Net Signed Area Db3C3A

1. **Problem statement:** We want to find the net signed area between the graph of $f(x)$ and the x-axis on the given intervals using sigma notation and summation formulas. The right endpoint of the $k$th subinterval is $x_k$, and all subintervals have equal width. --- ### (a) $f(x) = x - 3$ on $[1,5]$ 2. **Step 1: Define the partition and width** The interval length is $5 - 1 = 4$. Let the number of subintervals be $n$. Then the width of each subinterval is: $$\Delta x = \frac{5 - 1}{n} = \frac{4}{n}$$ 3. **Step 2: Define the right endpoints** The right endpoint of the $k$th subinterval is: $$x_k = 1 + k\Delta x = 1 + \frac{4k}{n}$$ 4. **Step 3: Write the Riemann sum for the net signed area** $$S_n = \sum_{k=1}^n f(x_k) \Delta x = \sum_{k=1}^n \left(x_k - 3\right) \Delta x = \sum_{k=1}^n \left(1 + \frac{4k}{n} - 3\right) \frac{4}{n} = \sum_{k=1}^n \left(-2 + \frac{4k}{n}\right) \frac{4}{n}$$ 5. **Step 4: Simplify the sum** $$S_n = \sum_{k=1}^n \left(-2 \cdot \frac{4}{n} + \frac{4k}{n} \cdot \frac{4}{n}\right) = \sum_{k=1}^n \left(-\frac{8}{n} + \frac{16k}{n^2}\right) = \sum_{k=1}^n -\frac{8}{n} + \sum_{k=1}^n \frac{16k}{n^2}$$ 6. **Step 5: Use summation formulas** Recall: $$\sum_{k=1}^n 1 = n$$ $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ So, $$S_n = -\frac{8}{n} \cdot n + \frac{16}{n^2} \cdot \frac{n(n+1)}{2} = -8 + \frac{16}{n^2} \cdot \frac{n(n+1)}{2} = -8 + \frac{16(n+1)}{2n} = -8 + \frac{8(n+1)}{n}$$ 7. **Step 6: Take the limit as $n \to \infty$** $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(-8 + \frac{8(n+1)}{n}\right) = -8 + 8 \cdot \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) = -8 + 8 \cdot 1 = 0$$ **Answer (a):** The net signed area is $0$. --- ### (b) $f(x) = \frac{x^2}{3}$ on $[2,5]$ 8. **Step 1: Define the partition and width** Interval length: $5 - 2 = 3$ $$\Delta x = \frac{3}{n}$$ 9. **Step 2: Right endpoints** $$x_k = 2 + k\Delta x = 2 + \frac{3k}{n}$$ 10. **Step 3: Riemann sum** $$S_n = \sum_{k=1}^n f(x_k) \Delta x = \sum_{k=1}^n \frac{(x_k)^2}{3} \Delta x = \sum_{k=1}^n \frac{\left(2 + \frac{3k}{n}\right)^2}{3} \cdot \frac{3}{n} = \sum_{k=1}^n \frac{\left(2 + \frac{3k}{n}\right)^2}{n}$$ 11. **Step 4: Expand the square** $$\left(2 + \frac{3k}{n}\right)^2 = 4 + 2 \cdot 2 \cdot \frac{3k}{n} + \left(\frac{3k}{n}\right)^2 = 4 + \frac{12k}{n} + \frac{9k^2}{n^2}$$ 12. **Step 5: Substitute and separate sums** $$S_n = \sum_{k=1}^n \frac{4 + \frac{12k}{n} + \frac{9k^2}{n^2}}{n} = \sum_{k=1}^n \left(\frac{4}{n} + \frac{12k}{n^2} + \frac{9k^2}{n^3}\right) = \sum_{k=1}^n \frac{4}{n} + \sum_{k=1}^n \frac{12k}{n^2} + \sum_{k=1}^n \frac{9k^2}{n^3}$$ 13. **Step 6: Use summation formulas** Recall: $$\sum_{k=1}^n 1 = n$$ $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ So, $$S_n = \frac{4}{n} \cdot n + \frac{12}{n^2} \cdot \frac{n(n+1)}{2} + \frac{9}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = 4 + 6 \cdot \frac{n+1}{n} + \frac{3}{2} \cdot \frac{(n+1)(2n+1)}{n^2}$$ 14. **Step 7: Take the limit as $n \to \infty$** $$\lim_{n \to \infty} S_n = 4 + 6 \cdot 1 + \frac{3}{2} \cdot 2 = 4 + 6 + 3 = 13$$ **Answer (b):** The net signed area is $13$. --- ### (c) $f(x) = x^3 - 1$ on $[0,2]$ 15. **Step 1: Define the partition and width** Interval length: $2 - 0 = 2$ $$\Delta x = \frac{2}{n}$$ 16. **Step 2: Right endpoints** $$x_k = 0 + k\Delta x = \frac{2k}{n}$$ 17. **Step 3: Riemann sum** $$S_n = \sum_{k=1}^n f(x_k) \Delta x = \sum_{k=1}^n \left(\left(\frac{2k}{n}\right)^3 - 1\right) \frac{2}{n} = \sum_{k=1}^n \left(\frac{8k^3}{n^3} - 1\right) \frac{2}{n} = \sum_{k=1}^n \left(\frac{16k^3}{n^4} - \frac{2}{n}\right)$$ 18. **Step 4: Separate sums** $$S_n = \sum_{k=1}^n \frac{16k^3}{n^4} - \sum_{k=1}^n \frac{2}{n} = \frac{16}{n^4} \sum_{k=1}^n k^3 - \frac{2}{n} \cdot n = \frac{16}{n^4} \sum_{k=1}^n k^3 - 2$$ 19. **Step 5: Use summation formula for cubes** Recall: $$\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$$ So, $$S_n = \frac{16}{n^4} \cdot \left(\frac{n(n+1)}{2}\right)^2 - 2 = \frac{16}{n^4} \cdot \frac{n^2 (n+1)^2}{4} - 2 = \frac{16}{4} \cdot \frac{n^2 (n+1)^2}{n^4} - 2 = 4 \cdot \frac{(n+1)^2}{n^2} - 2$$ 20. **Step 6: Take the limit as $n \to \infty$** $$\lim_{n \to \infty} S_n = 4 \cdot 1 - 2 = 2$$ **Answer (c):** The net signed area is $2$. --- **Summary of answers:** - (a) $0$ - (b) $13$ - (c) $2$