Intuition in the relation between force of mortality and survival function I understand that The force of mortality $ \mu (x)$ can be interpreted as the conditional density of failure at age x given survival to age x, while f(x) is the unconditional density of failure at age x.
I also understand that  $\mu_x=\frac {-S'(x)}{S(x)}$
I know that mathematically we can easily find $S_X(x)=e^{-\int_0^x \mu_y dy}$ but I can't find the intuition behind the formula ( wich is very impotant to me to be able to learn ).  
In fact, it's especially the bounds of integration that intrigue me. Why is it from 0 to x  and not from x to $\infty$ while the survival function is for (X>x)? The second thing is why the exponential function keeps showing up in everything related to a force. (Force of mortality, force of interest...).
My questions may seem a little bit out of nowhere but I'm asking perhaps  someone can give an explanation that will also improve my understanding of the exponential function.
Thank you  
 A: The terminology "force of mortality" is not arbitrarily chosen.  It is analogous to a concept in financial mathematics called "force of interest," and an actuary should be aware of the way they are mathematically similar models--the former is applied to the notion of survivorship, and the latter is applied to the notion of the time value of money.
To recap, the force of interest is a natural consequence of modeling the time value of money in the case where the frequency of compounding approaches infinity--i.e., continuous compounding.  If the effective annual rate of interest is $i$, then after $1$ year, a unit amount of money invested in a fund will become $1+i$.  If this sum is compounded at an annual rate, then the amount at the end of $n$ years is $(1+i)^n$.
When the fund is compounded $m$ times per year in such a way as to be equivalent to an effective annual rate of $i$, then we must have a nominal rate of interest $i^{(m)}$ such that $$\left(1 + \frac{i^{(m)}}{m}\right)^m = 1 + i.$$  Solving yields $$i^{(m)} = m((1+i)^{1/m} - 1).$$  As the frequency of compounding becomes greater, essentially tending to infinity, we obtain an expression for the "force" of interest that represents in some sense an instantaneous rate of compounding:  $$\delta = \lim_{m \to \infty} i^{(m)} = \log(1+i).$$  Another way to understand this relationship is that the accumulated value of the fund at time $t = 1$ is $$A(1) = 1+i = e^\delta,$$ and in general, $$A(t) =(1+i)^t = e^{\delta t}.$$  This of course relates to a constant interest rate throughout the year, but as is frequently the case in mathematics, the continuous case is easier to generalize than the discrete:  if $\delta(t)$ is a function of $t$, then $$A(t) = e^{\int_{u=0}^t \delta(u) \, du}$$ for a starting value of $A(0) = 1$.
Now, you might ask why, if the concepts are similar, the force of mortality has a negative sign but the force of interest doesn't.  The reason is because the former deals with survivorship--where a quantity (the proportion surviving) decreases with time--and the latter deals with accumulation, where a quantity increases with time.  In fact, if you simply reversed the direction of time, you would have $$A(0) = A(t)e^{-\int_{u=0}^t \delta(u) \, du},$$ which expresses the amount present at the beginning of investment as a function of the accumulated amount at time $t$ and the force of interest experienced by the fund over that duration of time.
So if we go back to the survivorship model, $S(t)$ is a random variable that expresses the probability of surviving after time $t$, but it can also be understood deterministically as modeling the fraction of lives surviving at a given time.  As already explained by other answers, this is captured by integrating the force of mortality experienced by the cohort over the time they are exposed to mortality, thus we can understand why the limits of integration are from $0$ to $t$, and why the negative sign is there.  The $e$ has to do with the idea that we are in essence looking at a "continuously compounding" process, just like we did with the force of interest.
A: You effectively start with 
$$\mu(y) = \dfrac{-\frac{dS}{dy}}{S}$$
then with separation of variables and integrating over $y$ from $0$ to $x$ and so $S$ from $S(0)=1$ to $S(x)$ (you in fact have $1 \gt S(x)$ so need to be careful with signs)
$$\int\limits_0^{x}\mu(y) \,dy = \int\limits_{1}^{S(x)}\frac{-1}{S}\,{dS}$$ to give 
$$\int\limits_0^{x}\mu(y) \,dy =  -\log_e(S(x))$$ and so 
$$S(x) = e^{-\int\limits_0^{x}\mu(y) \,dy}$$
The intuition is that if $\mu$ were constant, you would have $S(x) = e^{-\mu \,x}$, an exponential distribution.  If $\mu$ is not constant then the rectangle of $\mu \, x$ becomes an integral  $\int\limits_0^{x}\mu(y) \,dy$
Personally I would use $\lambda$ rather than $\mu$ and say hazard function rather than force of mortality because I am not an  actuary, but it makes no difference
A: Think of it this way: the force of mortality $\mu_y$ is roughly a measure of the fraction of people between age $y$ and $y+dy$ that will die. If you integrate this up to age $x$ you have
$$\int_0^{x}\mu_y \; dy \; ,$$
the fraction of people between age $0$ and age $x$ that will die. So, these people will not contribute to the number of people that will survive, so you have
$$S(x) \approx 1 - \int_0^{x}\mu_y \; dy \; .$$
And indeed $\exp(-x)\approx 1-x + O(x)$.
