What's the trouble? I assume you know how to integrate, so you just want to solve
∫[1,a] 1/x^3 dx = ∫[a,2] 1/x^3 dx
-1/2x^2 [1,a] = -1/2x^2 [a,2]
-1/2a^2 + 1/2 = -1/8 + 1/2a^2
1/a^2 = 5/8
a^2 = 8/5
a = √(8/5)
Now do a similar integral using x = 1/∛y to find b.
Sketch the region R defined by 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1/x3.
a) Find (exactly) the number a such that the line x = a divides R into two
parts of equal area
b) Then find (to 3 places) the number b such that the line y = b divides R
into two parts of equal area
1 answer
1 answer