Q. 55 - 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. 

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

Step-by-Step Solution

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 1. Given information is:

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

Step 2. Definite integral

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