Area Limit Integral 329226

1. **State the problem:** Find the exact area bounded by the curve $$x = g(y) = 4y^2 - y^3$$, the y-axis (i.e., $$x=0$$), and the lines $$y=1$$ and $$y=3$$ using the limit definition of the definite integral. 2. **Formula and concept:** The area bounded by the curve $$x = g(y)$$ and the y-axis from $$y=a$$ to $$y=b$$ is given by the definite integral: $$\text{Area} = \int_a^b g(y) \, dy$$ Here, $$a=1$$ and $$b=3$$. The limit definition of the definite integral is: $$\int_a^b g(y) \, dy = \lim_{n \to \infty} \sum_{i=1}^n g(y_i^*) \Delta y$$ where $$\Delta y = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}$$ and $$y_i^*$$ is a sample point in the $$i^{th}$$ subinterval. 3. **Set up the Riemann sum:** Divide the interval $$[1,3]$$ into $$n$$ equal parts: $$y_i = 1 + i \Delta y = 1 + \frac{2i}{n}$$ for $$i=0,1,2,...,n$$. Choose right endpoints for sample points: $$y_i^* = y_i = 1 + \frac{2i}{n}$$. Then the Riemann sum is: $$S_n = \sum_{i=1}^n g\left(1 + \frac{2i}{n}\right) \cdot \frac{2}{n}$$ 4. **Evaluate the sum inside:** Calculate $$g\left(1 + \frac{2i}{n}\right)$$: $$g\left(1 + \frac{2i}{n}\right) = 4\left(1 + \frac{2i}{n}\right)^2 - \left(1 + \frac{2i}{n}\right)^3$$ Expand: $$\left(1 + \frac{2i}{n}\right)^2 = 1 + \frac{4i}{n} + \frac{4i^2}{n^2}$$ $$\left(1 + \frac{2i}{n}\right)^3 = 1 + 3\cdot \frac{2i}{n} + 3\cdot \left(\frac{2i}{n}\right)^2 + \left(\frac{2i}{n}\right)^3 = 1 + \frac{6i}{n} + \frac{12i^2}{n^2} + \frac{8i^3}{n^3}$$ So: $$g\left(1 + \frac{2i}{n}\right) = 4\left(1 + \frac{4i}{n} + \frac{4i^2}{n^2}\right) - \left(1 + \frac{6i}{n} + \frac{12i^2}{n^2} + \frac{8i^3}{n^3}\right)$$ $$= 4 + \frac{16i}{n} + \frac{16i^2}{n^2} - 1 - \frac{6i}{n} - \frac{12i^2}{n^2} - \frac{8i^3}{n^3}$$ $$= 3 + \frac{10i}{n} + \frac{4i^2}{n^2} - \frac{8i^3}{n^3}$$ 5. **Write the Riemann sum explicitly:** $$S_n = \sum_{i=1}^n \left(3 + \frac{10i}{n} + \frac{4i^2}{n^2} - \frac{8i^3}{n^3}\right) \cdot \frac{2}{n} = \frac{2}{n} \sum_{i=1}^n \left(3 + \frac{10i}{n} + \frac{4i^2}{n^2} - \frac{8i^3}{n^3}\right)$$ Distribute the sum: $$S_n = \frac{2}{n} \left( \sum_{i=1}^n 3 + \sum_{i=1}^n \frac{10i}{n} + \sum_{i=1}^n \frac{4i^2}{n^2} - \sum_{i=1}^n \frac{8i^3}{n^3} \right)$$ $$= \frac{2}{n} \left( 3n + \frac{10}{n} \sum_{i=1}^n i + \frac{4}{n^2} \sum_{i=1}^n i^2 - \frac{8}{n^3} \sum_{i=1}^n i^3 \right)$$ 6. **Use formulas for sums:** $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$ Substitute: $$S_n = \frac{2}{n} \left( 3n + \frac{10}{n} \cdot \frac{n(n+1)}{2} + \frac{4}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} - \frac{8}{n^3} \cdot \left(\frac{n(n+1)}{2}\right)^2 \right)$$ Simplify each term: $$= \frac{2}{n} \left( 3n + 5(n+1) + \frac{4(n+1)(2n+1)}{6n} - \frac{8(n+1)^2}{4n} \right)$$ $$= \frac{2}{n} \left( 3n + 5n + 5 + \frac{4(n+1)(2n+1)}{6n} - 2 \frac{(n+1)^2}{n} \right)$$ $$= \frac{2}{n} \left( 8n + 5 + \frac{4(n+1)(2n+1)}{6n} - 2 \frac{(n+1)^2}{n} \right)$$ 7. **Simplify the fractions:** Calculate: $$\frac{4(n+1)(2n+1)}{6n} = \frac{4(2n^2 + 3n + 1)}{6n} = \frac{8n^2 + 12n + 4}{6n}$$ $$= \frac{8n^2}{6n} + \frac{12n}{6n} + \frac{4}{6n} = \frac{4n}{3} + 2 + \frac{2}{3n}$$ Calculate: $$2 \frac{(n+1)^2}{n} = 2 \frac{n^2 + 2n + 1}{n} = 2 \left(n + 2 + \frac{1}{n}\right) = 2n + 4 + \frac{2}{n}$$ 8. **Substitute back:** $$S_n = \frac{2}{n} \left(8n + 5 + \left(\frac{4n}{3} + 2 + \frac{2}{3n}\right) - \left(2n + 4 + \frac{2}{n}\right) \right)$$ $$= \frac{2}{n} \left(8n + 5 + \frac{4n}{3} + 2 + \frac{2}{3n} - 2n - 4 - \frac{2}{n} \right)$$ $$= \frac{2}{n} \left( (8n + \frac{4n}{3} - 2n) + (5 + 2 - 4) + \left(\frac{2}{3n} - \frac{2}{n}\right) \right)$$ $$= \frac{2}{n} \left( \left(8n - 2n + \frac{4n}{3}\right) + 3 + \left(-\frac{4}{3n}\right) \right)$$ $$= \frac{2}{n} \left( \left(6n + \frac{4n}{3}\right) + 3 - \frac{4}{3n} \right)$$ $$= \frac{2}{n} \left( \frac{18n}{3} + \frac{4n}{3} + 3 - \frac{4}{3n} \right) = \frac{2}{n} \left( \frac{22n}{3} + 3 - \frac{4}{3n} \right)$$ 9. **Distribute:** $$S_n = \frac{2}{n} \cdot \frac{22n}{3} + \frac{2}{n} \cdot 3 - \frac{2}{n} \cdot \frac{4}{3n} = \frac{44}{3} + \frac{6}{n} - \frac{8}{3n^2}$$ 10. **Take the limit as $$n \to \infty$$:** $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{44}{3} + \frac{6}{n} - \frac{8}{3n^2} \right) = \frac{44}{3} + 0 - 0 = \frac{44}{3}$$ **Final answer:** $$\boxed{\text{Area} = \frac{44}{3}}$$ This is the exact area bounded by the curve, the y-axis, and the lines $$y=1$$ and $$y=3$$ using the limit definition of the definite integral.