Q. 59 - 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. 

${\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}$

Step-by-Step Solution

Answer

${\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{1-{\left(0.5\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right)$

Step 1. Given information is:

${\int }_{0.5}^1 {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{dx}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From} \mathrm{Q} 49. \\ \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{f}\left(\mathrm{x}\right)= {\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{is} \\ {\mathrm{x}}^2{\mathrm{e}}^{-3{\mathrm{x}}^2}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left(\mathrm{x}\right)}^{2\mathrm{k}+2} \\ \mathrm{Also}, \mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{{\left(\mathrm{x}\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right) \\ \mathrm{Adding} \mathrm{the} \mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right]}_{0.5}^1 \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{{\left(1\right)}^{2\mathrm{k}+3}-{\left(0.5\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right) \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{1-{\left(0.5\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right) \end{gathered}$$