Why does this relationship hold (integrals)

Question

If I define

$$P(t,T) = \exp\{ \int _t^Tf(t,s)ds\}$$

then why is it true that

$$\ln P(0,T) - \ln P(0,t) = - \int_t^T f(0,s)ds \tag{1}$$

We would have

$$P(0,T) = \exp \{ - \int_0^T f(0,s)ds\}$$

but I don't quite see how $$\ln P(0,T) - \ln P(0,T) \dots (1) $$

edit: heh woops... #_#

Answer 1

$$\ln(P(0,t))-\ln(P(0,T))=\int_0^t f(0,s)ds-\int_0^T f(0,s)ds=\int_0^t f(0,s)ds+\int_T^0 f(0,s)ds=\int_T^0 f(0,s)ds+\int_0^t f(0,s)ds=\underset{Chasles}{=}\int_T^t f(0,s)ds=-\int_t^T f(0,s)ds$$

Answer 2

$\ln(P(t,T))=\int_t^T f(t,s)ds$

$\ln(P(0,T))-\ln(P(0,t))=\int_0^T f(0,s)ds-\int_0^tf(0,s)ds=\int_t^Tf(0,s)ds$ by the property of the integral:

$\int_a^bf+\int_b^cf=\int_a^c f$

and the property:

$\int_b^af=-\int_a^b f$