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ggrree
 3 years ago
Let R be the region in the first quadrant enclosed between the xaxis 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.
ggrree
 3 years ago
Let R be the region in the first quadrant enclosed between the xaxis 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.

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dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1330592696055:dw First solve for the point where line intersects the curve xx^2 = mx x(1x) = mx 1x = m x = 1m 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 1m \[A = \int\limits_{0}^{1m}(xx^{2}) mx = \frac{1}{8}\int\limits_{0}^{1}xx^{2}\] combine x terms in 1st integral \[A = \int\limits\limits_{0}^{1m}(1m)xx^{2} = \frac{1}{8}\int\limits\limits_{0}^{1}xx^{2}\] integrate and evaluate to solve for m
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