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October 1, 2014

October 1, 2014

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

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 8:49amSo 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 3:23amTo 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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