Displacement is the definite integral of velocity over the time interval $[0, 3]$:

Evaluate the double integral over the rectangular region $R = [1, 2] \times [0, 3]$:

• Inner integral with respect to $x$:
  $$\int_1^2 (4xy - y^2)\, dx = \left[ 2x^2 y - y^2 x \right]_{x=1}^2$$
  $$= (2(4)y - 2y^2) - (2(1)y - y^2) = (8y - 2y^2) - (2y - y^2) = 6y - y^2$$
• Outer integral with respect to $y$:
  $$\int_0^3 (6y - y^2)\, dy = \left[ 3y^2 - \frac{y^3}{3} \right]_0^3 = 3(9) - \frac{27}{3} = 27 - 9 = 18$$

Substitute $t = -\log_e x \implies x = e^{-t} \implies dx = -e^{-t}\, dt$.

When $x = 0$, $t \to +\infty$.
When $x = 1$, $t = 0$.

The integral transforms to:

By definition of the Gamma function $\Gamma(n) = \int_0^\infty t^{n-1} e^{-t}\, dt$ with $n = 1/2$:

Split the integral at $x = 0$ where $|x|$ changes definition:

Evaluating each part:

Total value: