Why do we integrate the modulating function in FM? I've been reading about FM and I keep coming to an expression, where the instantaneous phase

$$\Theta=2\pi f_it+\phi\text, \tag1$$ where 
$$f_i=f_c+k_fm(t)\text,\tag2$$
($f_c$ being the base carrier frequency and $k_f$ being some proportionality constant) is written as some sort of integral expression, like here:  

$$\Theta_i\{t\}=2\pi f_ct+2\pi k_f\int_0^t m\{t\}\,dt$$

No clear explanation is given, apart from:  


  The instantaneous phase of the modulated wave at any instant can be obtained by substituting 12.6 into 12.3 and integrating to get [the integral expression]


(12.3 and 12.6 being my equations (1) and (2) respectively). 

Why do we need to integrate the modulator signal?

Why is the following not an accurate expression for the phase: $\Theta =2\pi f_ct+2\pi k_fm(t)t+\phi$, (simply plugging equation (2) into equation (1))?
You need to integrate the modulating signal because frequency is the time derivative of phase. Therefore, the typical relationship from introductory calculus holds:

$$
\phi_i(t) = \int_{-\infty}^{t} \frac{d\phi_i(\tau)}{d\tau} d\tau = \int_{-\infty}^{t} 2\pi f_i(\tau) d\tau
$$

For causal signals, the lower limit on the integral changes to zero:

$$
\phi_i(t) = \int_{0}^{t} 2\pi f_i(\tau) d\tau
$$

$$
\phi_i(t) = \int_{0}^{t} 2\pi\left(f_c + k_fm(\tau)\right) d\tau
$$

$$
\phi_i(t) = 2\pi f_ct +\int_{0}^{t} 2\pi k_f m(\tau) d\tau
$$

which is the answer you were trying to get to.