Can convolution be written as $(f \ast g)(t)=\int_{\tau = -\infty}^{\infty}g(t-\tau)f(\tau)\,\mathrm{d}\tau$? In every textbook that I read the definition of convolution is almost always given by 
$$(f \ast g)(t)=\int_{\tau = -\infty}^{\infty} f(\tau)\,g(t-\tau)\,\mathrm{d}\tau\tag{1}$$ 
apart from a significant number of definitions having a factor of $1/\sqrt{2 \pi}$ out front.
In the course that I am taking I am required to use the definition given by $(1)$.
So the question that I have is; can I re-write the $(1)$ as 
$$(f \ast g)(t)=\int_{\tau = -\infty}^{\infty}g(t-\tau)f(\tau)\,\mathrm{d}\tau\tag{2}?$$
I have simply switched the order of the factors in the integrand and since muliplication is commutative by my logic this equation $(2)$ is plausible.
The reason I asked this question is because it just makes more sense for several reasons: $f(\tau)$ is usually a signal and $g(t- \tau)$ is the resolution function that 'sweeps' across the signal $f(\tau)$; so the statement '$f$ convolves with $g$' or 'convolution of $f$ with $g$' now makes sense intuitively. Also, $f$ comes before $g$ alphabetically so it's easier to remember the formula $(2)$ (provided it is indeed correct). 
I thought carefully before asking this, as the answer may be a trivial 'yes of course! STUPID question'. But, I have only just started learning convolution so it is not entirely obvious to me.

So to summarise,
Could someone please confirm that equation $(2)$ is correct (and if not why)? 
Thanks.

Yes, this follows from change of variables.  Consider the case when $f(t) = g(t) = 0$ if $t < 0$.  Then, for each $t \geq 0$, using the substitution $s = t - \tau$, we obtain
\begin{equation*}
(f*g)(t) = \int_{0}^{t} f(\tau)g(t - \tau) \, d\tau = \int_{0}^{t} f(t - s) g(s) \, ds.
\end{equation*}
The same idea works in the general case when the integral is an improper (Riemann) integral (or a Lebesgue integral).
