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}$

Short 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-by-step solution

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

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