Find the value of k for which the line y=kx bisects the area enclosed by the curve 4y (1 Viewer)

[fabl]

How do we

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

I tried to find the area enclosed by the curve which was 8/3 unit squared, then I halved that to get 4/3.

I then try integrating the y=kx and equating that to 4/3 but the answer i got is not the right answer.

The answer i got was 1/6..

The right answer was 1-1/cuberoot of 2

 

[FrankXie]

Re: Find the value of k for which the line y=kx bisects the area enclosed by the curv

I then try integrating the y=kx and equating that to 4/3 but the answer i got is not the right answer.

The answer i got was 1/6..

The right answer was 1-1/cuberoot of 2

did you draw a diagram? you should not integrate y=kx, but between the curve and the line

 

[amazing_graseu]

Re: Find the value of k for which the line y=kx bisects the area enclosed by the curv

First you need to find the point where both the graphs meet by solving simultaneously. When you do, the x-value at the point of intersection will be 4(1-k).

The equation should be:
y=x-(x^2/4)-kx
which you factorise to
y=x(1-k)-(x^2/4)

So integrate for 0 to 4(1-k) of the equation y=x(1-k)-(x^2/4) and equate that to 4/3
Here is the working out if you want it: