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I was solving a problem today, and it appears my approach is at serious odds with the provided solution.

The Problem

Find the area between $y = x^3 - 3x^2$, the $x$-axis and the lines $x = 2$, $x = 4$.

My Solution

I drew a sketch. It showed that between $x = 2$ and $x = 3$, the area bounded by the curve was below the $x$-axis, and the area between $x = 3$ and $x = 4$ above the $x$-axis.

Earlier in the chapter, my textbook explicitly stated to take care in situations like this.

I therefore wrote down: $$- \int_2^3 \! (x^3-3x^2) \, \mathrm{d}x + \int_3^4 \! (x^3-3x^2) \, \mathrm{d}x$$

I integrated $(\frac{x^4}{4}-x^3$ - correct as per provided solution), and following the arithmetic, arrived at an answer of:

$$= \frac{38}{4}$$

Provided Solution

Textbook simply states:

$$Area = \int_2^4 \! (x^3-3x^2) \, \mathrm{d}x$$

And arrives at:

$$= 4$$

I'm confused as to why my approach to take care of the bounded area below the $x$-axis is wrong and would therefore welcome your guidance and input.

ETA: Screen shot of textbook solution, as requested by Aditya Agarwal:

Screen shot of textbook solution

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  • $\begingroup$ Textbooks (and teachers!) are wrong sometimes :) $\endgroup$ – Miguel Sep 5 '15 at 6:15
  • $\begingroup$ @MiguelAtencia Granted. As I'm self-studying it can sometimes be a real dent to my confidence. So I am correct in this case? $\endgroup$ – Bangkockney Sep 5 '15 at 6:16
  • $\begingroup$ I think so. At this level I cannot figure out any sensible definition of area that allows for negative areas. $\endgroup$ – Miguel Sep 5 '15 at 6:22
  • $\begingroup$ @MiguelAtencia: On the contrary, negative areas are far more sensible! For example in the velocity-time graph the signed area under the graph is the displacement, and can easily be added (if say you are walking on a train). $\endgroup$ – user21820 Sep 5 '15 at 7:24
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    $\begingroup$ But in this case since the textbook explicitly asks students to deal with area below the graph separately, then it should be done if the graph goes below the $x$-axis and one follows the textbook. Unsigned area corresponds to distance rather than displacement, but addition now does not work nicely. $\endgroup$ – user21820 Sep 5 '15 at 7:26
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I think your textbook states $$\int^4_2|x^3-3x^2|dx$$ which is equal to $\frac{38}4$.
Edit: Yes, it is definitely wrong.

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  • $\begingroup$ I'm out right now but happy to when I get back. $\endgroup$ – Bangkockney Sep 5 '15 at 6:48
  • $\begingroup$ Have edited my question and added that for you. $\endgroup$ – Bangkockney Sep 5 '15 at 10:44

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