Confusion with displacement of parabolas. First I would just like to introduce my self and give you the extent of my mathematical background. I am currently a high school student that decided to self teach calculus. I am currently exploring applications of integration. 
I ran into some confusion when I started to think of the displacement of a parabolic function. Say for example we have a ball that followed the path of $$f(x)= -t^2 + 4t - 3.$$ I wanted to focus on when $1<t<3$ inclusive. If I were to integrate this function from $1$ to $3$ I get $4/3$. I don't understand why this is. If our $x$ intercepts of this function are $1$ and $3$ then wouldn't the displacement just be $2$?
 A: Integrating position with respect to time does not give you displacement. To get displacement, you want to integrate velocity with respect to time. 
The integral of velocity with respect to time is position.
If you want to find the displacement of the ball between time $t_{{1}}$ and $t_{{2}}$, just compute (position at $t_{{2}}$) - (position at $t_{{1}}$). It sounds like you may have already done that when you took the difference of the x-intercepts!
A: When one integrate a function $f(x)$ to get its displacement, $f(x)$ is taken to be a function of speed wrt time variable $x$. However, your $f(x)$ is a function of location wrt time $x$.
that's why integrating $f$ would give you nothing meaningful. You intuition was correct. Displacement is indeed just $2$.
When solving real life problems, you have to be careful about what you define as $f(x)$.
A: If $x$ is the position at time $t$, which would be normal, the displacement over $x_0$ to $x_1$ is $f(t_1)-f(t_0)$.  Note that $f$ should be a function of $t$ in your equation.  No need to integrate.  Dimensional analysis can help you.  The units of $f$ are length, so if you integrate $f \ dt$ you get length*time-what is that?
A: Let me first define displacement as "a vector quantity that refers to 'how far out of place an object is'; it is the object's overall change in position."
By integrating $f$ from $1$ to $3$, you are effectively computing the area under the curve, not the displacement.
From the definition, it is clear that you do not need calculus to compute displacement; what you need is the coordinates of both the starting and ending points of motion. Therefore, you are right that, in this case, your displacement is $2$ rightward, or in vector notation: $\left \langle 2,0 \right \rangle$.
