(1) approximate \n\n\\( \\int _ { 0 } ^ { 1 } \\cos ( \\sqrt { x } ) d x \\)\n\nwith error less than \n\n\\(…

Explanation:

Step 1: Use the Taylor series for (\cos t)

The Taylor series for (\cos t) is (\cos t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots) Let (t = \sqrt{x}), then (\cos(\sqrt{x})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{n}) So (\int_{0}^{1}\cos(\sqrt{x})dx=\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{n}dx) By the power - rule for integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{n}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!(n + 1)})

Step 2: Estimate the error

The series (\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!(n + 1)}) is an alternating series. 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}) of the partial sum (S_{N}=\sum_{n = 0}^{N}(-1)^{n}a_{n}) is bounded by (|E_{N}|\leq a_{N + 1}) Let (a_{n}=\frac{1}{(2n)!(n + 1)}) We want (|E_{N}|\lt\frac{1}{100}) When (n = 1), (a_{1}=\frac{1}{2!(1 + 1)}=\frac{1}{4}) When (n = 2), (a_{2}=\frac{1}{4!(2+1)}=\frac{1}{24\times3}=\frac{1}{72}) When (n = 3), (a_{3}=\frac{1}{6!(3 + 1)}=\frac{1}{720\times4}=\frac{1}{2880}\lt\frac{1}{100})

Step 3: Calculate the partial sum

(S_{2}=\sum_{n = 0}^{2}\frac{(-1)^{n}}{(2n)!(n + 1)}=\frac{1}{0!(0 + 1)}-\frac{1}{2!(1 + 1)}+\frac{1}{4!(2+1)}) (=1-\frac{1}{4}+\frac{1}{72}) (=\frac{72-18 + 1}{72}=\frac{55}{72}\approx0.764)

Answer:

(\int_{0}^{1}\cos(\sqrt{x})dx\approx\frac{55}{72}\approx0.764)