We need to find the area under the curve of the function $y = 9x^3 + 5x$ from $x = 1$ to $x = 2$. To do this, we use the definite integral:

$\text{Area} = \int_1^2 (9x^3 + 5x) \, dx$

Step 1: Integrate the function

We can integrate each term separately:

1. The integral of $9x^3$ is $\frac{9x^4}{4}$.
2. The integral of $5x$ is $\frac{5x^2}{2}$.

Thus, the indefinite integral of $9x^3 + 5x$ is:

$\int (9x^3 + 5x) \, dx = \frac{9x^4}{4} + \frac{5x^2}{2}$

Step 2: Evaluate the definite integral

Now, evaluate the integral from $x = 1$ to $x = 2$:

$\left[ \frac{9x^4}{4} + \frac{5x^2}{2} \right]_1^2$

Substitute $x = 2$:

$\frac{9(2)^4}{4} + \frac{5(2)^2}{2} = \frac{9(16)}{4} + \frac{5(4)}{2} = \frac{144}{4} + \frac{20}{2} = 36 + 10 = 46$

Substitute $x = 1$:

$\frac{9(1)^4}{4} + \frac{5(1)^2}{2} = \frac{9(1)}{4} + \frac{5(1)}{2} = \frac{9}{4} + \frac{5}{2} = \frac{9}{4} + \frac{10}{4} = \frac{19}{4}$

Step 3: Compute the final result

Now, subtract the two results:

$46 - \frac{19}{4} = \frac{184}{4} - \frac{19}{4} = \frac{165}{4} = 41.25$

Final Answer:

The area under the curve from $x = 1$ to $x = 2$ is $\boxed{41.25}$.

Would you like further details or explanations? Here's some related content to explore:

1. How do definite integrals represent areas under curves?
2. What happens if we calculate the area between curves?
3. What other techniques exist for solving definite integrals?
4. How does the Fundamental Theorem of Calculus apply here?
5. What are some real-world applications of definite integrals?

Tip: Remember that evaluating definite integrals requires finding the antiderivative and then applying the bounds!