How to solve an integral equation for $x$ Could someone explain to me how to solve this integration? Maybe use Laplace transform or Volterra or something else.
$$\int_{0}^{x}F(z)\,dz+ax=b,$$
where $a$ and $b$ are constant values. $F(z)$ is a cumulative distribution function, which is between $0$ to $1$. 
We can not differentiate on both sides, because the result is different from original solution like $x^2+2x=0;$ after differentiating both sides we get $2x+2=0$, the solutions are different.
 A: Letting $G(x) = \int_0^x F(z)\; dz$, your equation says
$$ G(x) + a x = b $$
In general there will not be a closed-form solution, however numerical
methods can be used.  For example, you might use Newton's method, which in this case is the iteration
$$ x_{n+1} = x_n - \frac{G(x_n) + a x_n - b}{F(x_n) + a} =  \frac{x_n F(x_n) - G(x_n) + b}{F(x_n) + a}$$
A: If $a>0$ and $b>0,$ we can prove that there is a solution. We know that 
$$\lim_{x\to-\infty}F(x)=0\quad\text{and}\quad\lim_{x\to\infty}F(x)=1, $$
since $F$ is a cumulative distribution function. Let $1>\varepsilon>0.$ Then there exists $N>0$ such that if $x>N,$ then $1\ge F(x)>1-\varepsilon.$ Note that if $x=0,$ then
$$\int_0^x F(z)\,dz+ax=0<b. $$
But now, if $x\to\infty,$ then if $x>N,$ we can write
\begin{align*}
\int_0^xF(z)\,dz+ax&=\underbrace{\int_0^NF(z)\,dz}_{=C}+\int_N^xF(z)\,dz+ax \\
&> C+\int_N^x(1-\varepsilon)\,dz+ax \\
&=C+(1-\varepsilon)(x-N)+ax.
\end{align*}
Since $a>0,$ this expression goes to $\infty$ as $x\to\infty.$ Therefore, there is some $M>N$ such that if $x>M,$ then $C+(1-\varepsilon)(x-N)+ax>b.$ 
Note that cumulative distribution functions are monotonically increasing, therefore continuous almost everywhere, therefore Riemann integrable. It follows that the integral of a cumulative distribution function is continuous, and $ax$ is continuous. By the Intermediate Value Theorem, therefore, there is an $x$ that solves the original equation. There can be only one solution, incidentally, since the integral and $ax$ are both monotonically increasing.
Since there is a solution in this case, it might make sense to use a binary search algorithm to find it. Binary search algorithms are extremely stable, and it can be predicted in advance how many runs you would need to achieve a desired accuracy. The only problem with the binary search algorithm is that you would need to bracket the solution in the interval $[0,N],$ first. I would find such an $N$ by simply trying powers of $10$ until you get the first one that makes the LHS greater than $b$.
I think you can also make a similar argument if $b>0$ and $-1<a<0,$ since eventually the integral should "beat" the $ax$ term, making the LHS increase without bound. Similarly if $b<0$ and $0<a<1.$ However, you could have significant problems, with perhaps no solution, if these conditions are not met.
A: As your $F$ is said to be a $\text{cdf}$ and the integration is from $z=0$, I infer that the distribution is that of a positive random variable. Hence the antiderivative will be a smooth monotonic curve starting at the origin and asymptotic to a line of slope $1$. You intersect it with the straight line $b-ax$, starting from $(0,b)$, with slope $-a$.
If $a>0$, you can expect a single intersection, somewhere between $x=0$ and $x=\dfrac ba$.
If $a<0$, you can expect a single solution if $b>0,a<-1$, and zero, one or two solutions if $0<a<-1$. More precisely, the tangent of slope $a$ to the antiderivative curve separates the domain $b<0$ in two areas with $0$ or $2$ roots, and there is a single root if $b>0$. This tangent is obtained by solving $F(x)=-a$ and plugging $x$ in the antiderivative.

If you can compute the intercept of the asymptote, 
$$c=\lim_{x\to\infty}\left(\int_0^x F(z)\,dz-x\right)$$
the intersection of the asymptote and the straight line gives you a first approximation. In the case of two roots, the tangency point separates them (in $[0,x]$ and $[x,\infty]$).
You can refine the roots with regula falsi, which converges fast while being reliable.
