To evaluate the double integral $\int_0^1\int_0^{y} (2xy - x) \, dx \, dy$, we will integrate step-by-step.

First, consider the inner integral:

$\int_0^{y} (2xy - x) \, dx$

We can split this into two separate integrals:

$\int_0^{y} 2xy \, dx - \int_0^{y} x \, dx$

Evaluate each integral separately.

For the first integral: $\int_0^{y} 2xy \, dx = 2y \int_0^{y} x \, dx = 2y \left[\frac{x^2}{2}\right]_0^{y} = 2y \left(\frac{y^2}{2} - 0\right) = y^3$

For the second integral: $\int_0^{y} x \, dx = \left[\frac{x^2}{2}\right]_0^{y} = \frac{y^2}{2} - 0 = \frac{y^2}{2}$

Now, subtract the results of the two integrals: $y^3 - \frac{y^2}{2}$

This result is the evaluation of the inner integral. Now, we proceed to the outer integral: $\int_0^1 \left(y^3 - \frac{y^2}{2}\right) \, dy$

We can split this into two integrals as well:

$\int_0^1 y^3 \, dy - \int_0^1 \frac{y^2}{2} \, dy$

Evaluate each integral separately.

For the first integral: $\int_0^1 y^3 \, dy = \left[\frac{y^4}{4}\right]_0^1 = \frac{1}{4} - 0 = \frac{1}{4}$

For the second integral: $\int_0^1 \frac{y^2}{2} \, dy = \frac{1}{2} \int_0^1 y^2 \, dy = \frac{1}{2} \left[\frac{y^3}{3}\right]_0^1 = \frac{1}{2} \left(\frac{1}{3} - 0\right) = \frac{1}{6}$

Now, subtract the results of the two integrals: $\frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$

Thus, the value of the double integral is:

$\boxed{\frac{1}{12}}$

Do you want details or have any questions?

Here are 5 numbered questions for you to ask:

1. How do you solve a definite integral with variable limits?
2. What are the applications of double integrals in real-life problems?
3. Can you explain the concept of changing the order of integration in double integrals?
4. How do you handle triple integrals?
5. What is the geometric interpretation of double integrals?

Tip: When splitting integrals, always ensure each part is integrated correctly before combining results to avoid errors.