To evaluate the double integral $\iint_R (x + y^2) \, dx \, dy$ over the region $R$ enclosed by the curves $y = x^2$, $x = 2$, and $y = 1$, let's break down the problem step by step.

1. Identify the Region $R$:

The region $R$ is bounded by:

• The curve $y = x^2$ (a parabola).
• The vertical line $x = 2$.
• The horizontal line $y = 1$.

Intersection Points:

• At $y = 1$, substitute into $y = x^2$ to find $x$. $1 = x^2 \quad \Rightarrow \quad x = \pm 1.$ So the parabola intersects the line $y = 1$ at the points $(-1, 1)$ and $(1, 1)$.

Thus, the region $R$ is bounded by the parabola from $x = -1$ to $x = 2$, and the limits on $y$ are determined by the parabola and the line $y = 1$.

2. Set up the Double Integral:

Since the region is simpler to describe in terms of $y$, we can set up the integral in the form: $\iint_R (x + y^2) \, dx \, dy = \int_{y=0}^{1} \int_{x=-\sqrt{y}}^{2} (x + y^2) \, dx \, dy.$

3. Evaluate the Integral:

Step 1: Integrate with respect to $x$:

$\int_{-\sqrt{y}}^{2} (x + y^2) \, dx = \left[ \frac{x^2}{2} + y^2x \right]_{x=-\sqrt{y}}^{x=2}.$

Substitute the limits:

• When $x = 2$: $\frac{2^2}{2} + y^2(2) = 2 + 2y^2.$
• When $x = -\sqrt{y}$: $\frac{(-\sqrt{y})^2}{2} + y^2(-\sqrt{y}) = \frac{y}{2} - y^2\sqrt{y}.$

So, the integral becomes: $\left( 2 + 2y^2 \right) - \left( \frac{y}{2} - y^2\sqrt{y} \right) = 2 + 2y^2 - \frac{y}{2} + y^2\sqrt{y}.$

Step 2: Integrate with respect to $y$:

Now, integrate the result with respect to $y$ from 0 to 1: $\int_0^1 \left( 2 + 2y^2 - \frac{y}{2} + y^2\sqrt{y} \right) \, dy.$

Let's evaluate each term separately:

1. $\int_0^1 2 \, dy = 2 \times 1 = 2.$
2. $\int_0^1 2y^2 \, dy = \frac{2y^3}{3} \Big|_0^1 = \frac{2}{3}.$
3. $\int_0^1 -\frac{y}{2} \, dy = -\frac{1}{2} \times \frac{y^2}{2} \Big|_0^1 = -\frac{1}{4}.$
4. $\int_0^1 y^2\sqrt{y} \, dy = \int_0^1 y^{5/2} \, dy = \frac{y^{7/2}}{7/2} \Big|_0^1 = \frac{2}{7}.$

Final Answer:

Adding up all the results: $2 + \frac{2}{3} - \frac{1}{4} + \frac{2}{7}.$

To combine these fractions: [ = 2 + \frac{28}{42} - \frac{10.5