We need to evaluate the double integral $\int_{1}^{2} \int_{0}^{x^2} \frac{y^2}{x} \, dy \, dx$.

1. Problem Description

We need to evaluate the double integral $\int_{1}^{2} \int_{0}^{x^2} \frac{y^2}{x} \, dy \, dx$.

2. Solution Steps

First, we integrate with respect to $y$:

$\int_{0}^{x^2} \frac{y^2}{x} \, dy = \frac{1}{x} \int_{0}^{x^2} y^2 \, dy$

$= \frac{1}{x} \left[ \frac{y^3}{3} \right]_{0}^{x^2} = \frac{1}{x} \left( \frac{(x^2)^3}{3} - \frac{0^3}{3} \right) = \frac{1}{x} \left( \frac{x^6}{3} \right) = \frac{x^5}{3}$.

Now, we integrate with respect to $x$:

$\int_{1}^{2} \frac{x^5}{3} \, dx = \frac{1}{3} \int_{1}^{2} x^5 \, dx$

$= \frac{1}{3} \left[ \frac{x^6}{6} \right]_{1}^{2} = \frac{1}{3} \left( \frac{2^6}{6} - \frac{1^6}{6} \right) = \frac{1}{3} \left( \frac{64}{6} - \frac{1}{6} \right)$

$= \frac{1}{3} \left( \frac{63}{6} \right) = \frac{1}{3} \left( \frac{21}{2} \right) = \frac{21}{6} = \frac{7}{2}$.

3. Final Answer

$\frac{7}{2}$