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.
I have tried to solve this and the solution seems odd... the solution that came out was $k=1 - \sqrt [3]{2}$
The step I took was to find the interval of the curve $4y=4x-x^2$ (which I got 0 to 4) then integrated after subtracting the line from the curve but the solution I got was $4/3$ or something similar to that.
Any help would be appriciated.
Simple. If the area is bisected by a line, there will be 2 sections. The area above the line but under the curve, and the rest of the area. And since its being bisected, one area will always be half of the whole thing. So to get the area of the first section, we must set up an integral from 0 to the intersection point of the 2 lines.
To get the intersection point, we will have to solve this equation for x $$x-\frac{x^2}4=kx$$ You should get this as your answer: $$x=4-4k$$
Then we set up the integral: $$\int_0^{4-4k} (x-\frac{x^2}4-kx) \,dx= \frac12\int_0^4(x-\frac{x^2}4) \,dx$$ Then solving the specific integral we are left with: $$\frac{(4-4k)^2}2-\frac{(4-4k)^3}{12}-\frac{k(4-4k)^2}2=\frac43$$ From there, we can solve for k:$$k=1-\sqrt[3]{\frac12}$$
The area to be bisected lies above the $x$-axis, but $1-\sqrt[3]2\lt0$, so that can’t be correct.
The line $y=kx$ intersects the parabola at the origin and at $x=4(1-k)$. The area bounded by a chord of a parabola is equal to two-thirds the area of the enclosing paralellogram. Using the fact that the midpoints of parallel chords lie on a line parallel to the parabola’s axis, we can find the height of this paralellogram to be $$\frac14(4(2-2k)-(2-2k)^2)-k(2-2k)=(1-k)^2,$$ so its area is equal to $\frac83(1-k)^3$. The area between the parabola and $x$-axis is equal to $8/3$ (which appears to be in agreement with the value you obtained from the integral), so $k$ is equal to the real solution to $(1-k)^3=\frac12$, i.e., $k=1-\sqrt[3]{1/2}$. It looks like you might’ve forgotten a reciprocal or division somewhere along the way.
Hint:
The abscissa of intersections of $y=0, 4y=x^2-4x$ are the roots of $$x^2-4x=0$$
Total area $$\int_0^4\dfrac{4x-x^2}4dx=?$$
The abscissa of intersections of $y=kx, 4y=x^2-4x$ are the roots of $$x^2-4x=-4kx\iff x=0,x=4-4k$$
$$\int_0^{4-4k}\dfrac{4x-x^2}4dx=?$$
