Find autocorrelation function of a signal
The variable \$T\$ is the independent variable of the autocorrelation function. So

$$R(T)=\int_{-\infty}^{\infty}g(t)g(t+T)dt$$

is the autocorrelation function. Substituting the given \$g(t)\$ gives

$$R(T)=e^{-2T}\int_{\max\{0,-T\}}^{\infty}e^{-4t}dt=e^{-2T}\frac14 e^{-4\max\{0,-T\}}=\frac14 e^{-2|T|}\tag{1}$$

which is a symmetric function with its maximum at \$T=0\$, as it should be the case for an autocorrelation function.

EDIT:

And now with all details of the integration:

$$g(t)g(t+T)=e^{-2t}u(t)e^{-2(t+T)}u(t+T)=e^{-2T}e^{-4t}u(t)u(t+T)$$

Now look at the product \$u(t)u(t+T)\$:

$$u(t)u(t+T)=\begin{cases}u(t),&amp;T&gt;0\\
u(t+T),&amp;T&lt;0\end{cases}$$

(draw \$u(t)\$ and \$u(t+T)\$ if you don't see this). So we get

$$T&gt;0:\quad e^{-2T}\int_{-\infty}^{\infty}e^{-4t}u(t)dt=e^{-2T}\int_0^{\infty}e^{-4t}dt=e^{-2T}\left(-\frac14\right)e^{-4t}\Big|_0^{\infty}=\frac14 e^{-2T}$$

$$T&lt;0:\quad e^{-2T}\int_{-\infty}^{\infty}e^{-4t}u(t+T)dt=e^{-2T}\int_{-T}^{\infty}e^{-4t}dt=e^{-2T}\left(-\frac14\right)e^{-4t}\Big|_{-T}^{\infty}=\frac14 e^{2T}$$

We can combine these two expressions if we use \$e^{-2|T|}\$, which results in Eq. (1) as given above.