You set up was almost accurate. Remember the arc length formula:
If f'(y) is continuous on the interval [a,b], then the length of the curve x = f(y), a ≤ y ≤ b should be;
L = ∫ᵇ ₐ √1 + [f'(y)]^2 * dy
We have to find the length of the curve given x = √y - 2y, and 1 ≤ y ≤ 4. You can tell your limits would be 1 to 4, and you are right on that part. But f'(y) would be rather...
f'(y) = 1/(2√y) - 2
So the integral would be:
∫⁴₁ √1 + (1/(2√y) - 2)² dy
Using a calculator we would receive the solution 5.832. Their is a definite curve, as represented below;