Question

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

${\int }_{0.1}^{0.2}{\tan }^{-1}\left({\mathrm{x}}^2\right)\mathrm{dx}$

Short answer

${\int }_{0.1}^{0.2}{\tan }^{-1}\left({\mathrm{x}}^2\right)\mathrm{dx}=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left[\frac{0.2^{4k+3}-0.1^{4k+3}}{4k+3}\right]$

Step-by-step solution

Step 1. Given information is: 

${\int }_{0.1}^{0.2}{\tan }^{-1}\left({\mathrm{x}}^2\right)\mathrm{dx}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From}\mathrm{Q}44. \\ \mathrm{Maclaurin}\mathrm{series}\mathrm{for}\mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1}\left({\mathrm{x}}^2\right)\mathrm{is} \\ {\tan }^{-1}\left(x^2\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{4k+2} \\ \mathrm{Also},\mathrm{F}=\int \mathrm{f} \\ F\left(x\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left[\frac{x^{4k+3}}{4k+3}\right] \\ \mathrm{Adding}\mathrm{the}\mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{\left[\frac{x^{4k+3}}{4k+3}\right]}_{0.1}^{0.2} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left[\frac{0.2^{4k+3}-0.1^{4k+3}}{4k+3}\right] \end{gathered}$$