Q. 60 - 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. 

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

Step-by-Step Solution

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 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

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