Integration question involving Area and f(t) Well I am doing a question and a link of the image is provided here:

I am wondering about my answers for a few parts.
$\textbf{Part A:}$ Can you just check if I'm correct on these
$$F(0)= 0$$
$$F(2)= 8$$
$$F(4)= 12$$
$$F(6)= 16$$
$$F(10)= 24$$
$\textbf{Part B:}$ Can you just check if I'm correct on this
Minimum at $y=-4$, $x=6$, and local maximum from $x[0,2]$ at $y = 4$.
$\textbf{Part C:}$ I'm not sure about this
I think $F$ is increasing from $(0,4)$ and decreasing from $(4,10)$. Is this right?
Part B: no idea, help please?
 A: Part A: 
$$F(0) = 0$$ $$F(2) = 8$$ $$F(4) = 12$$ $$F(6) = 8$$ $$F(10) = 0$$
because $F$ below the $x$-line has to be calculated negative.
Part B: local maximum is at $x=4$ and local minimums are at $x=0$ and $x=10$
Part C: Your are right. It's increasing for $x(0,4)$ and decreasing for $x(4,10)$
Part D: If we look at the graph we know that: 
$$f(x) = 4 \ ,\ x\in(0,2)$$ 
$$f(x) = -2x + 4 \ ,\ x\in(2,6)$$ 
$$f(x) = x - 4 \ ,\ x\in(6,10)$$
So if we integrate $f$ we get:
$$F(x) = 4x \ ,\ x\in(0,2)$$ 
$$F(x) = -x^2 + 4x \ ,\ x\in(2,6)$$ 
$$F(x) = \dfrac{x^2}{2} - 4x \ ,\ x\in(6,10)$$
I think you should be able to draw the graph with this information, but don't forget that f.e. for $F(4)$ don't calculate $-4^2 +4*4$ but $-2^2 + 4*2$ because these formulas are shifted as if they started at $(0,0)$
If you have any further questions feel free to ask. 
Hopefully this helps.
A: To sketch the function, note that


*

*On $[0,2],$ $f$ is constant and so its integral will be of the form $kx,$ and the graph of $F$ will be a straight line between $(0,0)$ and $(2,8)$ (you have already shown that $F(0) = 0$ and $F(2) = 8$).

*On $[2,6]$, $f$ is of the form $cx+d$ for negative $c$ and so the integral of $f$ will be a quadratic, and so the graph of $F$ will be a downward-facing parabola taking its maximum at $x=4.$ It will also be symmetrical on the interval $[2,6].$

*On $[6,10]$, $f$ is of the form $ax + b$ for positive $a,$ and so $f$ will be an upwards-facing parabola taking its minimum value at $x=10.$ You will only see the decreasing part of this parabola (we don't know what $f$ does after $x=10$).

*Finally, your pen should never come away from the page. There is a technical reason for this (functions $F$ of this form are always continuous) but it can also be justified intuitively by thinking about how the area of the shape under $f$ changes as $f$ increases.

