All good - just express k in terms of b, by dividing the-height-of-the-curve-at-b by b.
Ok so the problem is to find a line ex y=kx + m. That splits f(x)=4x-x^2 into two equal areas and the area is restricted by the x axis and the y axis. so its basically the area from x=0 to x=4 and splitting it. You are not allowed to split it parallel to any of the axises. so u cant make a line that is X=2.
My attempt to the solution was using y=kx So i integrated (4x-x^2)-kx from 0 to b. and i set it equal to half of area of 4x-x^2. The integral of f(x) is equal to F(x)=(2x^2)-(x^3)/3. and the area is 32/3 Area units. So half the area is 16/3 area units. The other region is integral of kx from 0 to b, + integral of 4x-x^2 from b to 4. So i end up with the following equations.
16/3 = (2x^2)-(x^3)/3 -(kx^2)/2 = (kx^2)/2 + 32/3 - 2b^2 + (b^3)/3.
And i get stuck there and dont know how to continue.
Im thankful for any help that i can get.