To evaluate the given double integral:

$I = \int_R x^2y \, dA$

where $R$ is the triangular region defined as: $R = \{(x, y) \mid 0 \leq x \leq 1, \, 0 \leq y \leq x\},$

we will set up the integral in terms of the region bounds.

Step 1: Set up the integral

The region $R$ suggests that $x$ ranges from $0$ to $1$, and for each $x$, $y$ ranges from $0$ to $x$. Therefore, the integral becomes: $I = \int_{x=0}^1 \int_{y=0}^x x^2y \, dy \, dx.$

Step 2: Integrate with respect to $y$

First, treat $x^2$ as a constant while integrating with respect to $y$: $\int_{y=0}^x x^2y \, dy = x^2 \int_{y=0}^x y \, dy = x^2 \left[ \frac{y^2}{2} \right]_0^x = x^2 \cdot \frac{x^2}{2} = \frac{x^4}{2}.$

So, the integral becomes: $I = \int_{x=0}^1 \frac{x^4}{2} \, dx.$

Step 3: Integrate with respect to $x$

Now integrate $\frac{x^4}{2}$ with respect to $x$: $\int_{x=0}^1 \frac{x^4}{2} \, dx = \frac{1}{2} \int_{x=0}^1 x^4 \, dx = \frac{1}{2} \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10}.$

Final Answer:

$I = \frac{1}{10}.$

Would you like additional explanation or verification of any step? Here are some related questions for deeper exploration:

1. How do you determine the bounds of integration for a triangular region?
2. Could this integral be evaluated using polar coordinates? Why or why not?
3. How does the order of integration affect the computation in such problems?
4. What is the geometric interpretation of the result $I = \frac{1}{10}$?
5. How would the solution change if $R$ were a different triangular region?

Tip: Always visualize the region $R$ and confirm its bounds before setting up the integral!