1. The problem is to find the expected value $E(X)$ given by the integral: $$E(X) = \int_{0}^{10} x \cdot \left[ \frac{1}{80}(x + 3) \right] \, dx$$ 2. The formula for expected value when given a probability density function $f(x)$ is: $$E(X) = \int x f(x) \, dx$$ Here, $f(x) = \frac{1}{80}(x + 3)$ for $0 \leq x \leq 10$. 3. Substitute $f(x)$ into the integral: $$E(X) = \int_0^{10} x \cdot \frac{1}{80}(x + 3) \, dx = \frac{1}{80} \int_0^{10} x(x + 3) \, dx$$ 4. Expand the integrand: $$x(x + 3) = x^2 + 3x$$ So, $$E(X) = \frac{1}{80} \int_0^{10} (x^2 + 3x) \, dx$$ 5. Integrate term-by-term: $$\int_0^{10} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{10} = \frac{10^3}{3} = \frac{1000}{3}$$ $$\int_0^{10} 3x \, dx = 3 \left[ \frac{x^2}{2} \right]_0^{10} = 3 \cdot \frac{100}{2} = 150$$ 6. Combine the results: $$\int_0^{10} (x^2 + 3x) \, dx = \frac{1000}{3} + 150 = \frac{1000}{3} + \frac{450}{3} = \frac{1450}{3}$$ 7. Multiply by $\frac{1}{80}$: $$E(X) = \frac{1}{80} \cdot \frac{1450}{3} = \frac{1450}{240} = \frac{145}{24} \approx 6.04$$ Final answer: $$E(X) = \frac{145}{24} \approx 6.04$$