Convolution of two exponential signals
Just in case if you look for a complete solution based on integration:

$$y(t)=\int_{-\infty}^{+\infty}e^{-4\tau}u(\tau)\times 6e^{-(t-\tau)}u(t-\tau)d\tau$$
Consider these facts:

Both step functions need to be equal to $1$ to have both terms (otherwise the corresponding term, and consequently $y(t)$, become zero). This implies:  


$\tau&gt;0$
$t-\tau&gt;0\Rightarrow \tau&lt;t$
From the two above, $t&gt;0$. Otherwise the integral is zero. 


We need $1$ and $2$ for the boundaries of integral. They say $\tau \in (0,t)$. Also we need $3$ to know where the output exists. It simply says the integral should be multiplied by $u(t)$. Now that we applied all the given information from the two step functions, we remove them and the integral becomes:
$$\begin{align}
y(t)&amp;=\left(\int_{0}^{t}e^{-4\tau}\times 6e^{-(t-\tau)}d\tau\right)u(t)\\[10pt]
&amp;=\left(6e^{-t}\int_{0}^{t}e^{-3\tau}d\tau\right)u(t)\\[10pt]
&amp;=\left(6e^{-t}\left(\frac{e^{-3\tau}}{-3}\biggr\vert^t_0\right)\right)u(t)\\[10pt]
&amp;=\left(-2e^{-t}\left(e^{-3t}-1\right)\right)u(t)\\[10pt]
&amp;=2(e^{-t}-e^{-4t})u(t)
\end{align}$$