The region R is defined by 1(</=)x(</=)2 and 0(</=)y(</=)1/(x^3).
Find the number 'b' such that the line y=b divides R into two parts of equal area.
Evaluate the integral of 1/x^3 from x = 1 to x = 2. That is the enclosed area of the region. I get 3/8. See what you get.
Then pick the upper limit of integration, b, such that the integral of 1/x^3 from 1 to b is half the number you got for the full integral.
To find the number 'b' such that the line y = b divides region R into two parts of equal area, we need to calculate the area of region R and then find the value of b that splits the area in half.
First, let's find the boundaries of region R. We are given that 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1/(x^3).
To find the area of region R, we need to integrate the function with respect to y and then integrate the resulting expression with respect to x over the given bounds.
The equation for the line y = b can be rewritten as y = b(x) = b.
To find the area of region R, we need to calculate the double integral of 1 with respect to y and x, over the given bounds. The integrand is 1 because we want to find the area of region R.
∬R 1 dA = ∫(1 ≤ x ≤ 2) ∫(0 ≤ y ≤ 1/(x^3)) 1 dydx
∬R 1 dA = ∫(1 ≤ x ≤ 2) ∫(0 ≤ y ≤ 1/(x^3)) dydx
Now, we need to find the value of b such that the area of region R divided by the line y = b is equal for both parts. Let's call the area of region R A.
A = 1/2 * ∫(1 ≤ x ≤ 2) ∫(0 ≤ y ≤ b) 1 dydx
To solve this equation for b, we need to express the double integral in terms of b and then solve for b.
A = 1/2 * ∫(1 ≤ x ≤ 2) ∫(0 ≤ y ≤ b) 1 dydx
We can evaluate the inner integral first:
A = 1/2 * ∫(1 ≤ x ≤ 2) [y] from 0 to b dx
A = 1/2 * ∫(1 ≤ x ≤ 2) (b - 0) dx
A = 1/2 * b * ∫(1 ≤ x ≤ 2) dx
A = 1/2 * b * [x] from 1 to 2
A = 1/2 * b * (2 - 1)
A = 1/2 * b
Now, we have the equation A = 1/2 * b. Since we want to split the area of region R in half, we can solve this equation for b:
A = 1/2 * b
1/2 * b = 1/2 * A
b = A
Therefore, the value of b that divides the area of region R into two equal parts is equal to the area of region R.