1. **Problem Statement:** Evaluate the definite integral $$\int_3^6 g(x) \, dx$$ where the graph of $$g$$ from $$x=2$$ to $$x=6$$ is a line segment rising linearly from approximately $$(2,0)$$ to $$(6,6)$$. 2. **Understanding the graph segment:** From $$x=2$$ to $$x=6$$, $$g(x)$$ is a line segment. We want the integral from $$3$$ to $$6$$, so we focus on this linear part. 3. **Find the equation of the line segment:** The line passes through points $$(2,0)$$ and $$(6,6)$$. Slope $$m = \frac{6-0}{6-2} = \frac{6}{4} = \frac{3}{2}$$. Using point-slope form with point $$(2,0)$$: $$ y - 0 = \frac{3}{2}(x - 2) \\ y = \frac{3}{2}x - 3 $$ So, $$g(x) = \frac{3}{2}x - 3$$ for $$2 \leq x \leq 6$$. 4. **Set up the integral:** $$ \int_3^6 g(x) \, dx = \int_3^6 \left( \frac{3}{2}x - 3 \right) dx $$ 5. **Integrate:** $$ \int \left( \frac{3}{2}x - 3 \right) dx = \frac{3}{2} \cdot \frac{x^2}{2} - 3x + C = \frac{3}{4}x^2 - 3x + C $$ 6. **Evaluate definite integral:** $$ \int_3^6 g(x) \, dx = \left[ \frac{3}{4}x^2 - 3x \right]_3^6 = \left( \frac{3}{4} \cdot 6^2 - 3 \cdot 6 \right) - \left( \frac{3}{4} \cdot 3^2 - 3 \cdot 3 \right) $$ Calculate each term: $$ \frac{3}{4} \cdot 36 = 27, \quad 3 \cdot 6 = 18 $$ $$ \frac{3}{4} \cdot 9 = \frac{27}{4} = 6.75, \quad 3 \cdot 3 = 9 $$ So, $$ (27 - 18) - (6.75 - 9) = 9 - (-2.25) = 9 + 2.25 = 11.25 $$ 7. **Final answer:** $$\int_3^6 g(x) \, dx = 11.25$$ This represents the area under the line segment of $$g(x)$$ from $$x=3$$ to $$x=6$$.