# Simple integration for an area problem

• yecko
In summary, the conversation involves a student seeking help with an integration problem. They have made a mistake in their solution, which another user points out and offers a different method for solving the problem. The student asks for clarification on a specific part of the solution.
yecko
Gold Member

## Homework Equations

Integration of graph is the area.

## The Attempt at a Solution

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:
Problems involving integration should NOT be posted in the Precalc section.

yecko said:

## Homework Equations

Integration of graph is the area.

## The Attempt at a Solution

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.

Buffu
yecko said:
http://i.imgur.com/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:

## 1. What is simple integration for an area problem?

Simple integration for an area problem is a mathematical method used to find the area under a curve by breaking it down into smaller, simpler shapes and summing their areas.

## 2. How is simple integration used in real life?

Simple integration is used in many fields, including physics, engineering, and economics, to find the area under a curve and solve various problems. For example, it can be used to calculate the distance traveled by an object over time or to find the total cost of a production process.

## 3. What are the basic steps of simple integration for an area problem?

The basic steps of simple integration for an area problem are: 1) identify the function that represents the curve, 2) break the curve into smaller sections, 3) find the area of each section using basic geometry, 4) sum the areas of all sections to find the total area under the curve.

## 4. Can simple integration be used for any type of curve?

No, simple integration can only be used for curves that have a known equation or can be approximated by a known function. It cannot be used for irregular curves or curves with infinite points, such as fractals.

## 5. Are there any limitations to using simple integration for an area problem?

Yes, there are limitations to using simple integration. It can only provide an approximate solution, as the smaller the sections used, the more accurate the result will be. Additionally, it may not be able to accurately calculate the area of curves with sharp turns or discontinuities.

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