1. **Problem statement:** We are given a differential equation $$y' = \frac{10.2x^2 + 15.6x + 2}{x^3 + 3x^2 + 2x}$$ for $$x > 0$$ with initial condition $$y(0.5) = 2.5$$. We need to evaluate numerically the definite integral $$\int_{0.5}^{2} y(x) \, dx$$. 2. **Step 1: Understand the problem.** We have the derivative of $$y$$, and an initial value for $$y$$ at $$x=0.5$$. To find $$y(x)$$, we integrate $$y'$$ from 0.5 to $$x$$ and add the initial value: $$y(x) = y(0.5) + \int_{0.5}^x y'(t) \, dt$$ 3. **Step 2: Express the integral to evaluate.** The integral we want is: $$\int_{0.5}^2 y(x) \, dx = \int_{0.5}^2 \left(y(0.5) + \int_{0.5}^x y'(t) \, dt \right) dx$$ 4. **Step 3: Swap the order of integration.** Using Fubini's theorem: $$\int_{0.5}^2 y(x) \, dx = \int_{0.5}^2 y(0.5) \, dx + \int_{0.5}^2 \int_{0.5}^x y'(t) \, dt \, dx$$ $$= y(0.5)(2 - 0.5) + \int_{0.5}^2 \int_{0.5}^x y'(t) \, dt \, dx$$ Swap integration order in the double integral: $$\int_{0.5}^2 \int_{0.5}^x y'(t) \, dt \, dx = \int_{0.5}^2 y'(t) \int_t^2 dx \, dt = \int_{0.5}^2 y'(t)(2 - t) \, dt$$ 5. **Step 4: Substitute values and simplify:** $$\int_{0.5}^2 y(x) \, dx = 2.5 \times 1.5 + \int_{0.5}^2 (2 - t) \frac{10.2 t^2 + 15.6 t + 2}{t^3 + 3 t^2 + 2 t} \, dt$$ 6. **Step 5: Simplify the integrand:** Factor denominator: $$t^3 + 3 t^2 + 2 t = t(t+1)(t+2)$$ Expand numerator times $$2 - t$$: $$ (2 - t)(10.2 t^2 + 15.6 t + 2) = -10.2 t^3 - 4.8 t^2 + 31.2 t + 4$$ 7. **Step 6: Perform partial fraction decomposition:** Set $$\frac{-10.2 t^3 - 4.8 t^2 + 31.2 t + 4}{t(t+1)(t+2)} = \frac{A}{t} + \frac{B}{t+1} + \frac{C}{t+2}$$ Solving gives: $$A = 2, B = 3, C = -15.2$$ 8. **Step 7: Integrate term-by-term:** $$\int_{0.5}^2 y(x) \, dx = 3.75 + \int_{0.5}^2 \left( \frac{2}{t} + \frac{3}{t+1} - \frac{15.2}{t+2} \right) dt$$ $$= 3.75 + \left[ 2 \ln|t| + 3 \ln|t+1| - 15.2 \ln|t+2| \right]_{0.5}^2$$ 9. **Step 8: Evaluate the logarithms numerically:** At $$t=2$$: $$2 \ln 2 + 3 \ln 3 - 15.2 \ln 4$$ At $$t=0.5$$: $$2 \ln 0.5 + 3 \ln 1.5 - 15.2 \ln 2.5$$ Calculate difference and add 3.75: Numerical evaluation yields approximately $$-18.3$$. **Final answer:** $$\int_{0.5}^2 y(x) \, dx \approx -18.3$$