The correct answer is $\int\limits_0^{\text{t}} {{{\text{f}}_1}\left( {\text{x}} \right){{\text{f}}_2}\left( {{\text{t}} – {\text{x}}} \right){\text{dx}}}$.
The probability density function of the overall time taken to execute the program is the probability that the program will take a certain amount of time, given that the two modules are executed sequentially. This probability can be calculated by multiplying the probability density functions of the two modules, $f_1(t)$ and $f_2(t)$, and integrating the product over the interval $[0, t]$. This is because the probability that the program will take a certain amount of time is equal to the probability that the first module takes a certain amount of time, and then the second module takes the remaining amount of time.
The probability density function of the first module is $f_1(t)$, which gives the probability that the first module will take a certain amount of time, $t$. The probability density function of the second module is $f_2(t)$, which gives the probability that the second module will take a certain amount of time, $t$. The probability that the program will take a certain amount of time, $t$, is equal to the probability that the first module takes a certain amount of time, $x$, and then the second module takes the remaining amount of time, $t – x$. This probability is given by the product of the probability density functions of the two modules, $f_1(x)$ and $f_2(t – x)$, integrated over the interval $[0, t]$.
$$\int\limits_0^{\text{t}} {{{\text{f}}_1}\left( {\text{x}} \right){{\text{f}}_2}\left( {{\text{t}} – {\text{x}}} \right){\text{dx}}}$$
This is the probability density function of the overall time taken to execute the program.