1. **Problem Statement:** Evaluate the integral $$\int_{2}^{4} \int_{x}^{4} 6x^{2} \, dy \, dx$$ and reverse the order of integration. 2. **Original Integral Setup:** The integral is $$\int_{2}^{4} \int_{y=x}^{4} 6x^{2} \, dy \, dx$$ where the inner integral integrates with respect to $y$ from $y=x$ to $y=4$, and the outer integral integrates with respect to $x$ from $2$ to $4$. 3. **Region of Integration:** The region is bounded by $x=2$, $x=4$, $y=x$, and $y=4$. 4. **Reversing the Order of Integration:** - Original limits: $x$ from 2 to 4, $y$ from $x$ to 4. - For fixed $y$, $x$ varies from 2 up to $y$ (since $y \geq x$). - $y$ varies from 2 (lowest $y$ on $x=2$) to 4. So the reversed integral is: $$\int_{y=2}^{4} \int_{x=2}^{y} 6x^{2} \, dx \, dy$$ 5. **Evaluate the inner integral:** $$\int_{2}^{y} 6x^{2} \, dx = 6 \int_{2}^{y} x^{2} \, dx = 6 \left[ \frac{x^{3}}{3} \right]_{2}^{y} = 6 \left( \frac{y^{3}}{3} - \frac{8}{3} \right) = 2(y^{3} - 8)$$ 6. **Evaluate the outer integral:** $$\int_{2}^{4} 2(y^{3} - 8) \, dy = 2 \int_{2}^{4} (y^{3} - 8) \, dy = 2 \left[ \frac{y^{4}}{4} - 8y \right]_{2}^{4}$$ Calculate the values: $$= 2 \left( \left( \frac{4^{4}}{4} - 8 \times 4 \right) - \left( \frac{2^{4}}{4} - 8 \times 2 \right) \right) = 2 \left( \left( \frac{256}{4} - 32 \right) - \left( \frac{16}{4} - 16 \right) \right)$$ $$= 2 \left( (64 - 32) - (4 - 16) \right) = 2 \left( 32 - (-12) \right) = 2 \times 44 = 88$$ **Final answer:** $$88$$