We know $F(0) = 0$
Okay, I just need some intuition on how to do this. So far, all I know is that the slope of $F(x)$ is always positive, therefore I should not have a negative slope anywhere on my graph. Other than that, I have no idea what I can do to graph $F(x)$ given $F'(x)$.
On
Some intuition about the shape:
Starting from $x = 0$ and going to the right of the $y$-axis, you begin with your maximum slope which begins to decrease immediately; so $F(x)$ is increasing and concave until the slope flattens at $F'(x) = 0$ at around $x = 2.2$. At this point, the $F'(x)$ begins to increase so $F(x)$ enters a region where it is convex with an increasingly steep slope until about $x = 4$, at which point $F(x)$ enters a region where its slope is decreasing making it concave again until just after $x = 5$, etc.
This cycle between concave and convex repeats itself so that $F(x)$ should be a monotonically increasing function whose slope behavior causes it to repeat an S shape as the function increases. This behavior is reflected about the origin.
In the following figure the red line represents $f$. $f$ is non-decreasing since the derivative is non-negative. $f$ has $0$ slope and slope $2$ as show (red numbers). $f$ has inflection points where ever $f'$ has local minima.