So you want the
Integral[1/x^3] from 1 to a = integral[1/x^3] from a to 2
(the integral of 1/x^3 is -1/(2x^2) )
then
-1/(2a^2) - (-1/2) = -1/(2(4)) - (-1/2a^2)
-1/(2a^2) + 1/2 = -1/8 + 1/(2a^2)
1/2 + 1/8 = 2/(2a^2)
5/8 = 1/a^2
5a^2 = 8
a^2 = 8/5 = 1.6
a = √1.6
The region R is defined by 1(</=)x(</=)2 and 0(</=)y(</=)1/(x^3).
a) Find the number 'a' such that the line x=a divides R into two parts of equal area.
b) Then find the number 'b' such that the line y=b divides R into two parts of equal area.
2 answers
To find the answer for b) though, do you have to find the area of the smaller rectangular portion within the region R, and then find the halves of the remaining area of the region? But then how would you work that out to find what b equals? I am confused.
1 answer