1. **State the problem:** We are given the function $f(x,y) = x^2 y - 3xy^3$ and asked to evaluate the definite integral $$\int_1^2 f(x,y) \, dx$$ treating $y$ as a constant. 2. **Recall the integral rule:** When integrating with respect to $x$, treat $y$ as a constant. The integral of $x^n$ with respect to $x$ is $$\frac{x^{n+1}}{n+1}$$ plus a constant. 3. **Set up the integral:** $$\int_1^2 (x^2 y - 3xy^3) \, dx = y \int_1^2 x^2 \, dx - 3y^3 \int_1^2 x \, dx$$ 4. **Compute each integral:** - $$\int_1^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$ - $$\int_1^2 x \, dx = \left[ \frac{x^2}{2} \right]_1^2 = \frac{2^2}{2} - \frac{1^2}{2} = 2 - \frac{1}{2} = \frac{3}{2}$$ 5. **Substitute back:** $$y \cdot \frac{7}{3} - 3y^3 \cdot \frac{3}{2} = \frac{7y}{3} - \frac{9y^3}{2}$$ 6. **Find common denominator and simplify:** $$\frac{14y}{6} - \frac{27y^3}{6} = \frac{14y - 27y^3}{6}$$ 7. **Final answer:** $$\int_1^2 f(x,y) \, dx = \frac{14y - 27y^3}{6}$$ This matches **Option A**.