Question

use a series to estimate the following integrals value with an error of magnitude less than (10^{-5}).
int_{0}^{0.2} sin x^{2} d x
int_{0}^{0.2} sin x^{2} d x approx square
(do not round until the final answer. then round to five decimal places as needed.)

Explanation:

Step1: Recall the Maclaurin series for \(\sin t\)

The Maclaurin series for \(\sin t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n+1)!}t^{2n + 1}=t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\frac{t^{7}}{7!}+\cdots\). Let \(t=x^{2}\), then \(\sin(x^{2})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n + 1)!}(x^{2})^{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{4n + 2}=x^{2}-\frac{x^{6}}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\cdots\)

Step2: Integrate the series term - by - term

\(\int_{0}^{0.2}\sin(x^{2})dx=\int_{0}^{0.2}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{4n+2}dx\). By the power - rule for integration \(\int x^{m}dx=\frac{x^{m + 1}}{m+1}+C(m
eq - 1)\), we have \(\int_{0}^{0.2}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{4n+2}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(4n + 3)}x^{4n+3}\big|_{0}^{0.2}\)

Step3: Evaluate the series

When \(n = 0\): \(\frac{(-1)^{0}}{(0+1)!(0 + 3)}(0.2)^{3}=\frac{1}{1\times3}\times(0.2)^{3}=\frac{0.008}{3}\approx0.0026667\)

When \(n = 1\): \(\frac{(-1)^{1}}{(2\times1+1)!(2\times4+3)}(0.2)^{7}=\frac{-1}{3!\times11}(0.2)^{7}=\frac{-1}{6\times11}\times1.28\times10^{-5}\approx\frac{-1.28\times10^{-5}}{66}\approx - 1.94\times10^{-7}\)

When \(n = 2\): \(\frac{(-1)^{2}}{(2\times2+1)!(2\times8+3)}(0.2)^{11}=\frac{1}{5!\times19}(0.2)^{11}=\frac{1}{120\times19}\times2.048\times10^{-8}\approx\frac{2.048\times10^{-8}}{2280}\approx9.0\times10^{-12}\)

Step4: Apply the Alternating Series Estimation Theorem

For an alternating series \(\sum_{n = 0}^{\infty}(-1)^{n}a_{n}(a_{n}\gt0,a_{n+1}\lt a_{n},\lim_{n
ightarrow\infty}a_{n}=0)\), the error \(E_{N}\) for the sum \(S_{N}=\sum_{n = 0}^{N}(-1)^{n}a_{n}\) is bounded by \(|E_{N}|\leq a_{N + 1}\).

Since \(a_{1}\approx1.94\times10^{-7}\lt10^{-5}\), we can use the sum \(S=a_{0}+a_{1}\)

\(\int_{0}^{0.2}\sin(x^{2})dx\approx\frac{(0.2)^{3}}{3}-\frac{(0.2)^{7}}{6\times11}\)

\(\int_{0}^{0.2}\sin(x^{2})dx\approx\frac{0.008}{3}-\frac{1.28\times10^{-5}}{66}\)

\(\int_{0}^{0.2}\sin(x^{2})dx\approx0.0026667-0.00000194\)

Answer:

\(0.00266\)