mathrm{d}x = \displaystyle\int_{0}^{\infty} \! \tfrac{1}{\sqrt{y}}e^{-y} \delta(y - 2) \, \, \mathrm{d}y$

Finally, using the basic properties of the delta function, we can directly obtain $\displaystyle\int_{0}^{\infty} \! \tfrac{1}{\sqrt{y}}e^{-y} \delta(y - 2) \, \, \mathrm{d}y= \frac{e^{-2}}{\sqrt{2}}$

giving $A = 1$ , $B = 2$ , $C = 1$ , $D = 2$ , and $F = 2$ for a result of $\boxed{10}$ .