Asked by Martha
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.
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.
Answers
Answered by
Reiny
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
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
Answered by
Scott
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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