The convolution product $y(t) = u(t) * h(t)$ is given by:
$$ \newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \bbx{y(t) = \int_{-\infty}^{+\infty}u(\tau)h(t-\tau) \ d\tau}$$
I must calculate it with with $u(t) = \exp(2t)\nu(-t)$ and $h(t) = \nu(t-3)$, $\nu$ being the Heaviside function. Here is my attempt:
$$\int_{-\infty}^{+\infty}u(\tau)h(t-\tau) \ d\tau$$ $$=\int_{-\infty}^{+\infty}\exp(2\tau)\nu(-\tau)\nu(t-\tau-3) \ d\tau$$ $$=\int_{-\infty}^{0}\exp(2\tau)\nu(t-\tau-3) \ d\tau$$ $$=\int_{-\infty}^{t-3}\exp(2\tau) \ d\tau$$ $$=\frac{\exp(2t-6)}{2}$$ I know the right answer is $\frac{1}{2}$, but I don't understand my mistake. My course uses $\int_{-\infty}^{+\infty}h(\tau)u(t-\tau) \ d\tau$ instead of $\int_{-\infty}^{+\infty}u(\tau)h(t-\tau) \ d\tau$ to get that result. Thank you in advance.
You write that $1/2$ must be the expected result. However I doubt this is correct.
Essentially you do the proper approach and it looks good to me. So I would say your solution is correct.
Simply specking only with trivial arguments one can achieve the constant function as result of the convolution. Imagine $u(t) == 1$ and $h(t)$ is a gaussian function with area of 1/2. If one has a Heavyside function in one of the convolution arguments I doubt it is even possible to select a suitable other function so that the convolution of both is the constant function.
The convolution is commutative, so the order of functions does not matter. One can exchange $u$ and $h$. This is proven by a simple subsitution in the integral.