use a series to estimate the following integrals value with an error of magnitude less than 10^(-5).\n0.4\n∫…

Explanation:

Step1: Recall the Maclaurin series of $\sin t$

The Maclaurin series of $\sin t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n+1)!}t^{2n + 1}=t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\frac{t^{7}}{7!}+\cdots$. 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

$\int_{0}^{0.4}\sin x^{2}dx=\int_{0}^{0.4}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{4n + 2}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\int_{0}^{0.4}x^{4n + 2}dx$. Since $\int_{0}^{0.4}x^{4n + 2}dx=\left[\frac{x^{4n+3}}{4n + 3}\right]{0}^{0.4}=\frac{(0.4)^{4n+3}}{4n + 3}$, then $\int{0}^{0.4}\sin x^{2}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\cdot\frac{(0.4)^{4n+3}}{4n + 3}$.

Step3: Determine the number of terms needed for the error bound

The series $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\cdot\frac{(0.4)^{4n+3}}{4n + 3}$ 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 $|E_N|\lt10^{-5}$. Let's calculate the terms of the series: When $n = 0$: $a_{0}=\frac{1}{1!}\cdot\frac{(0.4)^{3}}{3}=\frac{(0.4)^{3}}{3}=\frac{0.064}{3}\approx0.02133$. When $n = 1$: $a_{1}=\frac{(-1)^{1}}{3!}\cdot\frac{(0.4)^{7}}{7}=-\frac{1}{6}\cdot\frac{0.0016384}{7}\approx - 0.000039$. When $n = 2$: $a_{2}=\frac{1}{5!}\cdot\frac{(0.4)^{11}}{11}=\frac{1}{120}\cdot\frac{0.0000268435456}{11}\approx2.03\times10^{-8}\lt10^{-5}$.

Step4: Calculate the sum of the series with sufficient terms

$\int_{0}^{0.4}\sin x^{2}dx\approx\frac{(0.4)^{3}}{3}-\frac{(0.4)^{7}}{6\times7}$ $=\frac{0.064}{3}-\frac{0.0016384}{42}$ $= 0.021333 - 0.000039$ $=0.021294\approx0.02129$.

Answer:

$0.02129$