Calculus and minimum values This is a simple question but I think I don't understand exactly what the question is asking.

 A: Hint: use the second fundamental theorem of calculus. What is $F'(x)$? At what point(s) is $F'(x) = 0$? Now plug in the appropriate critical points and endpoints to $F(x)$ and evaluate them. The minimum value you obtain will be your answer. Your result should be easy verify by looking at the picture. 
A: We want the (signed) area under the graph from the left end to an $x$ of our choice to be as small as possible.
At any point where $f$ is negative, moving $x$ further to the right decreases $F$. Likewise wehre $f$ is positve, moving $x$ further to the left decreases $F$. So it should be clear that only the two roots $-3.5$ and $3.5$ are candidate positions for the absolute minimum.
The (absolute) area of the triangle bounded by teh $x$-axis and the graph from $0$ to $3.5$ is bigger than the area of the triangle from $-3.5$ to $0$ because they have the same base length but different height. Therefore at the root $x=3.5$ the function $F$ is smaller than at $-3.5$, i.e. the absolute minimum is assumed at $x=3.5$.
A: The function $F(x)$ is the area "below" the curve $y=f(t)$, "above" the $x$-axis, from $t=-5$ to $t=x$.  
Let us first go through an informal description of $F(x)$. We have $F(-5)=0$, the area from $-5$ all the way to $-5$ is $0$.
Now let's start moving to the right. Up to about $t=-3.5$, the curve $y=f(t)$ lies below the $t$-axis. So the "area" is getting more and more negative. At $x=-3.5$, by the usual method for finding the area of a triangle, we have $F(x)=-(3.5)(3)/2=-2.25$.
Now as we continue to move to the right, $f(t)$ is positive. so the area is increasing. It keeps on increasing until we reach the origin. Then for a while the function $f(t)$ is negative, so our area starts to decrease. It keeps decreasing until we reach $x=3.5$.
We calculate the value at $3.5$. The "area" up to $-3.5$ was $-2.25$. The area from $-3.5$ to $0$ is $(3.5)(1)/2=1.75$, and the "area" from $0$ to $3.5$ is $-(3.5)(2)/2$. So the total area $F(x)$ up to $x=3.5$ is $-2.25+1.75-3.5$, so it is $-4$.
At $x=3.5$, the total area starts increasing again. 
The only candidates for absolute min at at $x=-3.5$ and at $x=3.5$. And $x=3.5$ "wins" by quite a lot. The function value there is $-4$, which is less than $-2.25$.
Remark: I told the story from the point of view of area, mainly because it is the expected story. But I think a better way is to talk about it in terms of motion.
 So $f(t)$ is our velocity at time $t$, and $F(x)$ is our displacement from the position we occupied at $t=-5$. More physical, and one can make funny gestures as one describes the motion.
From time $t=-5$ to $t=-3.5$ we are moving with negative velocity, so backwards. Then until time $0$ we move forwards, but then start moving backwards again until time $t=3.5$. The question is at what time are we most to the left of our starting position. The only reasonable candidates, if you think about the motion, are at the end of a backwards moving period, or at an endpoint. So the only real candidates are $-3.5$ and $3.5$. 
