1. The problem is to evaluate the definite integral $$\int_1^6 x \, dx$$ which represents the area under the curve $y = x$ from $x=1$ to $x=6$. 2. The formula for the integral of $x$ is $$\int x \, dx = \frac{x^2}{2} + C$$ where $C$ is the constant of integration. 3. For definite integrals, we evaluate the antiderivative at the upper and lower limits and subtract: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. 4. Applying this to our problem: $$\int_1^6 x \, dx = \left. \frac{x^2}{2} \right|_1^6 = \frac{6^2}{2} - \frac{1^2}{2}$$ 5. Calculate the values: $$\frac{6^2}{2} - \frac{1^2}{2} = \frac{36}{2} - \frac{1}{2} = 18 - 0.5 = 17.5$$ 6. Therefore, the value of the integral is $$17.5$$ which is the area under the curve $y=x$ from $x=1$ to $x=6$.