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The given expression represents a double integral:

\[
\int_{0}^{1} \int_{0}^{y} (xy + y^2) \, dx \, dy
\]

### Step 1: Evaluate the inner integral (with respect to \(x\))
We first compute the integral of \((xy + y^2)\) with respect to \(x\), treating \(y\) as a constant.

\[
\int_0^y (xy + y^2) \, dx
\]

Separate the terms inside the integral:

\[
= \int_0^y xy \, dx + \int_0^y y^2 \, dx
\]

Now integrate each term:
1. \(\int_0^y xy \, dx = \frac{x^2 y}{2} \bigg|_0^y = \frac{y^3}{2}\)
2. \(\int_0^y y^2 \, dx = y^2 x \bigg|_0^y = y^3\)

Thus, the inner integral becomes:

\[
\frac{y^3}{2} + y^3 = \frac{3y^3}{2}
\]

### Step 2: Evaluate the outer integral (with respect to \(y\))
Now, we compute the outer integral:

\[
\int_0^1 \frac{3y^3}{2} \, dy
\]

Factor out \(\frac{3}{2}\):

\[
\frac{3}{2} \int_0^1 y^3 \, dy
\]

The integral of \(y^3\) is:

\[
\int_0^1 y^3 \, dy = \frac{y^4}{4} \bigg|_0^1 = \frac{1}{4}
\]

Thus, the outer integral becomes:

\[
\frac{3}{2} \times \frac{1}{4} = \frac{3}{8}
\]

### Final Answer:
The value of the double integral is \(\frac{3}{8}\).

Would you like any further explanation on this process?

Here are 5 related questions for you:
1. What is the geometric interpretation of a double integral?
2. How do you determine the limits of integration for a given region?
3. Can you generalize this technique for integrals with different limits?
4. What happens if you switch the order of integration in a double integral?
5. How would the process change if the integrand involved more complex functions?

**Tip**: Always check whether the region of integration allows for switching the order of integration to simplify the calculation.

Evaluate the double integral: \(\int_{0}^{1} \int_{0}^{y} (xy + y^2) \, dx \, dy\).

Learn how to evaluate the double integral \(\int_{0}^{1} \int_{0}^{y} (xy + y^2) \, dx \, dy\) with a detailed step-by-step solution. This guide covers essential calculus concepts, including evaluating double integrals and integration by parts. Suitable for advanced high school and college-level students.

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