Q. 70 - Calculus

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. 

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

Step-by-Step Solution

Answer

${\int }_0^{0.3}{\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \mathrm{dx}\approx 0$

Step 1. Given information is:

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

Step 2. Approximating the values

$$\begin{gathered} \mathrm{From} \mathrm{Q} 60. \\ {\int }_0^{0.3}{\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} \left[\frac{{(0.3)}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right] \\ {\int }_0^{0.3}{\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \mathrm{dx}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1} \left[\frac{{\left(3\right)}^{8k+9}}{{10}^{6k+8}}\right] \\ \mathrm{This} \mathrm{implies} \mathrm{that} \mathrm{to} \mathrm{approximate} \mathrm{the} \mathrm{values} \mathrm{of} \mathrm{integrals}, \mathrm{put} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{k} \\ {\int }_0^{0.3}{\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \mathrm{dx}=\frac{3^9}{8{\left(10\right)}^8}-\frac{3^{17}}{3·14{\left(10\right)}^{14}}+... \\ =0.0000246 - ... \\ \approx 0 \end{gathered}$$