$F$ is an increasing function and $F(0) > 0$. For $x \geq 0$, define $G(x) = \int_{0}^{x} F$. Prove that $G$ is one-to-one I am given a practice question where I am supposed to show that the function of G is one-to-one. Where $G(x) = \int_{0}^{x} F$, where $F$ is an increasing function and $F(0)>0$.
I thought about thinking about the second part of Fundamental Theorem of Calculus, but I don't know if that is the right method to show that the function is one to one. Also, I don't know if I can just say

Suppose that $F(0) = 1$ since $F(0) > 0$, then $G(x) = F(x) - 1$.

Any help will be great.
Edit: It was clarified that $x \geq 0$
 A: I would think this is a proof by contradiction:
Let's assume $G(x)$ is not one-to-one. Therefore there exists two different $x$ values, $x_1 < x_2$ such that $G(x_1) = G(x_2)$:
\begin{align*}
G(x_1) =&\ \int\limits_0^{x_1}F(t)dt \\
G(x_2)=&\ \int\limits_0^{x_2}F(t)dt = \int\limits_0^{x_1}F(t)dt + \int\limits_{x_1}^{x_2}F(t)dt = G(x_1) + \int\limits_{x_1}^{x_2}F(t)dt
\end{align*}
If $G(x_1) = G(x_2)$ then $\int\limits_{x_1}^{x_2}F(t)dt = 0$. You need to show that this isn't possible if $x_2 > x_1$ and $F(t)$ is increasing.
Edit (clarification of my comment):
This question was edited (which is fine and is good), but I do want to explain for anyone curious why we need $x \geq 0$ for the given problem statement.
If we are only given that $F(t)$ is monotonically increasing and $F(0) > 0$, that's not enough to conclude that $G(x) = \int\limits_0^xF(t)dt$ is one-to-one.
Consider the following:
$$
F(t) = 4t^3 + 2t + 1
$$
Clearly $F(0) = 1 > 0$ and it's clearly monotonically increasing (not even just monotonically non-decreasing--$F'(t) = 12t^2 + 2 > 0$ for all $t$)
$$
G(x) = \int\limits_0^xF(t)dt = x^4 + x^2 + x
$$
A simple graph of this function will show that it is one-to-one when $x\geq0$, in fact it's one-to-one for all $x \geq \xi$ such that $4\xi^3 + 2\xi + 1 = 0$, but if we allow a domain such that $\xi - \delta < x < \infty$, where $\delta > 0$, then it's clearly not one-to-one.
Which brings me to my ultimate criticism (of the question): we don't need $F(0) > 0$, we only need $F(0) \geq 0$ and $x \geq 0$. With that constraint I can easily explain the requirement for $x \geq 0$:
$$
F(t) = 4t^3 + 2t \rightarrow G(x) = \int\limits_0^xF(t)dt = x^4 + x^2
$$
This is $F(t)$, an odd function (and is clearly monotonically increasing), results in a even function: cannot possibly be one-to-one unless we restrict to $x \geq 0$.
Please Note: I'm talking about a "counterexample" as to why we must have $x \geq 0$ (i.e. a function, $F(t)$, that is monotonically increasing that does not result in a one-to-one function if $x < 0$)! This does not mean $F(t)$ needs to be odd--only monotonically increasing on the interval $0 < t < \infty$, with $F(0) \geq 0$ and $x \geq 0$.
Second Note: $F(t)$ is monotonically increasing but $F(0) < 0$: (assuming $x \geq 0$)
Simple counterexample:
$$
F(t) = 4(t - 1)^3 \rightarrow G(x) = \int\limits_0^xF(t)dt = (x - 1)^4 - 1
$$
Here $G(x)$ is not one-to-one because you get $G(0) = G(2) = 0$
A: Use the fondamental theorm of aclculus to obtain
$$G'(x)=F(x).$$ Since $F$ is increasing, we have $G(x)=F(x)>0$ for all $x\geq0$. That means $G'>0$ and so $G$ is strictly increasing (meaning $a>b$ implies $G(a)>G(b)). A strictly increasing function is one to one.
A: Differentiating under integral sign we have $G'(x)=f(x)>0$ as f(x) is an increasing function. Thus G(x) is an increasing function. We need to prove its montonically increasing for itto become one-to-one. Let's say for some interval $x_{0}-x_{1}$ f(x) has same value then $G(x_{0})=\int_0^{x_0} f(x)=H(x_{0})$. Now $G(x_{1})=H(x_{0})+\int_{x_{0}}^{x_{1}} f(x)$ but as we have assumed that here f(x) is constant and let it be equal to c.
we have $G(x_{1})=H(x_{0})+(x_{1}-x_{0})c$. Now I have assumed $x_1>x_0$ also c is >0 as f(x) is increasing and f(0)>0 . So $(x_1-x_0)c>0$. Hence $G(x)$ is one to one function.
