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The integral to evaluate is:

\[
\int_{0}^{3} \int_{1}^{2} x^2 \, dy \, dx
\]

To solve this, we proceed as follows:

1. First, evaluate the inner integral with respect to \( y \), treating \( x^2 \) as a constant:

\[
\int_{1}^{2} x^2 \, dy = x^2 \cdot \left[ y \right]_{1}^{2} = x^2 \cdot (2 - 1) = x^2
\]

So, the expression simplifies to:

\[
\int_{0}^{3} x^2 \, dx
\]

2. Now, evaluate the outer integral:

\[
\int_{0}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} = 9
\]

Thus, the result of the integral is:

\[
\int_{0}^{3} \int_{1}^{2} x^2 \, dy \, dx = 9
\]

Would you like a breakdown of each step or have any questions?

Here are five related questions you might find useful:
1. How would the answer change if the limits of \( x \) were different?
2. What happens if \( x^2 \) is replaced with another function, say \( x^3 \)?
3. How would you approach this problem if the integration limits for \( y \) were functions of \( x \)?
4. How do we evaluate double integrals in polar coordinates?
5. What techniques can simplify more complex iterated integrals?

**Tip:** Always check if the order of integration can simplify the problem before calculating.

Evaluate \( \int_{0}^{3} \int_{1}^{2} x^2 \, dy \, dx \)

Learn how to evaluate the double integral \( \int_{0}^{3} \int_{1}^{2} x^2 \, dy \, dx \) with a step-by-step solution, using iterated integration and the Fundamental Theorem of Calculus. Ideal for calculus students.

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