## ggrree 3 years ago 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.

|dw:1330592696055:dw| 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