I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

"Looking at the graph, find the value of k so that areas of region A and region B are equal"

I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

"Looking at the graph, find the value of k so that areas of region A and region B are equal"
0.2

No, I am sorry to say that is incorrect. I will leave it up until tomorrow evening to see if anyone else wants a go.

How about: k=0.724611?

Yes, indeed, ebaines. I knew you could get it. :)

Yes, indeed, ebaines. I knew you could get it. :)

Thanks! However, I must admit that I was unable to find a closed-form solution for this problem, but rather used an approximation method to solve a nasty equation:

1 = b*sin(b) + cos(b)

where b = arcsin(k)-pi.

Is there a more elegant way?

Yes, that's exactly what I done. I used Newton's method.

SOLUTION:


Let (a,k) be the point of intersection of y=k with y=sin(x).

Then, k=sin(a) and if the areas are equal:

\int_{0}^{a}(k-sin(x))dx=\int_{0}^{a}(sin(a)-sin(x))dx=a\cdot{sin(a)}+cos(a)-1=0

Solve for 'a' using Newton's method or some other estimation techniques and get: a=2.331122, so k=sin(a)=0.724611

The horizontal line which intersects sin(x) creating two equal regions is y=0.724611