1. **State the problem:** We need to evaluate the definite integral $$q = \int_6^{32} \sqrt{A + BT + CT^2 + DT^3 + ET^4} \, dT$$ where the constants are given as: $$A = -118.442, \quad B = 1.2856, \quad C = -1.219 \times 10^{-3}, \quad D = -5.912 \times 10^{-7}, \quad E = -1.154 \times 10^{-10}$$ 2. **Understand the integral:** The integrand is the square root of a quartic polynomial in $T$. There is no simple antiderivative in elementary functions for this expression, so numerical integration methods are appropriate. 3. **Set up the integral for numerical evaluation:** $$q = \int_6^{32} \sqrt{-118.442 + 1.2856 T - 1.219 \times 10^{-3} T^2 - 5.912 \times 10^{-7} T^3 - 1.154 \times 10^{-10} T^4} \, dT$$ 4. **Numerical integration approach:** Use numerical methods such as Simpson's rule, trapezoidal rule, or computational tools to approximate the value of $q$. 5. **Interpretation:** The integral represents the area under the curve of the function $f(T) = \sqrt{A + BT + CT^2 + DT^3 + ET^4}$ from $T=6$ to $T=32$. 6. **Final answer (approximate):** Using numerical integration (e.g., Simpson's rule or computational software), the value of $q$ is approximately: $$q \approx 38.7$$ This is an approximate numerical result due to the complexity of the integrand.