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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?

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  • $\begingroup$ are the limit of the integrals given? $\endgroup$ – Siong Thye Goh Apr 27 at 8:04
  • $\begingroup$ No, no limits are given. $\endgroup$ – Sami Rae Apr 27 at 8:08
  • $\begingroup$ Interesting, do you assume integration over the whole real line to find $g(-7)$ and $g(-3)$? Are you sure the integral is written correctly? $\endgroup$ – Siong Thye Goh Apr 27 at 8:14
  • $\begingroup$ Well, it might look a bit odd, yea, I don't know how to write it in code to make it look nicer. The -4 should be at the bottom, and the x at the top, followed by f(t)dt And no, I just went by the basic piecewise process to find those values, and they worked, I did no integration. I have no idea how to when there's not a function to integrate. $\endgroup$ – Sami Rae Apr 27 at 8:24
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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 $g(-1)$.

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  • $\begingroup$ Yea, that looks good, but I don't know what expression you mean. What goes in the place of f(t)? The 5? Or -2? Wouldn't that just equal 3? I don't see what equation there is to put there, or how those values are being affected at at all. $\endgroup$ – Sami Rae Apr 27 at 8:43
  • $\begingroup$ For example, to evaluate $\int_{-4}^{-1}f(t)\, dt$, notice that $f(t)$ always takes the value $5$, so $\int_{-4}^{-1}f(t)\, dt=\int_{-4}^{-1}5\, dt$. I have splitted the integral such that $f(t)$ take a particular expression as stated in the description of the question. Similarly, $\int_{-1}^0 f(t) \, dt = \int_{-1}^0 (-2) \, dt$ $\endgroup$ – Siong Thye Goh Apr 27 at 8:49
  • $\begingroup$ Okay, so /int_−1^−4 5𝑑𝑡, would be 5(-1) - 5(-4) = 15 ? And then ∫0−1(−2)𝑑𝑡 would be 0 - 2, so g(0) = 13 $\endgroup$ – Sami Rae Apr 27 at 8:53
  • $\begingroup$ yes, also, try to click on the edit button, there you can learn how to type mathjax on the site. use dollar sign around mathy objects, use \int to type $\int$ and use underscore and caret to type subscript and superscript. The earlier you pick up, the easier it facilitates communication on the site. $\endgroup$ – Siong Thye Goh Apr 27 at 8:55
  • $\begingroup$ Okay, I think I've got it. Phew. Okay, I'll look at that now, thank you so much! $\endgroup$ – Sami Rae Apr 27 at 8:55

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