Posted by **Anonymous** on Sunday, January 6, 2008 at 8:39am.

find the value of m so that the line y = mx divides the region enclosed by y = 2x-x^2 and the x-axis into two regions with equal area

i know that the area enclosed by y=2x-x^2 is 4/3 so half of that would be 2/3

however, i have no idea how to find the value of m!

- calculus -
**drwls**, Sunday, January 6, 2008 at 9:56am
The f(x) curve is above the x-axis from x=0 to the point where 2x = x^2, which is x = 2. The area is

INTEGRAL(2x - x^2) dx = x^2 -x^3/3 @x=2

0 to 2

= 4/3. That agrees with what you obtained.

You want the area under y = mx to be 2/3. That area will be (1/2)xy. where x and y are the coordinates were the curves intersect.

(1/2) xy = 2/3 or xy = 4/3

y = mx

y = 2x - x^2

You have three equations in three unknowns. That is enough for a solution.

- calculus (correction) -
**drwls**, Sunday, January 6, 2008 at 10:20am
The trangular area beneath the y = mx from 0 to the intersection point is not all of the region you need to integrate. Another portion, from the intersection point to x=2, lies below the y= 2x-x^2 parabola. You can require instead that the area between y=mx and y = 2x -x^2 from 0 to the intersection point (x) be 2/3. That will give you an equation in m and x, that can be combined with the two line equations for an answer.

- calculus -
**Reiny**, Sunday, January 6, 2008 at 11:08am
I looked at the area between y = 2x-x^2 and y = mx from 0 to a, where a is the x coordinate of their intersection point

then first, ma = 2a-a^2

m = 2 - a (equ. #1)

Agreeing with the total area to be 4/3, as you both found, then

2/3 = integral(2x - x^2 - mx)dx from 0 to a

2/3 = a^2 - a^3/3 - (ma^2)/2 or

4 = 6a^2 - 2a^3 - 3ma^2

sub equ #1 into and simplifying that gives us

4 = a^3

so a = 4^(1/3) and

m = 2 - 4^(1/3)

- calculus -
**nehal**, Thursday, February 4, 2016 at 6:38pm
reiny is correct

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