Question

Use the limit process to find the area of the region. y=16-x^2. Interval: [1,3]. Taking the integral of the function proves that the area is 70/3. However, when using the limit process, I am making some dumb math error somewhere and getting the wrong answer (88/3). I will show my work and maybe someone will be able to spot my dumb math error, it's really bothering me. (adding in next post)

Answer 1

you from chicago

Answer 2

\[\lim_{n \rightarrow infinity}\sum_{1}^{n}\frac{2}{n}(f(\frac{2i}{n})\]
= \[\lim_{n \rightarrow infinity}\sum_{1}^{n}\frac{2}{n}(16-(\frac{2i}{n})^2)\]
= \[\lim_{n \rightarrow infinity}\sum_{1}^{n}\frac{2}{n}(16-\frac{4i ^{2}}{n ^{2}})\]
= \[\lim_{n \rightarrow infinity}\sum_{1}^{n}(\frac{32}{n}-\frac{8i^{2}}{n^{3}})\]
= \[32 - \lim_{n \rightarrow infinity}(\frac{8}{n^{3}})(\frac{n(n+1)(2n+1)}{6})\]
=\[32 - \lim_{n \rightarrow infinity}(\frac{4}{3})\frac{(n+1)(2n+1)}{n^{2}})\]
=\[32 - \lim_{n \rightarrow infinity}(\frac{4}{3})\frac{(n+1)(2n+1)}{n^{2}})\]
= \[32 - \lim_{n \rightarrow infinity}(\frac{4}{3})(2+\frac{3}{n}+\frac{1}{n^{2}}))\]
= \[32 - \frac{4}{3}(2) = 32-\frac{8}{3} =\frac{88}{3}\]

Answer 3

do you have tthe orginal function

Answer 4

Yeah I posted it in my OP: y=16-x^{2}

Answer 5

i cant see anything wrong with your math