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 each subinterval are: $$-2, -1, 0, 1, 2, 3$$. 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:** - $$g(-2) \approx 0.5$$ - $$g(-1) \approx -0.5$$ - $$g(0) \approx -1.2$$ - $$g(1) \approx 0.2$$ - $$g(2) \approx 1.5$$ - $$g(3) \approx 2.7$$ 6. **Calculate the sum:** $$S = 1 \times (0.5 - 0.5 - 1.2 + 0.2 + 1.5 + 2.7) = 1 \times (3.2) = 3.2$$ 7. **Final answer:** $$\int_{-3}^{3} g(x) \, dx \approx 3.2$$ This means the approximate area under the curve from $$x=-3$$ to $$x=3$$ using six subintervals and right endpoints is 3.2.