Question

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

Answer 1

Find the value of h, such that the vertical line x=h divides the region R into two regions of equal area.

Answer 2

@FaiqRaees

Answer 3

I guess it's gonna be something like this: \[\int\limits_{1}^{h} \frac{ 1 }{ x ^{3} }dx=\int\limits_{h}^{3} \frac{ 1 }{ x ^{3} }dx\]

Answer 4

Thanks

Answer 5

May I ask how to solve for h?

Answer 6

Solve the integrals, then you'll have an equation of h.

Answer 7

OKay I'll try

Answer 8

Let me know =)

Answer 9

This is as far as I got

Answer 10

@math&ing001

Answer 11

Yeah, you're on the right track. Now you just plug in to get the equation.

Answer 12

Plug in what?

Answer 13

\[[-1/2x^{2}]_{1}^{h}=-\frac{ 1 }{ 2h ^{2} }+\frac{ 1 }{ 2 }\] Same goes for the RHS.

Answer 14

\[1 = -\frac{ 1 }{ 2h ^{2} }+\frac{ 1 }{ 2 }\]

Answer 15

?

Answer 16

Come ooon do I have to solve the whole thing
\[[-1/2x ^{2}]_{1}^{h} = [-1/2x ^{2}]_{h}^{3}\]\[=>-\frac{ 1 }{ 2h ^{2} }+\frac{ 1 }{ 2 }=-\frac{ 1 }{ 2*9 }+\frac{ 1 }{ 2h ^{2} }\] Now you multiply both sides by h² to get a second degree equation, you solve for h. One of the solutions should be outside the range [1,3] which you discard. That's it you have h.

Answer 17

equals about 1.64

Answer 18

Is that right?

Answer 19

I got h=sqrt(9/5)=1.34