Wednesday

March 4, 2015

March 4, 2015

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.

**Answer this Question**

**Related Questions**

calculus - The region R is defined by 1(</=)x(</=)2 and 0(</=)y(</=)...

CALCULUS problem - There are four parts to this one question, and would really ...

Calculus - Find the number b such that the line y = b divides the region bounded...

calculus - Find the number b such that the line y = b divides the region bounded...

calculus - Find the number b such that the line y = b divides the region bounded...

calc - 1. Let R be the region bounded by the x-axis, the graph of y=sqr(x) , and...

Calculus - The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3 A...

cal 2 - Find the number b such that the line y = b divides the region bounded by...

calculus II - We're doing areas by integrals now, with 2 eqns. I have a few ...

Calculus - Let R be the region in the first quadrant under the graph of y=1/sqrt...