91Ó°ÊÓ

The given integral, i.e.${∫}_0^t{e}^{-x^2}dx$

A Taylor series approximation describes a number as a polynomial with a comparable value to a number in a limit around a specific value.

Use the Maclaurin series of $e^x$and replace x by $-x^2$,

$$\begin{gathered} e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+..... \\ {e}^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+..... \\ {∫}_0^t{e}^{-x^2}dx={∫}_0^x\left[1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+.....\right]dx \\ ={\left[x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+.....\right]}_0^t \end{gathered}$$

Further, simplify for two-term approximation.

$$\begin{gathered} {∫}_0^t{e}^{-x^2}dx=t-\frac{t^3}{3}+\frac{t^5}{10}-\frac{t^7}{42}+....... \\ =t-\frac{t^3}{3} \end{gathered}$$

Use conditions of alternating series, i.e.,

$$\begin{gathered} \left|{\mathrm{a}}_{\mathrm{n}+1}\right|<\left|{\mathrm{a}}_{\mathrm{n}}\right|. \\ \underset{\mathrm{n}→∞}{\lim }{\mathrm{a}}_{\mathrm{n}}=0 \end{gathered}$$

Solve the error value.

$$\begin{gathered} Error=\left|S-\left(a_1+a_2+a_3+...+a_n\right)\right| \\ ≤\left|a_{n+1}\right| \\ \left|{∫}_0^t{e}^{-x^2}dx-\left(a_1+a_2\right)\right|=\left|{∫}_0^t{e}^{-x^2}dx-\left(t-\frac{t^3}{3}\right)\right| \\ \left|{∫}_0^t{e}^{-x^2}dx-\left(t-\frac{t^3}{3}\right)\right|≤\left|a_{n+1}\right| \end{gathered}$$

Substitute$a_{n+1}=a_3=\frac{t^5}{10}$ , and we get,

$$\begin{gathered} \left|{∫}_0^t{e}^{-x^2}dx-\left(t-\frac{t^3}{3}\right)\right|≤\left|\frac{t^5}{10}\right| \\ \left|{∫}_0^t{e}^{-x^2}dx-\left(t-\frac{t^5}{10}\right)\right|≤\left|\frac{0.1^5}{10}\right| \\ \left|{∫}_0^t{e}^{-x^2}dx-\left(t-\frac{t^3}{3}\right)\right|<1×{10}^{-6} \\ Error<1×{10}^{-6} \end{gathered}$$

Hence, a two-term approximation of the integral, i.e.,${∫}_0^t{e}^{-x^2}dx=t-\frac{t^3}{3}$.

And the value of the error bound is$<1×{10}^{-6}$.