Integration (1 Viewer)

[Lukybear]

Find the value of k for which the line y = kx bisects the area enclosed by the curve 4y=4x-x^2 and the x-axis.

How do you do this.

Towards the end i found the solution in this equation:

(1-2k+k^2)(1-k) = 1/2

But i dont know how to solve this equation.

 

[Revacious]

i got k = 1 - 1/cuberoot 2 tell me if thats right and ill tell you how i did it, cos its a bit annoying to type up

 

[kelllly]

I'll do it for you:

(1 - 2k + k^2)(1 - k) = 1/2
(1 - k)^2(1 - k) = 1/2
(1 - k)^3 = 1/2
1 - k = (1/2)^(1/3)
1 - (1/2)^(1/3) = k
k = 1 - 1/(cube root 2)

Is that right, Lukybear?

 

[untouchablecuz]

doesnt seem right

edit: oh wait, it is : )

also, correction: in the third line its meant to be When x=4-4k

 

[Lukybear]

yea you guys are right. Thxs

 

[Lukybear]

Untouch thxs for post up.

Could you explain in your method:
A1 = ....
A2 = .... and how they both equal half of area?

 

[untouchablecuz]

notice that A₁ represents the area enclosed above the line and parabola and that A₂ represents the area enclosed below the line, parabola and the x axis

because we want to find k such that the area is bisected, we LET A₁=A₂ and solve for k

get me?