Question

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

${\int }_0^{0.3}\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}$

Short answer

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

Step-by-step solution

Step 1. Given information is: 

${\int }_0^{0.3}\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}$

Step 2. Definite integral

$$\begin{gathered} \mathrm{From}\mathrm{Q}45. \\ \mathrm{Maclaurin}\mathrm{series}\mathrm{for}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{1+{\mathrm{x}}^3}\mathrm{is} \\ \frac{1}{1+{\mathrm{x}}^3}=\underset{\mathrm{k}=0}{\overset{\infty }{\sum {(-1)}^{\mathrm{k}}}}{\mathrm{x}}^{3\mathrm{k}} \\ \mathrm{Also},\mathrm{F}=\int \mathrm{f} \\ F\left(x\right)=\sum _{k=0}^{\infty }{\left(-1\right)}^k\left(\frac{x^{3k+1}}{3k+1}\right) \\ \mathrm{Adding}\mathrm{the}\mathrm{limits}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{k=0}^{\infty }{\left(-1\right)}^k{\left[\frac{x^{3k+1}}{3k+1}\right]}_0^{0.3} \\ =\sum _{k=0}^{\infty }{\left(-1\right)}^k\left[\frac{{(0.3)}^{3k+1}-0}{3k+1}\right] \\ =\sum _{k=0}^{\infty }{\left(-1\right)}^k\left[\frac{{(0.3)}^{3k+1}}{3k+1}\right] \end{gathered}$$