Simple integration for an area problem

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Homework Help Overview

The discussion revolves around a problem involving integration to find the area under a graph. Participants are examining the methods and reasoning behind their attempts to solve the problem, which appears to involve calculating areas represented by integrals.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants express uncertainty about their methods and seek clarification on potential mistakes. There are questions regarding the correct approach to integration and the interpretation of results. Some participants mention the need to properly present their work for review.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's attempts. Some guidance has been offered regarding the preferred method of presenting solutions, and there is acknowledgment of the correctness of certain area results. However, there is no explicit consensus on the overall approach or final answer.

Contextual Notes

There are indications that the problem may have been misposted in a section not intended for integration problems, and participants are addressing issues related to the presentation of their work, including the orientation of images shared for review.

yecko
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Homework Statement


ieKweIK.png

Homework Equations


Integration of graph is the area.

The Attempt at a Solution


KpUfOnC.jpg

I don't think my way should have any problem in it, but I can't get the right answer.
Are there any careless mistakes in it? Or any other problems?
And how is the true answer get? And what is the answer?
Thank you very much
 
Last edited:
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Problems involving integration should NOT be posted in the Precalc section.
Thread moved.
 
yecko said:

Homework Statement


ieKweIK.png

Homework Equations


Integration of graph is the area.

The Attempt at a Solution


KpUfOnC.jpg

I don't think my way should have any problem in it, but I can't get the right answer.
Are there any careless mistakes in it? Or any other problems?
And how is the true answer get? And what is the answer?
Thank you very much

I refuse to lie sideways to read your work. Anyway, the preferred method in this forum is for you to type out your solution; some people will look at photos of solutions, but only if they are properly oriented.
 
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yecko said:
http://i.imgur.com/WvL8be9.jpg
WvL8be9.jpg

sorry for the problems. i didn't aware of them. hope you may help with this. thank you

Your total area result is correct. Computing A2 instead of A1 seems easier, since you do not need to break up the integration region into two parts: you just have ##A_2 = \int_0^{x_m} (4x - 5 x^2 - mx) \, dx,## where ##x_m## is the positive intersection point. From the requirement ##A_2 = (1/2) A## I get the equation
$$ -\frac{1}{150} m^3+ \frac{2}{25} m^2-\frac{8}{25} m+ \frac{32}{75} = \frac{16}{75},$$
leading to a much different ##m## than yours.
 
Last edited:

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