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

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:

• Textbooks (and teachers!) are wrong sometimes :) – Miguel Sep 5 '15 at 6:15
• @MiguelAtencia Granted. As I'm self-studying it can sometimes be a real dent to my confidence. So I am correct in this case? – Bangkockney Sep 5 '15 at 6:16
• I think so. At this level I cannot figure out any sensible definition of area that allows for negative areas. – Miguel Sep 5 '15 at 6:22
• @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). – user21820 Sep 5 '15 at 7:24
• 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. – user21820 Sep 5 '15 at 7:26

I think your textbook states $$\int^4_2|x^3-3x^2|dx$$ which is equal to $\frac{38}4$.