Differentiation of integral of density function

Question

In the book "A First Course in Probability, by Ross", example 2d, chapter 5, there is the following equation:

$E[C_t(X)] = ct\int_0^tf(x)dx - c\int_0^t xf(x)dx + k\int_t^\infty xf(x)dx - kt\int_t^\infty f(x)dx$

Then, authors go to:

$\frac{d}{dt}E[C_t(X)] = ctf(t) + cF(t) -ctf(t) -ktf(t) + ktf(t) - k[1 -F(t)] = (k+c)F(t) -k$

In this example, f(x) is the density function and F(t) is the cumulative distribution function.

I did not understand how they passed from $\frac{d}{dt}c\int_0^t xf(x)dx$ to $cF(t)$.

I think the correct was pass from something like $\frac{d}{dt}c\int_0^t f(x)dx$ to something like $cF(t)$.

I am sorry if it is basic math, but I really did not understand it.

Thanks.

Answer 1

The derivative is $$[ctf(t)+c\int_0^{t} f(x)dx]- ctf(t)-ktf(t)-k\int_t^{\infty} f(x)dx+ktf(t).$$ Now write $\int_0^{t} f(x)dx$ as $F(t)$ and $\int_t^{\infty} f(x)dx$ as $1-F(t)$. So you get $(k+c)F(t)-k$