1. **State the problem:** We want to approximate the definite integral $$\int_{-3}^3 g(x) \, dx$$ using six subintervals and right endpoints. 2. **Determine the width of each subinterval:** The interval length is $$3 - (-3) = 6$$. Dividing into 6 subintervals, each subinterval width is $$\Delta x = \frac{6}{6} = 1$$. 3. **Identify the right endpoints:** Starting at $$x = -3$$, the right endpoints for the six subintervals are $$x = -2, -1, 0, 1, 2, 3$$. 4. **Approximate the integral using the right endpoint Riemann sum:** $$ \int_{-3}^3 g(x) \, dx \approx \sum_{i=1}^6 g(x_i) \Delta x = \Delta x \left[g(-2) + g(-1) + g(0) + g(1) + g(2) + g(3)\right] $$ 5. **Estimate the values of $$g(x)$$ at the right endpoints from the graph:** - $$g(-2) \approx 1.0$$ - $$g(-1) \approx 0.5$$ - $$g(0) \approx -0.5$$ - $$g(1) \approx -0.8$$ - $$g(2) \approx 0.5$$ - $$g(3) \approx 1.5$$ 6. **Calculate the sum:** $$ S = 1 \times (1.0 + 0.5 - 0.5 - 0.8 + 0.5 + 1.5) = 1 \times (2.2) = 2.2 $$ 7. **Final answer:** $$\int_{-3}^3 g(x) \, dx \approx 2.2$$ This means the approximate area under the curve from $$x=-3$$ to $$x=3$$ using six subintervals and right endpoints is about 2.2.