use a series to estimate the following integrals value with an error of magnitude less than 10^(-8).\n…

use a series to estimate the following integrals value with an error of magnitude less than 10^(-8).\n int_{0}^{0.3} e^{-x^{2}} dx \n int_{0}^{0.3} e^{-x^{2}} dx approx square \n(do not round until the final answer. then round to five decimal places as needed.)

use a series to estimate the following integrals value with an error of magnitude less than 10^(-8).\n int_{0}^{0.3} e^{-x^{2}} dx \n int_{0}^{0.3} e^{-x^{2}} dx approx square \n(do not round until the final answer. then round to five decimal places as needed.)

Answer

Explanation:

Step1: Recall the Maclaurin series for $e^u$

The Maclaurin series for $e^u=\sum_{n = 0}^{\infty}\frac{u^n}{n!}=1 + u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$. Let $u=-x^{2}$, then $e^{-x^{2}}=\sum_{n = 0}^{\infty}\frac{(-x^{2})^{n}}{n!}=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}x^{2n}}{n!}=1 - x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\frac{x^{8}}{4!}-\cdots$.

Step2: Integrate the series term - by - term

$\int_{0}^{0.3}e^{-x^{2}}dx=\int_{0}^{0.3}\sum_{n = 0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{n!}\int_{0}^{0.3}x^{2n}dx$. Since $\int_{0}^{0.3}x^{2n}dx=\left[\frac{x^{2n + 1}}{2n+1}\right]{0}^{0.3}=\frac{(0.3)^{2n + 1}}{2n+1}$, then $\int{0}^{0.3}e^{-x^{2}}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(0.3)^{2n + 1}}{n!(2n+1)}$.

Step3: Determine the number of terms for the error bound

The series $\sum_{n = 0}^{\infty}\frac{(-1)^{n}(0.3)^{2n + 1}}{n!(2n+1)}$ is an alternating series. For an alternating series $\sum_{n = 0}^{\infty}(-1)^{n}a_{n}$ ($a_{n}\gt0$, $a_{n+1}\leq a_{n}$, and $\lim_{n\rightarrow\infty}a_{n}=0$), the error bound $|E_N|\leq a_{N + 1}$. We want $|E_N|\lt10^{-8}$. Let $a_{n}=\frac{(0.3)^{2n + 1}}{n!(2n+1)}$. When $n = 3$: $a_{3}=\frac{(0.3)^{2\times3 + 1}}{3!(2\times3+1)}=\frac{(0.3)^{7}}{6\times7}=\frac{0.0002187}{42}\approx5.20714\times 10^{-6}$ When $n = 4$: $a_{4}=\frac{(0.3)^{2\times4+1}}{4!(2\times4 + 1)}=\frac{(0.3)^{9}}{24\times9}=\frac{0.000019683}{216}\approx9.1125\times10^{-8}$ When $n = 5$: $a_{5}=\frac{(0.3)^{2\times5 + 1}}{5!(2\times5+1)}=\frac{(0.3)^{11}}{120\times11}=\frac{0.00000177147}{1320}\approx1.342\times10^{-9}\lt10^{-8}$ So we use the sum of the first 5 terms of the series.

Step4: Calculate the sum of the first 5 terms

When $n = 0$: $\frac{(-1)^{0}(0.3)^{2\times0+1}}{0!(2\times0 + 1)}=0.3$ When $n = 1$: $\frac{(-1)^{1}(0.3)^{2\times1+1}}{1!(2\times1+1)}=-\frac{(0.3)^{3}}{3}=-\frac{0.027}{3}=- 0.009$ When $n = 2$: $\frac{(-1)^{2}(0.3)^{2\times2+1}}{2!(2\times2+1)}=\frac{(0.3)^{5}}{2\times5}=\frac{0.00243}{10}=0.000243$ When $n = 3$: $\frac{(-1)^{3}(0.3)^{2\times3+1}}{3!(2\times3+1)}=-\frac{(0.3)^{7}}{6\times7}\approx - 5.20714\times10^{-6}$ When $n = 4$: $\frac{(-1)^{4}(0.3)^{2\times4+1}}{4!(2\times4+1)}=\frac{(0.3)^{9}}{24\times9}\approx9.1125\times10^{-8}$

$S_5=0.3-0.009 + 0.000243-5.20714\times10^{-6}+9.1125\times10^{-8}$ $S_5=0.291237792875\approx0.29124$

Answer:

$0.29124$