Let: $$f(x)= \begin{cases} 0, & x < -4 \\ 5, \qquad\quad& \llap{-4 \le{}} x < -1 \\ -2, & \llap{-1 \le{}} x < 3 \\ 0,& x \ge 3\end{cases}$$
$$g(x) = \int_{-4}^x f(t)dt$$
I need to determine the value of each of the following: $g(-7) ,g(-3) ,g(0) ,g(4)$
I already know $g(-7) = 0$ and $g(-3) = 5$ I don't know how I know that, I just followed the basics of reading a piecewise, but the last two do not work the same way, and I don't understand how to find $g'(x)$ in order to make sense of the last two. There's no function to get a derivative from. I have spent hours trying to figure out how to do this, and need someone to help me start from scratch on this, because it makes absolutely no sense to me.
I also need to know how to find the absolute maximum, and what value of $x$ it occurs at. Again, no clue how to approach it. There's no function of $f(x)$ to derive anything from, so how is this done?
Guide:
Just split the integral over different region and integrate them.$$g(0) = \int_{-4}^{0}f(t) \, dt = \int_{-4}^{-1}f(t) \, dt + \int_{-1}^0 f(t) \, dt$$
$$g(4) = \int_{-4}^{4}f(t) \, dt = \int_{-4}^{-1}f(t) \, dt + \int_{-1}^3 f(t) \, dt + \int_3^4 f(t) \, dt$$
Try to complete the above computation by replacing $f(t)$ with the right expression and evalute them.
Notice that $g$ increases from $-4$ to $-1$ and then it decreases. Hence the maximum occur when $x=-1$. Find $g(-1)$.