Background
Two authors, Bob and Martha, are about to begin writing an epic co-authored book, The Adventures of Aaron the Aardwolf. It will take them A years to write. When they finish this book, they will immediately begin work on new, individually authored books. Bob will spend X years writing The Bilinear Bonanza of Bonnie the Butterfly, and Martha will spend Y years writing Memoirs of Maude the Magnificient Mangabey, independently. Suppose that A, X, Y are i.i.d. $Expo(\lambda)$. On a timeline where time $0$ is defined as the time when they begin their collaboration, consider the following quantities.
A: time at which The Adventures of Aaron the Aardwolf is completed;
B: time at which The Bilinear Bonanza of Bonnie the Butterfly is completed;
M: time at which Memoirs of Maude the Magnificient Mangabey is completed;
T: time at which the last to be completed of these three books is completed.
Excercise question:
Find the distribution of B (which is also the distribution of M).
My solution
Here, $B=A+X$, because A is the time it takes for the co-authors to finish the joint book and then X is the additional time to finish the individually co-authored book.
The CDF of B is then,
$$ \begin{aligned} F_B(b)&=P(B \leq b) \\ &=\int_0^{\infty} P(B \leq b \mid X=x) f_X(x) d x \\ &=\int_0^{\infty} P(A + X \leq b \mid X=x) f_X(x) d x \\ &=\int_0^{\infty} P(A \leq b-x) f_X(x) d x \end{aligned} $$
From here, we can go in two solution paths. One given by the book exercise solution and one that felt most natural to me. And my main question is why these paths are not producing the same result.
Solution path one (given by the book key)
Differentiating both sides with respect to $b$, we have that the PDF of $B$ is $$ f_B(b)=\int_0^{\infty} f_A(b-x) f_X(x) d x=\int_0^b \lambda e^{-\lambda(b-x)} \lambda e^{-\lambda x} d x $$
My first question is how did we get the second equality here?: More precisely, how come that the second integral limit changed from infinity to b?
Simplifying, $$ f_B(b)=\lambda^2 \int_0^b e^{-\lambda b} d x=\lambda^2 b e^{-\lambda b}, \text { for } b>0 $$
Solution path two (my path)
Instead of differentiating to obtain the PDF, we can just write out the $P(A \leq b-x)$ as that is the CDF of an exponential
$$ \begin{aligned} F_B(b)&=P(B \leq b) \\ &=\int_0^{\infty} P(B \leq b \mid X=x) f_X(x) d x \\ &=\int_0^{\infty} P(A + X \leq b \mid X=x) f_X(x) d x \\ &=\int_0^{\infty} P(A \leq b-x) f_X(x) d x \\ &=\int_0^{\infty} (1-e^{-\lambda(b-x)}) f_X(x) d x \\ &=\int_0^{\infty} (1-e^{-\lambda(b-x)}) \lambda e^{-\lambda(x)} d x \\ &=\int_0^{\infty} (\lambda e^{-\lambda(x)}-\lambda e^{-\lambda(b-x)}e^{-\lambda(x)}) d x \\ &=\int_0^{\infty} \lambda e^{-\lambda(x)}d x-\int_0^{\infty}\lambda e^{-\lambda(b-x)-\lambda(x)} dx \\ &=1-\int_0^{\infty}\lambda e^{-\lambda b+\lambda x-\lambda x} dx \\ &=1-\int_0^{\infty}\lambda e^{-\lambda b} dx \\ &=1-\lambda e^{-\lambda b}\int_0^{\infty} 1dx \\ &=1-\lambda e^{-\lambda b}[x]_0^{\infty} \\ &=-\infty \\ \end{aligned} $$
That term diverges to negative infinity which is clearly wrong. Not sure what error I am committing. And that is my second question.
Any input is appreciated.