To solve the double integral $\int_0^1 \int_0^x (2xy - x) \, dy \, dx$, we can proceed as follows:

1. Evaluate the inner integral with respect to $y$: $\int_0^x (2xy - x) \, dy$

   First, split the integral: $\int_0^x 2xy \, dy - \int_0^x x \, dy$

   Now, evaluate each integral separately:

   $\int_0^x 2xy \, dy = 2x \int_0^x y \, dy$ $2x \left[ \frac{y^2}{2} \right]_0^x = 2x \left( \frac{x^2}{2} - 0 \right) = 2x \cdot \frac{x^2}{2} = x^3$

   Next, $\int_0^x x \, dy = x \int_0^x 1 \, dy = x [y]_0^x = x (x - 0) = x^2$

   So, the inner integral becomes: $\int_0^x (2xy - x) \, dy = x^3 - x^2$

2. Evaluate the outer integral with respect to $x$: $\int_0^1 (x^3 - x^2) \, dx$

   Split the integral again: $\int_0^1 x^3 \, dx - \int_0^1 x^2 \, dx$

   Evaluate each integral separately:

   $\int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} - 0 = \frac{1}{4}$

   $\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$

   Subtract the results: $\int_0^1 (x^3 - x^2) \, dx = \frac{1}{4} - \frac{1}{3} = \frac{3}{12} - \frac{4}{12} = -\frac{1}{12}$

Thus, the value of the double integral is: $\boxed{-\frac{1}{12}}$

Do you want details or have any questions?

Here are 5 questions you can ask:

1. How do you perform a double integral with different limits?
2. Can you explain Fubini's theorem and its application in double integrals?
3. How do you change the order of integration in a double integral?
4. What are some practical applications of double integrals in physics?
5. How do you evaluate double integrals over more complex regions?

Tip: Always visualize the region of integration when dealing with double integrals to understand the limits and integrand better.