Is $F(2)-F(-2)$ the same thing as $f(2)-f(-2)$?  I thought that they were the same thing and that I would just use the graph numbers but the answer is not $4$ so apparently I'm wrong.
Based on the picture, we have $f$ to be the linear function on $[-2,-1]$ given by $f(-2)=-2$ and $f(-1)=0$. And $f$ to be the linear function on $[-1,2]$ given by $f(-1)=0$ and $f(2)=2$.
 A: Hint: Long way... Notice that $f$ is a linear function on $[-2,-1]$ and $[-1,2]$ so you can explicitly find it. Then you can antiderivate (integrate) $f$ on the desired interval.
So you have $$f(x)=\left\lbrace\begin{array}{rcl}2x+2 &\mathrm{if}& -2\leq x\leq -1 \\ \frac{2}{3}x+\frac{2}{3} &\mathrm{if}& -1\leq x\leq 2 \end{array}\right. $$
Then you can integrate (antiderivate) $f$ and evaluate as needed.
You could use the following to compute: $$F(2)-F(-2) = \int_{-2}^2 f(x)dx = \int_{-2}^{-1} 2x+2\ dx + \int_{-1}^{2} \frac{2}{3}x+\frac{2}{3} dx$$
A: There is no unique antiderivative of $f$, but for any number $a \in (-2,2)$ the quantity
$$F(x) = \int_a^x f(t)\,dt$$
is the area (with sign) of the portion of plane between the graph of $f$ and the horizontal axis. Thus
\begin{align*}
F(2)-F(-2) & = \int_a^2 f(t)\,dt - \int_a^{-2} f(t)\, dt \\
& = \int_a^2 f(t)\,dt + \int_{-2}^a f(t)\, dt \\
& = \int_{-2}^2 f(t)\, dt.
\end{align*}
This area is the sum of the areas (with sign) of two triangles and corresponds to $-1+3=2$. On the other hand, $f(2)-f(-2) = 4$.
