The given double integral is:

$I = \iint_R x y (x + y) \, dx \, dy$

where the region $R$ is bounded by the curves $y = x^2$ and $y = x$.

Step 1: Understand the region $R$

The curves $y = x^2$ and $y = x$ intersect at two points. To find the points of intersection, we set the equations equal to each other:

$x^2 = x$

This simplifies to:

$x(x - 1) = 0$

So the intersection points are $x = 0$ and $x = 1$. Therefore, the region $R$ is bounded by $y = x^2$ (the lower curve) and $y = x$ (the upper curve), for $x$ ranging from $0$ to $1$.

Step 2: Set up the limits of integration

The integral is a double integral with respect to $x$ and $y$. The bounds for $x$ range from 0 to 1, and for each $x$, $y$ ranges from $x^2$ to $x$. Therefore, the limits of integration are:

$I = \int_0^1 \int_{x^2}^{x} x y (x + y) \, dy \, dx$

Step 3: Simplify the integrand

Expand the integrand $x y (x + y)$:

$x y (x + y) = x^2 y + x y^2$

Thus, the integral becomes:

$I = \int_0^1 \int_{x^2}^{x} \left( x^2 y + x y^2 \right) \, dy \, dx$

Step 4: Integrate with respect to $y$

Now, we can integrate each term with respect to $y$:

1. For $\int_{x^2}^{x} x^2 y \, dy$: $\int_{x^2}^{x} x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_{x^2}^{x} = x^2 \left( \frac{x^2}{2} - \frac{x^4}{2} \right) = \frac{x^4}{2} - \frac{x^6}{2}$

2. For $\int_{x^2}^{x} x y^2 \, dy$: $\int_{x^2}^{x} x y^2 \, dy = x \left[ \frac{y^3}{3} \right]_{x^2}^{x} = x \left( \frac{x^3}{3} - \frac{x^6}{3} \right) = \frac{x^4}{3} - \frac{x^7}{3}$

Step 5: Combine the results

Now, combining both results:

$I = \int_0^1 \left( \frac{x^4}{2} - \frac{x^6}{2} + \frac{x^4}{3} - \frac{x^7}{3} \right) \, dx$

Step 6: Simplify the integrand

Combine like terms:

$I = \int_0^1 \left( \frac{5x^4}{6} - \frac{x^6}{2} - \frac{x^7}{3} \right) \, dx$

Step 7: Integrate with respect to $x$

Now, integrate each term:

1. For $\int_0^1 \frac{5x^4}{6} \, dx$: $\int_0^1 \frac{5x^4}{6} \, dx = \frac{5}{6} \left[ \frac{x^5}{5} \right]_0^1 = \frac{5}{6} \times \frac{1}{5} = \frac{1}{6}$

2. For $\int_0^1 \frac{x^6}{2} \, dx$: $\int_0^1 \frac{x^6}{2} \, dx = \frac{1}{2} \left[ \frac{x^7}{7} \right]_0^1 = \frac{1}{2} \times \frac{1}{7} = \frac{1}{14}$

3. For $\int_0^1 \frac{x^7}{3} \, dx$: $\int_0^1 \frac{x^7}{3} \, dx = \frac{1}{3} \left[ \frac{x^8}{8} \right]_0^1 = \frac{1}{3} \times \frac{1}{8} = \frac{1}{24}$

Step 8: Combine the results

Now, sum the results:

$I = \frac{1}{6} - \frac{1}{14} - \frac{1}{24}$

To combine these, find a common denominator. The least common denominator of 6, 14, and 24 is 168:

$I = \frac{28}{168} - \frac{12}{168} - \frac{7}{168} = \frac{28 - 12 - 7}{168} = \frac{9}{168} = \frac{3}{56}$

Final Answer:

$I = \frac{3}{56}$

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Would you like further details on any step, or do you have additional questions? Here are 5 related questions for exploration:

1. How would the result change if the bounds were altered to different functions?
2. Can this double integral be interpreted geometrically, and how?
3. What happens if we reverse the order of integration in this case?
4. How would you solve a similar problem using polar coordinates?
5. What are the benefits of using symbolic computation software to evaluate such integrals?

Tip: When performing multiple integrations, always check the limits of integration first to ensure they match the region described in the problem.