Question

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

${\int }_0^{0.3}{\mathrm{x}}^4{\tan }^{-1}\left(3{\mathrm{x}}^3\right)\mathrm{dx}$

Short answer

${\int }_0^{0.3}{\mathrm{x}}^4{\tan }^{-1}\left(3{\mathrm{x}}^3\right)\mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1}\left[\frac{{(0.3)}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right]$

Step-by-step solution

Step 1. Given information is: 

${\int }_0^{0.3}{\mathrm{x}}^4{\tan }^{-1}\left(3{\mathrm{x}}^3\right)\mathrm{dx}$

Step 2. Definite integral 

$$\begin{gathered} \mathrm{From}\mathrm{Q}50. \\ \mathrm{Maclaurin}\mathrm{series}\mathrm{for}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4{\tan }^{-1}\left(3{\mathrm{x}}^3\right)\mathrm{is} \\ {\mathrm{x}}^4{\tan }^{-1}\left(3{\mathrm{x}}^3\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1}{\mathrm{x}}^{6\mathrm{k}+7} \\ \mathrm{Also},\mathrm{F}=\int \mathrm{f} \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1}\left(\frac{{\mathrm{x}}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right) \\ \mathrm{Adding}\mathrm{the}\mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1}{\left[\frac{{\mathrm{x}}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right]}_0^{0.3} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1}\left[\frac{{(0.3)}^{6\mathrm{k}+8}-{\left(0\right)}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right] \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1}\left[\frac{{(0.3)}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right] \end{gathered}$$