b8-taylor and maclaurin series: problem (2 points) express the definite integral as an infinite series. \n$$…

b8-taylor and maclaurin series: problem (2 points) express the definite integral as an infinite series. \n$$ int _ { 0 } ^ { 1 } cos left( \frac { 8 } { 9 } x ^ { 4 } \right) d x = sum _ { n = 0 } ^ { infty } $$ \nfind the value of the definite integral to within an error of at most $ 10 ^ { - 4 } $. \n$$ int _ { 0 } ^ { 1 } cos left( \frac { 8 } { 9 } x ^ { 4 } \right) d x approx $$ \nnote: you can earn partial credit on this problem. \npreview my answers submit answers \nyou have attempted this problem 0 times. \nyou have 15 attempts remaining.

b8-taylor and maclaurin series: problem (2 points) express the definite integral as an infinite series. \n$$ int _ { 0 } ^ { 1 } cos left( \frac { 8 } { 9 } x ^ { 4 } \right) d x = sum _ { n = 0 } ^ { infty } $$ \nfind the value of the definite integral to within an error of at most $ 10 ^ { - 4 } $. \n$$ int _ { 0 } ^ { 1 } cos left( \frac { 8 } { 9 } x ^ { 4 } \right) d x approx $$ \nnote: you can earn partial credit on this problem. \npreview my answers submit answers \nyou have attempted this problem 0 times. \nyou have 15 attempts remaining.

Answer

Answer:

  1. For the infinite - series representation: (\sum_{n = 0}^{\infty}\frac{(- 1)^{n}(\frac{8}{9})^{2n}}{(2n)!}\cdot\frac{1}{8n + 1})
  2. For the approximation: (0.9054)

Explanation:

Step1: Recall the Maclaurin series of (\cos t)

The Maclaurin series of (\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}t^{2n}), where (t\in(-\infty,\infty)). Let (t = \frac{8}{9}x^{4}). Then (\cos(\frac{8}{9}x^{4})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(\frac{8}{9}x^{4})^{2n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{8}{9})^{2n}}{(2n)!}x^{8n})

Step2: Integrate term - by - term

We know that (\int_{0}^{1}\cos(\frac{8}{9}x^{4})dx=\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{8}{9})^{2n}}{(2n)!}x^{8n}dx). By the power - rule of integration (\int x^{m}dx=\frac{x^{m + 1}}{m+1}+C(m\neq - 1)), we have (\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{8}{9})^{2n}}{(2n)!}x^{8n}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{8}{9})^{2n}}{(2n)!}\int_{0}^{1}x^{8n}dx). And (\int_{0}^{1}x^{8n}dx=\left[\frac{x^{8n + 1}}{8n+1}\right]{0}^{1}=\frac{1}{8n + 1}). So (\int{0}^{1}\cos(\frac{8}{9}x^{4})dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{8}{9})^{2n}}{(2n)!}\cdot\frac{1}{8n + 1})

Step3: Use the Alternating Series Estimation Theorem

The series (\sum_{n = 0}^{\infty}a_{n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{8}{9})^{2n}}{(2n)!}\cdot\frac{1}{8n + 1}) is an alternating series with (a_{n}=\frac{(\frac{8}{9})^{2n}}{(2n)!}\cdot\frac{1}{8n + 1}). The Alternating Series Estimation Theorem states that for an alternating series (\sum_{n = 0}^{\infty}(-1)^{n}a_{n}(a_{n}\gt0,a_{n + 1}\leq a_{n},\lim_{n\rightarrow\infty}a_{n}=0)), the error (E_{N}) in approximating the sum (S) by the sum of the first (N) terms (S_{N}) satisfies (|E_{N}|\leq a_{N+1})

  • Calculate (a_{0}): (a_{0}=\frac{(\frac{8}{9})^{0}}{0!}\cdot\frac{1}{0 + 1}=1)
  • Calculate (a_{1}): (a_{1}=\frac{(\frac{8}{9})^{2}}{2!}\cdot\frac{1}{8 + 1}=\frac{64}{81\times2}\cdot\frac{1}{9}=\frac{32}{729}\approx0.0439)
  • Calculate (a_{2}): (a_{2}=\frac{(\frac{8}{9})^{4}}{4!}\cdot\frac{1}{16 + 1}=\frac{4096}{6561\times24}\cdot\frac{1}{17}=\frac{4096}{2670864}\approx0.0015)
  • Calculate (a_{3}): (a_{3}=\frac{(\frac{8}{9})^{6}}{6!}\cdot\frac{1}{24+1}=\frac{262144}{531441\times720}\cdot\frac{1}{25}=\frac{262144}{9565938000}\approx2.74\times10^{-5})

(S\approx a_{0}-a_{1}+a_{2}-a_{3}=1 - 0.0439+0.0015 - 0.0000274\approx0.9054)