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.
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!
4 answers
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.
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)
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)
reiny is correct
2 answers