\(F_X(t)=P(X_1+X_2\le t)\) (orden: \(dx_1\,dx_2\))
Sea \(X_1,X_2 \overset{}{\sim} \operatorname{Exp}(\lambda)\) con \[ f_{X_1,X_2}(x_1,x_2)=\lambda^2 e^{-\lambda(x_1+x_2)},\qquad x_1\ge0,\;x_2\ge0. \]
La región de integración es \(\{(x_1,x_2): x_1\ge0,\;x_2\ge0,\;x_1+x_2\le t\}\). Integrando primero en \(x_1\),
\[ F_X(t)=\iint_{x_1+x_2\le t}\lambda^2 e^{-\lambda(x_1+x_2)}\,dx_1\,dx_2 =\int_{x_2=0}^{t}\int_{x_1=0}^{\,t-x_2}\lambda^2 e^{-\lambda(x_1+x_2)}\,dx_1\,dx_2. \]
\[ \begin{aligned} \int_{0}^{t-x_2}\lambda^2 e^{-\lambda(x_1+x_2)}\,dx_1 &= \lambda^2 e^{-\lambda x_2}\int_{0}^{t-x_2} e^{-\lambda x_1}\,dx_1\\[6pt] &= \lambda^2 e^{-\lambda x_2}\left[\frac{-1}{\lambda}e^{-\lambda x_1}\right]_{0}^{t-x_2}\\[6pt] &= \lambda^2 e^{-\lambda x_2}\left(\frac{-1}{\lambda}e^{-\lambda(t-x_2)}+\frac{1}{\lambda}\right)\\[6pt] &= \lambda e^{-\lambda x_2}\big(1-e^{-\lambda(t-x_2)}\big). \end{aligned} \]
Sustituyendo: \[ \begin{aligned} F_X(t) &= \int_{0}^{t}\lambda e^{-\lambda x_2}\big(1-e^{-\lambda(t-x_2)}\big)\,dx_2\\[6pt] &= \lambda\int_{0}^{t} e^{-\lambda x_2}\,dx_2 \;-\; \lambda\int_{0}^{t} e^{-\lambda x_2}e^{-\lambda(t-x_2)}\,dx_2. \end{aligned} \]
Obsérvese que \(e^{-\lambda x_2}e^{-\lambda(t-x_2)}=e^{-\lambda t}\) (constante en \(x_2\)). Entonces:
\[ \lambda\int_{0}^{t} e^{-\lambda x_2}\,dx_2 = \lambda\left[\frac{-1}{\lambda}e^{-\lambda x_2}\right]_0^t = 1 - e^{-\lambda t}, \]
y
\[ \lambda\int_{0}^{t} e^{-\lambda x_2}e^{-\lambda(t-x_2)}\,dx_2 = \lambda e^{-\lambda t}\int_{0}^{t} 1\,dx_2 = \lambda t e^{-\lambda t}. \]
Por lo tanto, \[ \boxed{\,F_X(t)= (1 - e^{-\lambda t}) - \lambda t e^{-\lambda t} ,\quad t\ge0.\,} \]