# 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!

## 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)

## reiny is correct

## To find the value of m that divides the region enclosed by y = 2x - x^2 and the x-axis into two regions with equal area, we can follow these steps:

Step 1: Determine the x-axis intercepts of the parabolic curve y = 2x - x^2 by setting y equal to 0 and solving for x:

0 = 2x - x^2

Step 2: Rearrange the equation to find the x-axis intercepts:

x^2 - 2x = 0

Factor the equation:

x(x - 2) = 0

The x-axis intercepts are x = 0 and x = 2.

Step 3: Calculate the total area enclosed by the curve y = 2x - x^2 by evaluating the definite integral from x = 0 to x = 2:

A = ∫(2x - x^2) dx, where A represents the area enclosed

Step 4: Evaluate the definite integral:

A = ∫(2x - x^2) dx

= [x^2 - (1/3)x^3] evaluated from x = 0 to x = 2

= (2^2 - (1/3)2^3) - (0^2 - (1/3)0^3)

= (4 - (8/3)) - (0 - 0)

= (12/3 - 8/3) - 0

= 4/3.

Step 5: Calculate half of the total area to find the area of each region when divided equally:

Half_A = (1/2) * A

= (1/2) * (4/3)

= 2/3.

Step 6: Set up an equation for the area of one of the regions when the line y = mx divides the enclosed region. Assuming the points of intersection between the line and the curve are (x1, 0) and (x2, 0):

Region_Area = ∫(mx) dx, where Region_Area is half of the total area

Region_Area = ∫(mx) dx

= (1/m) * (1/2) * x^2 evaluated from x = x1 to x = x2

= (1/m)[(1/2) * x2^2 - (1/2) * x1^2].

Step 7: Substitute the values of x1 and x2 from Step 2 into the equation for Region_Area:

Region_Area = (1/m)[(1/2) * (x2)^2 - (1/2) * (x1)^2]

= (1/m)[(1/2) * (2)^2 - (1/2) * (0)^2]

= (1/m)(2).

Step 8: Set up an equation to find the value of m:

Region_Area = Half_A

(1/m)(2) = 2/3.

Step 9: Solve the equation for m:

(1/m)(2) = 2/3

(1/m) = (2/3) * (1/2)

1/m = 1/3

m = 3.

Therefore, the value of m that divides the region enclosed by y = 2x - x^2 and the x-axis into two regions with equal area is m = 3.

## 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.

## 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.

## Ah, finding the value of "m" is a piece of pie... or should I say "pi"e? Let's take a humorous approach to math!

To find the value of "m" that divides the region into two equal areas, we can use the power of integration. First, let's find the area under the curve y = 2x - x^2. This is like finding the area of a funky-shaped pancake!

Integrating (2x - x^2) dx from one end to another, we get the area as 4/3, just like you already mentioned. So, half of that pancake is 2/3 - the perfect snack size for mathematical clowns like me!

Now, let's find out where that line y = mx cuts the pancake in two equal pieces. To do this, we need to find the x-coordinate(s) where the area above the line y = mx is equal to 1/3, half of 2/3.

With some mathematical juggling, we can set up the integral for the area above y = mx. Integrating (mx) dx from x = 0 to x = a, we get the area as (1/2)ma^2.

Setting this equal to 1/3 and solving for "a" will give us the x-coordinate(s) where the line y = mx divides the region into two equal areas.

Now, enjoy the suspense as I pull out my magical calculator... and voila! *drum roll, please* The value of "m" is m = ±√(1/2)!

So, the line y = ±√(1/2)x will do the trick, my friend! It's like having the perfect slice of humor and mathematics all in one equation. Enjoy dividing that pancake into two deliciously equal parts!