calculus
posted by Stella .
The region R is defined by 1(</=)x(</=)2 and 0(</=)y(</=)1/(x^3).
Find the number 'b' such that the line y=b divides R into two parts of equal area.

Evaluate the integral of 1/x^3 from x = 1 to x = 2. That is the enclosed area of the region. I get 3/8. See what you get.
Then pick the upper limit of integration, b, such that the integral of 1/x^3 from 1 to b is half the number you got for the full integral.