Consider a system with the components $A,B,C$. The lifetime of a single component is noted by the independent random variables $T_k, k \in \{A,B,C\}$. Each of the random variables $T_k$ is exponentially distributed with $\lambda > 0$.
Let $$T = max(T_A+T_B, T_C)$$ be the random variable describing the lifetime of the system.
I want to calculate the pdf of the random variable $T_A+T_B$. Since each random variable is independent, can this be computed by: $$f_{T_A+T_B}(t) =\int_{-\infty}^\infty f_{T_A}(x)f_{T_B}(t-x)dx = \int_0^\infty \lambda e^{-\lambda x} \lambda e^{-\lambda (t-x)} dx$$
or is this incorrect?
You are correct.
$$F_{T_a+T_b}(t)=P(T_a+T_b \leq t)=\int_{\mathbb{R}^2}{1_{\{x+y \leq t\}}f_{T_a}(x)f_{T_b}(y)dxdy}$$ $$F_{T_a+T_b}(t)=\int_{\mathbb{R}}{f_{T_a}(x)dx\int_{-\infty}^{t-x}{f_{T_a}(x)dy}}$$ $$F_{T_a+T_b}(t)=\int_{\mathbb{R}}{f_{T_a}(x)F_{T_b}(t-x)dx}$$
By differentiating the last equation wrt $t$ , $$f_{T_a+T_b}(t)=\int_{\mathbb{R}}{f_{T_a}(x)f_{T_b}(t-x)dx}$$ Replacing now,
$$f_{T_a+T_b}(t)=\int_{\mathbb{R}}{\lambda e^{-\lambda x} 1_{x>0}\lambda e^{-\lambda (t-x)}1_{t-x>0}dx}$$
$$f_{T_a+T_b}(t)=\int_{0}^t{\lambda e^{-\lambda x}\lambda e^{-\lambda (t-x)}dx}$$