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R is the 4th quadrant region enclosed by the x-axis and the curve y = x^2 – 2kx, where k > 0. Find the value of k so that the area of the region R is 36 square units. 2 3 <-- 4 6
looks good to me unless i made an error this is the way I went about it: calculate intersections of y=0 and y=x^2-2kx those were my lower and upper limits (the intersections were) You should find that between these intersections that 0>x^2-2kx for x on that interval so your upper function is f(x)=0 and your lower function is g(x)=x^2-2kx did the whole integration and plug in limits and solve the equation for k
that's exactly what i did. Glad we got the same answer :D
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