Question

Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.

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

Short answer

${\int }_0^{0.3}\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}\approx 0.297975$

Step-by-step solution

Step 1. Given information is: 

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

Step 2. Approximating the values

$$\begin{gathered} \mathrm{From}\mathrm{Q}55. \\ {\int }_0^{0.3}\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}=\sum _{\mathrm{k}=0}^{\infty }{\left(-1\right)}^{\mathrm{k}}\left[\frac{{(0.3)}^{3\mathrm{k}+1}}{3\mathrm{k}+1}\right] \\ \mathrm{This}\mathrm{implies}\mathrm{that}\mathrm{to}\mathrm{approximate}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{integrals},\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k} \\ {\int }_0^{0.3}\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}=0.3-\frac{{\left(0.3\right)}^4}{4}+\frac{{\left(0.3\right)}^7}{7}-\frac{{\left(0.3\right)}^{10}}{10}+... \\ =0.3-\frac{0.0081}{4}+\frac{0.002187}{7} \\ \approx 0.297975 \end{gathered}$$