Is this the same question just expressed differently?

  1. Find

(a) $\displaystyle \int_2^4 \left(\frac{2}{x-1}-1\right)dx$

(b) The area between the curve $y = \left(\dfrac{2}{x-1}-1\right)dx$ and the $x$-axis over the interval $[2,4]$

  • $\begingroup$ Remove $dx$ from the equation for the curve. Looks like you copied without editing. $\endgroup$ – AHusain May 15 '18 at 5:02
  • $\begingroup$ I just copied it straight from the workbook and it's written with the $dx$ for the equation for the curve $\endgroup$ – Juan Pablo May 15 '18 at 5:07

The second one is really asking for:

$$\int_2^4 \ \left| \ \frac{2}{x-1} - 1 \ \right| \ \text{d}x$$

This is because the original integral subtracts area while the function is below the horizontal axis—the reason why, for instance:

$$\int_{-2\pi}^{2\pi} \sin(x) \ \text{d}x = 0$$

  • $\begingroup$ I get that the original integral subtracts area between [3,4] when it is below the horizontal axis and in some cases it will cancel out completely and return 0 like the example you gave, but how did you get $\displaystyle \int_2^4 |\displaystyle \frac{2}{x-1}|$ from (b) (Where did the -1 go?) $\endgroup$ – Juan Pablo May 15 '18 at 5:47
  • $\begingroup$ Oh wow, I didn't see it because it was so close to the $dx$. Will fix. $\endgroup$ – Kaj Hansen May 15 '18 at 6:15
  • $\begingroup$ OK that makes sense. So those lines on either side of the function just mean total area, whether it is above or below the x-axis? $\endgroup$ – Juan Pablo May 15 '18 at 6:30
  • $\begingroup$ The lines denote the absolute value function. But yeah, $\int_a^b |f(x)| \ dx$ gives the total amount of area between the function and the horizontal axis on the interval $[a,b]$, regardless of whether $f(x) < 0$ anywhere. $\endgroup$ – Kaj Hansen May 15 '18 at 6:38
  • $\begingroup$ Awesome got it! Thank you for your help $\endgroup$ – Juan Pablo May 15 '18 at 22:55

No. The function $\frac{2}{x-1}-1$ changes sign at $x = 3$. The integral, version (a), will partially cancel the positive (signed) area with the negative (signed) area. The area between $\frac{2}{x-1}-1$ and the $y$-axis, version (b), does not have a negative contribution for $x \in [3,4]$.

  • $\begingroup$ So just so I have this right. Part (b) is the same as part (a) except in part (b) we disregard the area below the x-axis? I.e. it is just the area between the curve and above the x-axis? $\endgroup$ – Juan Pablo May 15 '18 at 5:34
  • $\begingroup$ @JuanPablo : No. In (b) all the area contributions are positive, even the part from where the function is negative. The change is between accumulating signed area and accumulating unsigned area. $\endgroup$ – Eric Towers May 15 '18 at 13:47
  • $\begingroup$ Awesome, thanks $\endgroup$ – Juan Pablo May 15 '18 at 22:54

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