what is the of the curve 3y^(3/4)-(3/32)y^(2/3) from y=-64 to y=27?

Question

This is what I did:

1. x'=(4)x^(1/3)-(1/16)x^(-1/3)
2. integral of (sqrt(1+(x')^2)
3. integral of (sqrt(1+(4097/256)y^(1/9)-(1/2)y^(-1/9) )
4. integral of ((sqrt(1+(4097/256)y^(1/9)-(1/2)y^(-1/9) ) from 5 to 10

can anyone tell me what am I doing wron here???

Answer 1

x'² = 16 y^(2/3) - 1/2 + (1/256)y^(-2/3)

1 + x'² = 1 + 16 y^(2/3) - 1/2 + (1/256)y^(-2/3)
= 16 y^(2/3) + 1/2 + (1/256)y^(-2/3)
= [4 y^(1/3) + (1/16) y^(-1/3)]²

[y^a]^b = y^(a*b), not y^(a^b)