...but if the Heaviside function you mention is what I said, then the function $\,e^{-2t}u(t)\,$ is zero on the negative reals and thus $\int e^{-2t}u(t)\,dt=\int e^{-2t}\,dt=-\frac{1}{2}e^{-2t}+C\,,\,\,C=\,\text{constant}$ only when $\,t> 0\,$
I think that what you really had is something like $\,\,\displaystyle{\int_0^t e^{-2x}u(x)\,dx}\,\,$ , which gives exactly what you said, so that the integral isn't zero only if $\,t>0\,$ , but without the limits the integral is indefinite and you can't get what you say you did.