Posted by **Martha** on Thursday, January 21, 2010 at 3:21am.

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.

- calculus -
**Reiny**, Thursday, January 21, 2010 at 8:49am
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

- calculus -
**Scott**, Friday, January 22, 2010 at 3:23am
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.

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