Question

Find the equation of the line y=mx that divides the region \[\int_{0}^{1} x-x^{2}\] into 2 regions of equal area.

Answer 1

argh! stupid dx...
anyway, here's what I got:
\[\int_{0}^{1} (x-x^{2}) dx = 1/6\]
1/2 of 1/6 is 1/12 which equals \[\int_{0}^{1} (x-x^{2}) dx - \int_{0}^{1} mx dx\]in the end I got m to be 1/6.
How about it, everything look correct?

Answer 2

btw, m's a constant

Answer 3

It seems correct but when I plot y=x, y=x^2 and y=x/6, it doesn't seem to divide the area evenly.

Answer 4

ah...hmm

Answer 5

What I'd do is solve for m from this:
\[\int_0^1 x-mx\ dx=\int_0^1mx-x^2\ dx\]

Answer 6

wait, what's the y=x, the y=x^2, and the y=x/6 for?
and I'm not sure where you're getting that..

Answer 7

I plotted y=x and y=x^2 to check out the area between them and then y=x/6 to see whether this divides the area.
I created that equation so that I could make sure that the areas between our mystery line and the y=x line and between mystery line and y=x^2 are equal.

Answer 8

Is that just a rendering of the given equations? 'Cause I thought that you'd plot y=x-x^2 and y=x/6

Answer 9

first notice that:
\[\int\limits_{0}^{.5}x-x^2dx=1/12=\int\limits_{.5}^{1}x-x^2dx\]
and so we know that the line which will half our reqion will pass through .5 and is a vertical line.

so the line we are finding for is x=1/2