What is the arc length of x=(3y)^(4/3)- ((3y)^(2/3))/(32) from y= -125 to 512

Please Help! I don't understand

Assuming that you actually mean x = 3y^(4/3) - (3/32) t^(2/3):

dx/dy = 4y^(1/3) - (1/16) y^(-1/3).

So, √(1 + (dx/dy)^2)
= √[1 + (16y^(2/3) - 1/2 + (1/256) y^(-2/3))]
= √(16y^(2/3) + 1/2 + (1/256) y^(-2/3))
= √(4y^(1/3) + (1/16) y^(-1/3))^2
= |4y^(1/3) + (1/16) y^(-1/3)|.

So, the arc length equals
∫(y = -125 to 512) |4y^(1/3) + (1/16) y^(-1/3)| dy
= ∫(y = -125 to 0) -(4y^(1/3) + (1/16) y^(-1/3)) dy + ∫(y = 0 to 512) (4y^(1/3) + (1/16) y^(-1/3)) dy
= -[3y^(4/3) + (3/32) t^(2/3)] {for y = -125 to 0} + [3y^(4/3) + (3/32) t^(2/3)] {for y = 0 to 512}
= [3 * 5^4 + (3/32) * 5^2] + [3 * 8^4 + (3/32) * 8^2]
= 453483 / 32.

I hope this helps!