1. **State the problem:** Approximate the area under the curve $f(x) = 9 - x^2$ from $x = -3$ to $x = 3$ using right-hand sum (RHS) and left-hand sum (LHS) with 6 subintervals. 2. **Formula and setup:** The interval length is $6$ (from $-3$ to $3$). Divide into $6$ equal subintervals, so each subinterval width is $$\Delta x = \frac{3 - (-3)}{6} = 1.$$ The subintervals are $[-3,-2], [-2,-1], [-1,0], [0,1], [1,2], [2,3]$. 3. **Right-hand sum (RHS):** Use the right endpoint of each subinterval to evaluate $f(x)$: $$x_i = -2, -1, 0, 1, 2, 3.$$ Calculate each: $$f(-2) = 9 - (-2)^2 = 9 - 4 = 5,$$ $$f(-1) = 9 - 1 = 8,$$ $$f(0) = 9 - 0 = 9,$$ $$f(1) = 9 - 1 = 8,$$ $$f(2) = 9 - 4 = 5,$$ $$f(3) = 9 - 9 = 0.$$ Sum these values and multiply by $\Delta x$: $$\text{RHS}_6 = 1 \times (5 + 8 + 9 + 8 + 5 + 0) = 35.$$ 4. **Left-hand sum (LHS):** Use the left endpoint of each subinterval: $$x_i = -3, -2, -1, 0, 1, 2.$$ Calculate each: $$f(-3) = 9 - 9 = 0,$$ $$f(-2) = 5,$$ $$f(-1) = 8,$$ $$f(0) = 9,$$ $$f(1) = 8,$$ $$f(2) = 5.$$ Sum and multiply by $\Delta x$: $$\text{LHS}_6 = 1 \times (0 + 5 + 8 + 9 + 8 + 5) = 35.$$ 5. **Interpretation:** Both sums approximate the area under the curve and both equal 35 in this case. **Final answers:** $$\text{RHS}_6 = 35,$$ $$\text{LHS}_6 = 35.$$