Simple integration for an area problem

[yecko]

Homework Statement

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

 

[Mark44]

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

 

[Ray Vickson]

Homework Statement

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.

 

[yecko]

http://i.imgur.com/WvL8be9.jpg

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

 

[Ray Vickson]

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.

 

[SammyS]

http://i.imgur.com/WvL8be9.jpg[IMG ]http://i.imgur.com/WvL8be9.jpg[/PLAIN]
sorry for the problems. i didn't aware of them. hope you may help with this. thank you

What is $\ (4-m)^2\ ?$