To evaluate the double integral $\iint_R (x+y) \, dx \, dy$ over the region $R$ enclosed by the lines $x=1$, $x=2$, $y=-x+2$, and $y=x+1$, we need to determine the bounds for the region $R$ and set up the integral.

Step 1: Determine the Region of Integration

The region $R$ is bounded by the following lines:

1. $x = 1$ (vertical line)
2. $x = 2$ (vertical line)
3. $y = -x + 2$ (a line with a negative slope)
4. $y = x + 1$ (a line with a positive slope)

Intersection Points:

Let's find the intersection points of the lines $y = -x + 2$ and $y = x + 1$ to determine the exact bounds of $y$ for a given $x$.

Set $y = -x + 2$ equal to $y = x + 1$: $-x + 2 = x + 1$ $2 = 2x + 1$ $x = \frac{1}{2}$ Substitute $x = \frac{1}{2}$ into either equation to find $y$: $y = \frac{1}{2} + 1 = \frac{3}{2}$

So, the intersection point of the two lines is $\left(\frac{1}{2}, \frac{3}{2}\right)$.

Step 2: Set Up the Double Integral

For $x$ between 1 and 2, $y$ will range between the lines $y = x + 1$ (upper bound) and $y = -x + 2$ (lower bound).

Therefore, the double integral is: $\int_{x=1}^{2} \int_{y=-x+2}^{x+1} (x + y) \, dy \, dx$

Step 3: Evaluate the Inner Integral

First, evaluate the inner integral with respect to $y$: $\int_{y=-x+2}^{x+1} (x + y) \, dy$ The antiderivative of $x + y$ with respect to $y$ is: $\int (x + y) \, dy = xy + \frac{y^2}{2}$

Evaluate this from $y = -x + 2$ to $y = x + 1$: [ \left[ xy + \frac{y^2}{2} \right]_{y=-x+2}^{y=x+1} = \left[ x(x + 1) + \frac{(x + 1)^2}{2} \right] - \left[ x(-x + 2) + \frac{(-x + 2)^2}{2} \right] ]

Step 4: Simplify the Expression

Substitute and simplify each expression:

Upper bound $y = x + 1$: $x(x + 1) + \frac{(x + 1)^2}{2} = x^2 + x + \frac{x^2 + 2x + 1}{2} = x^2 + x + \frac{x^2}{2} + x + \frac{1}{2}$ $= \frac{3x^2}{2} + 2x + \frac{1}{2}$

Lower bound $y = -x + 2$: $x(-x + 2) + \frac{(-x + 2)^2}{2} = -x^2 + 2x + \frac{x^2 - 4x + 4}{2} = -x^2 + 2x + \frac{x^2}{2} - 2x + 2$ $= -\frac{x^2}{2} + 2$

Subtract the lower bound from the upper bound: $\frac{3x^2}{2} + 2x + \frac{1}{2} - \left(-\frac{x^2}{2} + 2\right) = 2x^2 + 2x - \frac{3}{2}$

Step 5: Integrate with Respect to $x$

Now, integrate this with respect to $x$ from 1 to 2: $\int_{1}^{2} \left(2x^2 + 2x - \frac{3}{2}\right) dx$

The antiderivative is: $\frac{2x^3}{3} + x^2 - \frac{3x}{2}$

Evaluate from $x = 1$ to $x = 2$: [ \left[\frac{2(2)^3}{3} + (2)^2 - \frac{3(2)}{2}\right] - \left[\frac{2(1)^3}{3} + (1)^2 - \frac{3(1)}{2}\right] ] [ = \left[\frac{16}{3} + 4 - 3\right] - \left[\frac{2}{3} + 1 - \frac{3}{2}\right] ] $= \left(\frac{16}{3} + 1\right) - \left(\frac{2}{3} + \frac{1}{2}\right)$ Simplifying: $= \frac{19}{3} - \frac{7}{6} = \frac{38}{6} - \frac{7}{6} = \frac{31}{6}$

Final Answer:

The value of the double integral is $\frac{31}{6}$.

Would you like further details or have any questions?

Here are some related questions you might find useful:

1. How would the integral change if the bounds for $x$ were different?
2. Can you compute the area of the region $R$ directly?
3. How would the integral be affected if the function inside the integral were more complex?
4. What if the region were bounded by different lines, such as curves?
5. How is the double integral interpreted geometrically?

Tip: When setting up double integrals, always sketch the region of integration to avoid errors in determining bounds.