Thread: Area- find the value of m

Find the value of m at which the area, S, of the region enclosed by the parabola y=X^2-X and the line y=MX is divided into two equal halves by the x-axis. (assume that m>0) Please help me! What do I need to do? find the intersection point? and the integral? but I don't know how... thanks for your help!

Hey hahalol.

Hint: Consider splitting the integral into two parts where you split the interval [a,b] where a and b correspond to the intersections of the two graphs for the x-values into [a,u] + [u,b] where the Integral over [a,u]f(x)dx = Integral over [u,b] f(x)dx. You need to solve for u.

Hello.
What I did is that I found out the intersectino points which are 0 and m+1, and I found out the integral of (X^2-X)dx into [0,1] and also the integral of (MX-X^2+X)dx into [0,m+1]. As the parabola is divided into two equal halves, I put -1/6 (the result of the integral of (X^2-X)dx) equal to 1/6(m+1)^3 (the result of the integral (MX-X^2+X)dx) and it gives me m=-1-cube root of -1..
I don't know if what I did is good..

What you need to do is solve Integral over [0,u](MX - X^2 + X)dX = Integral over [u,M+1] (MX - X^2 + X)dX. If you evaluate this expression, can you solve for u given M?

Thank you for your help!
The answer is m=1?
I'm not sure, maybe I made some mistakes, but I handled the concept.

Originally Posted by hahalol

Thank you for your help!
The answer is m=1?
I'm not sure, maybe I made some mistakes, but I handled the concept.