Compute the double integral: \( \int_{0}^{4} \int_{0}^{\frac{x}{2}} x y^{3} d x d y \) (5 Marks)

To calculate the double integral \[ \int_{0}^{4} \int_{0}^{\frac{x}{2}} x y^{3} \, dy \, dx, \] we first need to integrate with respect to \( y \) while treating \( x \) as a constant. 1. **Inner Integral**: Compute \[ \int_{0}^{\frac{x}{2}} x y^{3} \, dy. \] Since \( x \) is constant for this integral, we can factor it out: \[ = x \int_{0}^{\frac{x}{2}} y^{3} \, dy. \] Now, we compute the integral: \[ \int y^3 \, dy = \frac{y^4}{4}. \] Evaluating from \( 0 \) to \( \frac{x}{2} \): \[ = \left[ \frac{1}{4} \left( \frac{x}{2} \right)^4 \right] - \left[ \frac{1}{4} (0)^4 \right] = \frac{1}{4} \cdot \frac{x^4}{16} = \frac{x^4}{64}. \] Thus, the inner integral becomes: \[ x \cdot \frac{x^4}{64} = \frac{x^5}{64}. \] 2. **Outer Integral**: Now we need to compute the outer integral: \[ \int_{0}^{4} \frac{x^5}{64} \, dx. \] Factoring out the constant \( \frac{1}{64} \): \[ = \frac{1}{64} \int_{0}^{4} x^5 \, dx. \] Now we evaluate this integral: \[ \int x^5 \, dx = \frac{x^6}{6}. \] Evaluating from \( 0 \) to \( 4 \): \[ = \left[ \frac{4^6}{6} - \frac{0^6}{6} \right] = \frac{4096}{6} = \frac{2048}{3}. \] Thus, the outer integral becomes: \[ \frac{1}{64} \cdot \frac{2048}{3} = \frac{2048}{192} = \frac{64}{3}. \] In conclusion, the value of the double integral is \[ \boxed{\frac{64}{3}}. \]