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The problem you've shared asks us to evaluate the double integral of the function \( f(x, y) = x^2 + y^3 \) over the region \( R = [-1, 1] \times [0, 1] \).

The double integral to evaluate is:

\[
\iint_R f(x, y) \, dx \, dy = \int_{0}^{1} \int_{-1}^{1} (x^2 + y^3) \, dx \, dy
\]

Let's break this down and solve it step by step.

### Step 1: Evaluate the inner integral with respect to \( x \).
\[
\int_{-1}^{1} (x^2 + y^3) \, dx
\]

This integral can be separated into two parts:
\[
\int_{-1}^{1} x^2 \, dx + \int_{-1}^{1} y^3 \, dx
\]

#### First part:
\[
\int_{-1}^{1} x^2 \, dx
\]
The integral of \( x^2 \) with respect to \( x \) is:
\[
\frac{x^3}{3}
\]
Evaluating from \( -1 \) to \( 1 \):
\[
\left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}
\]

#### Second part:
\[
\int_{-1}^{1} y^3 \, dx
\]
Since \( y^3 \) is independent of \( x \), it can be factored out of the integral:
\[
y^3 \int_{-1}^{1} dx
\]
The integral of \( dx \) over \( -1 \) to \( 1 \) is:
\[
\int_{-1}^{1} dx = 1 - (-1) = 2
\]
Thus, the second part becomes:
\[
2y^3
\]

#### Putting it all together:
\[
\int_{-1}^{1} (x^2 + y^3) \, dx = \frac{2}{3} + 2y^3
\]

### Step 2: Evaluate the outer integral with respect to \( y \).
Now, we integrate the result with respect to \( y \):
\[
\int_{0}^{1} \left(\frac{2}{3} + 2y^3\right) \, dy
\]
This can also be separated into two integrals:
\[
\frac{2}{3} \int_{0}^{1} dy + 2 \int_{0}^{1} y^3 \, dy
\]

#### First part:
\[
\frac{2}{3} \int_{0}^{1} dy = \frac{2}{3} \times 1 = \frac{2}{3}
\]

#### Second part:
\[
2 \int_{0}^{1} y^3 \, dy
\]
The integral of \( y^3 \) with respect to \( y \) is:
\[
\frac{y^4}{4}
\]
Evaluating from \( 0 \) to \( 1 \):
\[
\left[\frac{y^4}{4}\right]_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}
\]
Thus, the second part becomes:
\[
2 \times \frac{1}{4} = \frac{1}{2}
\]

### Final result:
Adding both parts together:
\[
\frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6}
\]

Thus, the value of the double integral is:

\[
\iint_R f(x, y) \, dx \, dy = \frac{7}{6}
\]

Would you like more details or have any questions?

Here are some related questions:
1. How do you evaluate a double integral over a different region of integration?
2. What changes when the limits of integration are different?
3. How would the result change if the function were \( f(x, y) = x^2 \cdot y^3 \)?
4. Can you explain the geometric interpretation of the double integral?
5. How does Fubini's Theorem assist in evaluating double integrals?

**Tip:** When working with double integrals, it’s often helpful to separate the integrals if the function allows for easy integration with respect to one variable at a time.

Learn how to evaluate the double integral of f(x, y) = x^2 + y^3 over the rectangular region R = [-1, 1] × [0, 1]. Detailed step-by-step solution covering integration with respect to x and y, applying Fubini's Theorem. Suitable for advanced undergraduate mathematics students.

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