[THOUGHT] To compute the contour integral of $f(z) = z^2$ over the upper semicircular arc $C$, we can use two approaches: direct parametrization of $C$ or leveraging the closed contour $\Gamma = C \cup L$. Since $f(z)$ is entire, the integral over the closed contour $\Gamma$ is zero. Thus, the integral over $C$ is the negative of the integral over $L$.

First, compute the integral over $L$ (the real line segment from $-5/2$ to $5/2$):

Therefore, the integral over $C$ is $-\frac{125}{12}$.

To verify, parametrize $C$ as $z(\theta) = \frac{5}{2} e^{i\theta}$ for $0 \leq \theta \leq \pi$. Then:

The integral becomes:

Evaluating the integral:

Thus, the integral over $C$ is:

Both methods confirm the result. The fraction $-\frac{125}{12}$ has $p = 125$ and $q = 12$, which are coprime. Therefore, $p + q = 137$.

[/THOUGHT] [PYTHON]

Calculate the integral over L

from sympy import symbols, integrate

x = symbols('x') integral_L = integrate(x**2, (x, -5/2, 5/2)) result = -integral_L print(result) [/PYTHON]

[THOUGHT] The Python output is $-10.4166666666667$, which equals $-\frac{125}{12}$. Here, $p = 125$ and $q = 12$ are coprime, so $p + q = 137$. [/THOUGHT] $\boxed{137}$