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I have three functions :
$$y = 2x^2 + 2x \tag{1}$$ $$y = -x - 1 \tag{2}$$ $$x = 0 \tag{3}$$

Exercise asks for the area enclosed by three of those functions.

It is clear that the area from $x = \frac{-1}{2}$ to $x = 0$ is included, but the problem is $y = 2x^2 + 2x$ and $y = -x - 1$ also encloses another region on their own from $x = -1$ to $x = \frac{-1}{2}$.

So does the total region of the three functions accounts for that separately enclosed region?

Here's the graph :
https://www.desmos.com/calculator/fqqngyi1ur

Prem
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lptd
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  • Normally, it would not be counted as it isn't bounded by the other plot. Though I'd prefer to use inequalities instead to avoid confusion – ioveri Apr 03 '24 at 04:53
  • Is that how the question was actually worded? In most cases, it is probably just asking for the region from $x = -0.5$ to $x=0$, but it may be asking you to sum this area with the other region you mentioned. By the way, $x = 0$ is not a function. – Paul Ash Apr 03 '24 at 05:09
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    It is actually worded that way. That's why there's ambiguity in how we can calculate the total area. – lptd Apr 03 '24 at 05:58
  • I have updated my answer to high-light which area was intended & why. – Prem Apr 03 '24 at 07:14

1 Answers1

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I used your DESMOS Output to make this Image :

INTEGRATION AREA

I have shaded & marked 4 general regions.

In general , (1) is the area between 2 functions & the $X$ Axis.
It could be Positive (above the Axis) or negative (below the Axis) like in this Case.

Then (2) is the area between 2 functions , where we have to Identify "Exactly" 2 Points of Intersection.
When there are more than 2 Intersections , it will be ill-defined & we have to give additional Criterion.

Moving to (3) , it is the area between some $X$ Co-ordinate (here it is $x=0$ ) & nearest Point of Intersection , which we have to Identify.
When there are Intersections on both sides of that $x$ Co-ordinate , it will be ill-defined & we have to give additional Criterion.

Lastly , (4) is the area between 2 functions between 2 $x$ Co-ordinates.
There are certain complications when there are Intersections between these limits , which I will not get into at the moment.

Coming to OP Case , we should know what the Exact Wording was to know what was intended.

Assuming certain things , we can say that Area (3) was to be calculated.

ADDENDUM :

OP is wondering whether it is (3) or (3)+(2) or (3)+(2)+(1).
Well , it can not be (3)+(2) or (3)+(2)+(1) because :
(A) In general , there might be multiple Intersections when $x$ is negative , like with 2 functions involving $sin$ terms. We can not go indefinitely backwards looking for all Intersections to include in the area.
(B) At every Intersection , Area Polarity changes & accounting for that is not standard. Certain Situations require that we add up the magnitudes. Certain Situation require that we add & subtract accordingly.
(C) Such Complications are not handled in Intro-Level text-books , hence we are left with only (3) to Evaluate.

Prem
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  • It is your (A) in the addendum that clicks for me. It all makes sense now. – lptd Apr 03 '24 at 08:32
  • I did think that your confusion might be resolved with that , @DivineOrca , which was why I deliberately added that Section later ! There are more complications which I have avoided at the moment. – Prem Apr 03 '24 at 09:03