1. The problem is to evaluate the expression $$\int_3^9 x \, dx$$ which is the definite integral of the function $f(x) = x$ from $x=3$ to $x=9$. 2. The formula for the definite integral of $f(x) = x$ over $[a,b]$ is: $$\int_a^b x \, dx = \left[ \frac{x^2}{2} \right]_a^b = \frac{b^2}{2} - \frac{a^2}{2}$$ 3. Applying the formula with $a=3$ and $b=9$: $$\int_3^9 x \, dx = \frac{9^2}{2} - \frac{3^2}{2} = \frac{81}{2} - \frac{9}{2}$$ 4. Simplify the expression: $$\frac{81}{2} - \frac{9}{2} = \frac{81 - 9}{2} = \frac{72}{2} = 36$$ 5. Therefore, the value of the definite integral is $36$. This means the area under the curve $y=x$ from $x=3$ to $x=9$ is $36$ square units.