91Ó°ÊÓ

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

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

Use the Maclaurin series of $e^x$and replace x by -x as:

$$\begin{gathered} e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+.... \\ {e}^{=x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+..... \\ \sqrt{X}{e}^{-x}=x^{1/2}-x^{3/2}+\frac{x^{5/2}}{2!}-\frac{x^{7/2}}{3!}+\frac{x^{9/2}}{4!}+..... \\ {∫}_0^t\sqrt{X}{e}^{-x}dx={∫}_0^t\left[x^{1/2}-x^{3/2}+\frac{x^{5/2}}{2!}-\frac{x^{7/2}}{3!}+\frac{x^{9/2}}{4!}+.....\right]dx \end{gathered}$$

Further, simplify for two-term approximation as:

$$\begin{gathered} {∫}_0^t\sqrt{X}{e}^{-x}dx={\left[\frac{2}{3}{x}^{3/2}-\frac{2}{5}{x}^{5/2}+\frac{x^{712}}{7}+...\right]}_0^t \\ =\frac{2}{3}{t}^{3/2}-\frac{2}{5}t+\frac{t^{7/2}}{7}+.... \\ {∫}_0^t\sqrt{X}{e}^{-x}dx=\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2} \end{gathered}$$

Use conditions of alternating series, i.e.,

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

Solve the error value as:

role="math" localid="1658487890701" $$\begin{gathered} Error=\left|S-\left(a_1+a_2+a_3+...+a_n\right)\right| \\ ≤\left|a_{n+1}\right| \\ \left|{∫}_0^t\sqrt{X}{e}^{-x}dx-\left(a_1+a_2\right)\right|=\left|{∫}_0^t\sqrt{X}{e}^{-x}dx-\left(a_1+a_2\right)\right|=\left|{∫}_0^t\sqrt{x}{e}^{-x}dx-\left(\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2}\right)\right| \\ \left|{∫}_0^t\sqrt{X}{e}^{-x}dx-\left(\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2}\right)≤\left|a_{n+1}\right|\right| \end{gathered}$$

Substitute$a_{n+1}=a_3=\frac{t^{7/2}}{7}$ , and we get:

$$\begin{gathered} \left|{∫}_0^t\sqrt{X}{e}^{-x}dx-\left(\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2}\right)\right|≤\left|\frac{t^{7/2}}{7}\right| \\ \left|{∫}_0^t\sqrt{X}{e}^{-x}dx-\left(\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2}\right)\right|≤\frac{{\left(0.01\right)}^{7/2}}{7} \\ \left|{∫}_0^t\sqrt{X}{e}^{-x}dx-\left(\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2}\right)\right|≤\frac{1×{10}^{-7}}{7} \\ Error<\frac{1×{10}^{-7}}{7} \end{gathered}$$

Hence, a two-term approximation of the integral, i.e.,$\left|{∫}_0^t\sqrt{X}{e}^{-x}dx=\frac{2}{3}{t}^{3/2}-\frac{2}{5}{t}^{5/2}\right|$.

And the value of the error bound is$<\frac{1×{10}^{-7}}{7}$.