# Simple integration for an area problem

1. Feb 1, 2017

### yecko

1. The problem statement, all variables and given/known data

2. Relevant equations
Integration of graph is the area.

3. 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: Feb 1, 2017
2. Feb 1, 2017

### Staff: Mentor

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

3. Feb 1, 2017

### Ray Vickson

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.

4. Feb 1, 2017

5. Feb 1, 2017

### Ray Vickson

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: Feb 1, 2017
6. Feb 1, 2017

### SammyS

Staff Emeritus