Determine the volume of the region evaluating the double integral: \( \int_{0}^{1} \int_{x^{2}}^{x}(x y) d y d x \)

To solve the given double integral \( \int_{0}^{1} \int_{x^{2}}^{x}(x y) d y d x \), we first evaluate the inner integral with respect to \( y \). 1. **Inner Integral** calculation: \[ \int_{x^{2}}^{x} (x y) \, dy = x \int_{x^{2}}^{x} y \, dy \] The integral of \( y \) is: \[ \int y \, dy = \frac{y^2}{2} \] Evaluating from \( y = x^2 \) to \( y = x \): \[ = x \left[ \frac{y^2}{2} \right]_{x^2}^{x} = x \left( \frac{x^2}{2} - \frac{(x^2)^2}{2} \right) \] \[ = x \left( \frac{x^2}{2} - \frac{x^4}{2} \right) = x \left( \frac{x^2 - x^4}{2} \right) = \frac{x^3 - x^5}{2} \] 2. **Outer Integral** calculation: Now we evaluate the outer integral: \[ \int_{0}^{1} \frac{x^3 - x^5}{2} \, dx = \frac{1}{2} \int_{0}^{1} (x^3 - x^5) \, dx \] We can split this into two separate integrals: \[ = \frac{1}{2} \left( \int_{0}^{1} x^3 \, dx - \int_{0}^{1} x^5 \, dx \right) \] Calculating each integral: \[ \int x^3 \, dx = \frac{x^4}{4} \quad \text{and evaluated from 0 to 1 gives } \frac{1}{4} \] \[ \int x^5 \, dx = \frac{x^6}{6} \quad \text{and evaluated from 0 to 1 gives } \frac{1}{6} \] 3. **Final Calculation**: Putting it all together: \[ = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{6} \right) \] Finding a common denominator (which is 12): \[ = \frac{1}{2} \left( \frac{3}{12} - \frac{2}{12} \right) = \frac{1}{2} \left( \frac{1}{12} \right) = \frac{1}{24} \] Thus, the volume of the region evaluated by the double integral is \( \frac{1}{24} \).