Question

Let R be the region in the first quadrant enclosed between the x-axis and the curve y = x - x^2. There is a straight line passing through the origin which cuts the region R into two regions so that the area of the lower region is 7 times the area of the upper region.
Find the slope of this line.

Answer 1

First solve for the point where line intersects the curve
x-x^2 = mx
x(1-x) = mx
1-x = m
x = 1-m

Then define area A (upper region)
area of entire region R is 8A so A = 1/8*R
A is also area above line from 0 to 1-m
\[A = \int\limits_{0}^{1-m}(x-x^{2})- mx = \frac{1}{8}\int\limits_{0}^{1}x-x^{2}\]
combine x terms in 1st integral
\[A = \int\limits\limits_{0}^{1-m}(1-m)x-x^{2} = \frac{1}{8}\int\limits\limits_{0}^{1}x-x^{2}\]
integrate and evaluate to solve for m