TITLE: How to solve the equation for $x$ with integration?
QUESTION [0 upvotes]: Could someone explain to me how to solve this integration?
$$a-b\int_{0}^{x}[1-F(z)]dz+x=0$$
Where $a$ and $b$ are constants, $F(z)$ is a cumulative distribution function (the distribution is not specified), how can I solve for $x$ from this equation?

REPLY [2 votes]: Let $~b\neq 0~$
$$a-b\int_{0}^{x}[1-F(z)]dz+x=0$$
Differentiating with respect to $~x~$ using Leibniz integral rule, we have
$$-~b~\frac{d}{dx}~\int_{0}^{x}[1-F(z)]dz~+~1~=0$$
$$\implies -~b~[1-F(x)]+1=0$$
$$\implies F(x)=\frac{b-1}{b}\qquad \text{as $~b\neq 0~$}$$
$$\implies x=F^{-1}\left(\frac{b-1}{b}\right)$$


Leibniz Integral Rule (Differentiation under the integral sign):
Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$,
$$\frac{d}{dx}(\int_{a(x)}^{b(x)} f(x,t) dt)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$

REPLY [0 votes]: The given integral equation is $$a-b\int_{0}^{x} |1-f(z)|dz+x=0.$$ Differentiate w.r.t. $x$, then
$$-b |1-f(x)|+1=0 \Rightarrow |1-f(x)|=\frac{1}{b}\Rightarrow f(x)=1+\frac{1}{b},~\mbox{or}~ f(x)=1-\frac{1}{b} ~\mbox{b}>0.$$ No solution if $b<0.$
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