There are curves y=x^2 and y=mx, where m is some positive constant. No matter what positive constant m is, the two curves enclose a region in the first quadrant. Find the positive constant m such that the area of the region bounded by the curves y=x^2 and y=mx is equal to 8.

Question

anonymous · Answer

$$
\int_0^m \left(m x-x^2\right) \, dx=\frac{m^3}{6}=8
$$
Solve for m

anonymous · Answer

$$
m=2 \sqrt[3]{6}
$$

campbell_st · Answer

well find the points of intersection of the curves... so equate the  curves


$$x^2 = mx ... or ....x^2 - mx = 0$$

then 

$$x(x - m) = 0$$

so the 2 curves intesect at x = 0 and x = m

here is the picture

|dw:1398318250549:dw|

so you need to find the area between the curves and you know the area is 8

so integrating its 

$$A = \int\limits_{0}^{m}mx...dx - \int\limits_{0}^{m} x^2... dx$$

evaluate the integral then solve for m. 

hope it helps