Q. 53 - Calculus

Question

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60. 

${\int }_{-1}^1{\mathrm{e}}^{\raisebox{1ex}{$-{\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \mathrm{dx}$

Step-by-Step Solution

Answer

${\int }_{-1}^1{\mathrm{e}}^{\raisebox{1ex}{$-{\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }{\left(-\frac{1}{3}\right)}^{\mathrm{k}}\frac{1}{\mathrm{k}!}\left[\frac{2}{2\mathrm{k}+1}\right]$

Step 1. Given information is:

${\int }_{-1}^1{\mathrm{e}}^{\raisebox{1ex}{$-{\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \mathrm{dx}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From} \mathrm{Q} 43. \\ \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\raisebox{1ex}{$-{\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \mathrm{dx} \mathrm{is} \\ {\mathrm{e}}^{\raisebox{1ex}{$-{\mathrm{x}}^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\sum _{\mathrm{k}=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{\mathrm{k}!}{\mathrm{x}}^{2\mathrm{k}} \\ \mathrm{Also}, \mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right)==\sum _{\mathrm{k}=0}^{\infty }{\left(-\frac{1}{3}\right)}^{\mathrm{k}}\frac{1}{\mathrm{k}!}\left(\frac{{\mathrm{x}}^{2\mathrm{k}+1}}{2\mathrm{k}+1}\right) \\ \mathrm{Adding} \mathrm{the} \mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }{\left(-\frac{1}{3}\right)}^{\mathrm{k}}\frac{1}{\mathrm{k}!}{\left[\frac{{\mathrm{x}}^{2\mathrm{k}+1}}{2\mathrm{k}+1}\right]}_{-1}^1 \\ =\sum _{\mathrm{k}=0}^{\infty }{\left(-\frac{1}{3}\right)}^{\mathrm{k}}\frac{1}{\mathrm{k}!}\left[\frac{1^{2\mathrm{k}+1}-{(-1)}^{2\mathrm{k}+1}}{2\mathrm{k}+1}\right] \\ =\sum _{\mathrm{k}=0}^{\infty }{\left(-\frac{1}{3}\right)}^{\mathrm{k}}\frac{1}{\mathrm{k}!}\left[\frac{2}{2\mathrm{k}+1}\right] \end{gathered}$$