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?

  • $\begingroup$ 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


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 $g(-1)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.