To evaluate the double integral

$\int_{0}^{1} \int_{2}^{1} x(x + y) \, dx \, dy,$

let’s break it down step by step:

Step 1: Analyze the limits

The inner integral has limits from $2$ to $1$ for $x$, which are reversed, indicating that the direction of integration is from $x = 2$ to $x = 1$. When this happens, the bounds can be swapped, and the integral will gain a negative sign.

Thus, the integral becomes: $-\int_{0}^{1} \int_{1}^{2} x(x + y) \, dx \, dy.$

Step 2: Evaluate the inner integral

The inner integral is: $\int_{1}^{2} x(x + y) \, dx.$

Expand $x(x + y)$: $x(x + y) = x^2 + xy.$

The integral becomes: $\int_{1}^{2} (x^2 + xy) \, dx = \int_{1}^{2} x^2 \, dx + \int_{1}^{2} xy \, dx.$

(a) Compute $\int_{1}^{2} x^2 \, dx$:

$\int_{1}^{2} x^2 \, dx = \left[\frac{x^3}{3}\right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.$

(b) Compute $\int_{1}^{2} xy \, dx$:

Here, $y$ is treated as a constant. The integral is: $\int_{1}^{2} xy \, dx = y \int_{1}^{2} x \, dx = y \left[\frac{x^2}{2}\right]_{1}^{2}.$ $\left[\frac{x^2}{2}\right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}.$ Thus: $\int_{1}^{2} xy \, dx = y \cdot \frac{3}{2}.$

Combine the results of (a) and (b): $\int_{1}^{2} x(x + y) \, dx = \frac{7}{3} + \frac{3y}{2}.$

Step 3: Evaluate the outer integral

Now substitute the result of the inner integral into the outer integral: $-\int_{0}^{1} \left(\frac{7}{3} + \frac{3y}{2}\right) \, dy.$

Split into two separate integrals: $-\int_{0}^{1} \frac{7}{3} \, dy - \int_{0}^{1} \frac{3y}{2} \, dy.$

(a) Compute $-\int_{0}^{1} \frac{7}{3} \, dy$:

$-\int_{0}^{1} \frac{7}{3} \, dy = -\frac{7}{3} \int_{0}^{1} 1 \, dy = -\frac{7}{3} \cdot \left[y\right]_{0}^{1} = -\frac{7}{3} \cdot (1 - 0) = -\frac{7}{3}.$

(b) Compute $-\int_{0}^{1} \frac{3y}{2} \, dy$:

$-\int_{0}^{1} \frac{3y}{2} \, dy = -\frac{3}{2} \int_{0}^{1} y \, dy = -\frac{3}{2} \cdot \left[\frac{y^2}{2}\right]_{0}^{1}.$ $\left[\frac{y^2}{2}\right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.$ Thus: $-\int_{0}^{1} \frac{3y}{2} \, dy = -\frac{3}{2} \cdot \frac{1}{2} = -\frac{3}{4}.$

Step 4: Combine the results

$-\int_{0}^{1} \left(\frac{7}{3} + \frac{3y}{2}\right) \, dy = -\frac{7}{3} - \frac{3}{4}.$

Find a common denominator ($12$): $-\frac{7}{3} = -\frac{28}{12}, \quad -\frac{3}{4} = -\frac{9}{12}.$

Combine: $-\frac{28}{12} - \frac{9}{12} = -\frac{37}{12}.$

Final Answer:

$\boxed{-\frac{37}{12}}$

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Would you like further details or have any questions? Here are related topics to explore:

1. How do you determine if integration limits are reversed?
2. What are strategies for solving double integrals with variable limits?
3. How do constants affect integrals in nested integrations?
4. What is the geometric interpretation of a double integral?
5. How does reversing the order of integration affect the result?

Tip: Always check if the limits of integration are in the correct order; reversing them changes the sign of the integral.